9.2 – Chi-square contingency tables

Introduction

This page covers a lot of ground to extend our discussion of implementations of the chi-square test to experimental data and introduces several related statistics, including a correlation coefficient for categorical data and an introduction to agreement and reliability measurement statistics.

We begin with contingency tables.

We just completed a discussion about goodness of fit tests, inferences on categorical traits for which a theoretical distribution or expectation is available. We offered Mendelian ratios as an example in which theory provides clear-cut expectations for the distribution of phenotypes in the F₂ offspring generation. This is an extrinsic model — theory external to the study guides in the calculation of expected values — and it reflects a common analytical task in epidemiology.

But for many other kinds of categorical outcomes, no theory is available. Today, most of us accept the link between tobacco cigarette smoking and lung cancer. The evidence for this link was reported in many studies, some better designed than others.

Consider the famous Doll and Hill (1950) report on smoking and lung cancer. They reported (Table IV Doll and Hill 1950), for men with lung cancer, only two out of 649 were non-smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were smokers (Table 1).

Table 1. Doll and Hill (1950).

A more recent example from the study of the efficacy of St John’s Wort (Hypericum perforatum) as a treatment for major depression (Shelton et al 2001, Apaydin et al. 2016). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve (Table 2).

Table 2. St. John’s Wort and depression

These kinds of problems are direct extensions of our risk analysis work in Chapter 7. Now, instead of simply describing the differences between case and control groups by Relative Risk Reduction or odds ratios, we instruct how to do inference on the risk analysis problems.

Thus, faced with absence of coherent theory as guide, the data themselves can be used (intrinsic model), and at least for now, we can employ the test.

In this lesson we will learn how to extend the analyses of categorical data to cases where we do not have a prior expectation — we use the data to generate the tests of hypotheses. This would be an intrinsic model. One of the most common two-variable categorical analysis involves 2X2 contingency tables. We may have one variable that is the treatment variable and the second variable is the outcome variable. Some examples provided in Table 3.

Table 3. Some examples of treatment and outcome variables suitable for contingency table analysis.

Table 4 were examples of published studies returned from a quick PubMed search; there are many examples (meta-analysis opportunities!). However, while they all can be analyzed by contingency tables analysis, they are not exactly the same. In some cases, the treatments are fixed effects, where the researcher selects the levels of the treatments (see Ch12.3 and Ch14.3 for discussion about fixed vs random effects in experimental design).

In other cases, each of these treatment variables need not be actual treatments (in the sense that an experiment was conducted), but it may be easier to think about these as types of experiments. These types of experiments can be distinguished by how the sampling from the reference population were conducted. Before we move on to our main purpose, to discuss how to calculate contingency tables, I wanted to provide some experimental design context by introducing two kinds of sampling (Chapter 5.5).

Our first kind of sampling scheme is called unrestricted sampling. In unrestricted sampling, you collect as many subjects (observations) as possible, then assign subjects to groups. A common approach would be to sample with a grand total in mind; for example, your grant is limited and so you only have enough money to make 1000 copies of your survey and you therefore approach 1000 people. If you categorize the subjects into just two categories (eg, Liberal, Conservative), then you have a binomial sample. If instead you classify the subjects a number of variables, eg, Liberal or Conservative, Education levels, income levels, homeowners or renters, etc., then this approach is called multinomial sampling. The point in either case is that you have just utilized a multinomial sampling approach. The aim is to classify your subjects to their appropriate groups after you have collected the sample.

Logically what must follow unrestricted sampling would be restricted sampling. Sampling would be conducted with one set of “marginal totals” fixed. Margins refers to either the row totals or to the column totals. The sampling scheme is referred to as compound multinomial. The important distinction from the other two types is that the number of individuals in one of the categories is fixed ahead of time. For example, in order to determine if smoking influences respiratory health, you approach as many people as possible to obtain 100 smokers.

The contingency table analysis

We introduced the 2×2 contingency table (Table 4) in Chapter 7.

Table 4. 2X2 contingency table

Contingency tables are all of the form (Table 5)

Table 5.  Basic format of a 2X2 table

Regardless of how the sampling occurred, the analysis of the contingency table is the same; mechanically, we have a chi-square type of problem. In both contingency table and the “goodness of fit” Chi-Square analyses the data types are discrete categories. The difference between GOF and contingency table problems is that we have no theory or external model to tell us about what to expect. Thus, we calculate expected values and the degrees of freedom differently in contingency table problems. To learn about how to perform contingency tables we will work through an example, first the long way, and then using a formula and the a, b, c, and d 2X2 table format. Of course, R can easily do contingency tables calculations for us.

Doctors noticed that a new form of Hepatitis (HG, now referred to as GB virus C), was common in some HIV+ populations (Xiang et al 2001). Accompanying the co-infection, doctors also observed an inverse relationship between HG loads and HIV viral loads: patients with high HG titers often had low HIV levels. Thus, the question was whether co-infection alters the outcome of patients with HIV — do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? I’ve represented the Xiang et al (2001) data in Table 6.

Table 6. Progression of AIDS for patients co-infected with HG [GB virus C], Xiang et al data (2001)

Our question was: Do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? A typical approach to analyze such data sets is to view the data as discrete categories and analyze with a contingency table. So we proceed.

Setup the table for analysis

Rules of contingency tables (see Kroonenberg and Verbeek 2018). What follows is a detailed walk through setting up and interpreting a 2X2 table. Note that I maintain our a, b, c, d cell order, with row 1 referencing subjects exposed or part of treatment group and row 2 referencing subjects not exposed (or part of the control group), as introduced in Chapter 7.

The Hepatitis G data from Xiang et al 2001 data, arranged in 2X2 format (Table 7).

Table 7.  Format of 2X2 table Xiang et al (2001) dataset.

The data are placed into the cells labeled a, b, c, and d.

Cell a : The number of HIV+ individuals infected with HG that lived beyond the end of the study.

Cell b : The number of HIV+ individuals infected with HG that died during the study.

Cell c : The number of HIV+ individuals not infected with HG that lived.

Cell d : The number of HIV+ individuals not infected with HG that died.

This is a contingency table because the probability of living or dying for these patients may have been contingent on coinfection with hepatitis G.

State the examined hypotheses

H_O: There is NO association between the probability of living and coinfection with hepatitis G.

For a contingency table this means that there should be the same proportion of individuals with and without hepatitis G that either lived or died (1:1:1:1).

Now, compute the Expected Values in the Four cells (a, b, c, d)

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total / Total = Expected Proportion of those that lived beyond the study.
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total / Total = Expected Proportion that died during the study.
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (step 1 above) x Row 1 Total = Expected For Cell A.
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died x Row 1 Total = Expected For Cell B
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived x Row 2 Total = Expected For Cell C
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died x Row 2 Total = Expected For Cell D

Yes, this can get a bit repetitive! But, we are now done — remember, we’re working through this to review how the intrinsic model is applied to obtain expected values.

To summarize what we’ve done so far, we now have

Observations arranged in a 2X2 table

Calculated the expected proportions, ie, the Expected values under the Null Hypothesis.

Now, we proceed to conduct the chi-square test of the null hypothesis — The observed values may differ from these expected values, meaning that the Null Hypothesis is False.

Table 8. Copy Table 6 data set

Get the Expected Values

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total (198) / Total (362) = 0.5470
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total (164) / Total (362) = 0.4530
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (0.547) x Row 1 Total (218) = Cell A = 119.25
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died (0.453) x Row 1 Total (218) = Cell B = 98.75.
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived (0.547) x Row 2 Total (144) = Cell C = 78.768
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died 0(.453) X Row 2 Total = Cell D = 65.232

Thus, we have (Table 9)

Table 9. Expected values for Xiang et al (2001) data set (Table 6 and Table 8)

Now we are ready to calculate the Chi-Square Value

Recall the formula for the chi-square test

Table 10. Worked contingency table for Xiang et al (2001) data set (Table 6 and Table 8)

Cell
a	(103 – 87.768)2 = 78.768	7.4547
b	(41 – 65.232)2 = 65.232	9.002
c	(95 – 119.25)2 = 119.25	4.93134
d	(123 – 98.75)2 = 98.75	5.95506
		27.3452

Adding all the parts we have the chi-square test statistic . To proceed with the inference, we test using the chi-square distribution.

Determine the Critical Value of test and evaluate the null hypothesis

Now recall that we can get the critical value of this test in one of two (related!) ways. One, we could run this through our statistical software and get the p-value, the probability of our result and the null hypothesis is true. Second, we look up the critical value from an appropriate statistical table. Here, we present option 2.

We need the Type I error rate α and calculate the degrees of freedom for the problem. By convention, we set α = 0.05. Degrees of freedom for contingency table is calculated as

Degrees of Freedom = (# rows – 1) x (# columns – 1)

and for our example → DF = (2 – 1) x (2 – 1) = 1.

Note. How did we get the 1 degree of freedom? Aren’t there 4 categories and shouldn’t we therefore have k – 1 = 3 df? We do have four categories, yes, but the categories are not independent. We need to take this into account when we evaluate the test and this lack of independence is accounted for by the loss of degrees of freedom.

What exactly is not independent about our study?

1. 1. The total number of observations is set in advance from the data collection.
   2. We also have the column and row totals set in advance. The number of HIV individuals with or without Hepatitis G infection was determined at the beginning of the experiment. The number of individuals that lived or died was also determined before we we conducted the test.
   3. Then the first cell (let us make that cell A) can still vary anywhere from 0 to N _i (sample size of the first drug).
   4. Once the first cell (A) is determined then the next cell in that row (B) will have to add up to N _i.
   5. Also the other cell in the same column as the first cell (C) must add up to the Column 1 Total.
   6. Lastly, the last cell (D) must have the Row 1 Total add up to the correct number.

All this translates into there being only one cell that is FREE TO VARY, hence, only one degree of freedom.

Get the Critical Value from a chi-square distribution table: For our example, look up the critical value for DF = 1, α = 0.05, you should get 3.841. The 3.841 is the value of the chi-square we would expect at 5% and the null hypothesis is in fact true condition in the population.

Once we obtain the critical value we simply use the previous statement regarding the probability of the Null Hypothesis being TRUE. We Reject the Null Hypothesis when the calculated test statistic is greater than the Critical Value. We Accept the Null Hypothesis when the calculated  is less than the Critical Value. In our example we got 27.3452 for our calculated test statistic, thus we reject the null hypothesis and conclude, that, at least for the sample in this study, there was an association between the probability of patients living past the study period and the presence of hepatitis G.

from a, b, c, and d 2X2 table format formula

If a hand calculation is required, a simpler formula to use is

where N is the sum of all observations in the table, r₁ and r₂ are the marginal row totals, and c₁ and c₂ are the marginal column totals from the 2×2 contingency table (eg, Table 4). If you are tempted to try this in a spreadsheet, and assuming your entries for a, b, c, d, and N look like Table 11, then a straightforward interpretation requires referencing five spreadsheet cells no less than 13 times! Not to mention a separate call to the distribution to calculate the p-value (one tailed test).

Table 11. Example spreadsheet with formulas for odds ratio (OR), Pearson’s , and p-value from distribution.

	A	B	C	D	E	F	G
1
2	a	103			OR	=(2*5)/(3*4)
3	b	41
4	c	95
5	d	123			chisq, p	=(((2*5-3*4)^2)*6)/ (SUM(2:3)*SUM(4:5)* SUM(2,4)*SUM(3,5))	=CHIDIST(F5,1)
6	N	632

Now, let’s see how easy this is to do in R.

effect size

Inference on hypotheses about association, there’s statistical significance — evaluate p-value compared to Type I error, eg, 5% — and then there’s biological importance, the strength of the association. The concept of effect size is an attempt to communicate the likely importance of a result. Several statistics are available to communicate effect size: for that’s φ, phi for the φ (phi) coefficient.

where N is the total. Phi for our example is

R code

sqrt(27.3452/362)

and R returns

[1] 0.274844

Effect size statistics typically range from 0 to 1; Cohen (1992) suggested the following interpretation

Table 12. Cohen’s interpretation of d values.

For our example, effect size of the association between co-infection with GB virus C and mortality was weak in the patients with HIV infection. The function phi() in psych package will calculate phi coefficient from a 2X2 table.

Agreement, kappa, and inter-rater reliability

I imagine by now you have made the leap — can we use the contingency table to assess agreement between (inter-) two judges (raters) who evaluate the same object? A naive approach would be to calculate simple agreement, the percent of objects in the list of items evaluated for which both judges agree (concur). For a binomial Yes or No, then agreement (concordance) occurs when both judges report “Yes,” or both report “No,” displayed on the main diagonal of the 2X2 table. The other possible combinations (Yes vs No and No vs Yes) are disagreement (discordance) and would be displayed on the off-diagonal of the 2X2 matrix.

Unfortunately, simple agreement, while intuitive, cannot discriminate between true agreement and agreement by chance. Cohen’s kappa (Cohen 1960, McHugh 2012) is used to assess agreement, aka inter-rater reliability, for binomial and nominal data types for two judges (raters). Cohen’s 1960 paper has been enormously impactful, with more than 53K citations (Google Scholar, Fall 2024), and spawned a number of additional measures of agreement, eg, weighted kappa for more than binomial levels, Fleiss kappa for three or more raters, ICC for ratio scale, discussed in Ch12.3). It’s beyond the scope of this text to review the context of such terms as agreement, and reliability, for just two terms used to evaluate results from two independent observers of the “same” event or object, and leave that to the reader to confirm how to report such endeavors.

Off we go. Cohen’s kappa is defined as

where p_o is the proportion in agreement and p_c is proportion of agreement expected by chance. Essentially, then, interpretable kappa values fall between 0 and 1 (0% to 100%); negative kappa estimates are possible.  Cohen (1960) noted that for kappa to equal 1, the off diagonal elements would all have to be zero.

Cohen provided an estimate of the standard error for kappa as

and 95% confidence intervals are then .

Here, we review calculation of simple agreement and Cohen’s kappa, with just the two levels. Both simple agreement and Cohen’s kappa (κ), are available in R via the icc package. Cohen’s kappa is also available in the psych package, with the cohen.kappa() function providing a confidence interval estimate directly.

Use of both is illustrated here with a simulated coin toss example modified from a nice 2018 summary of Cohen’s kappa at r-bloggers. The R code

library(irr)
# simulate some data
N = 24
flips1 <- rbinom(n = N, size = 1, prob = 0.5)
flips2 <- rbinom(n = N, size = 1, prob = 0.5)
coins<-cbind(flips1, flips2)

Confirm the data set.

head(coins)  
flips1 flips2 
[1,] 0 1 
[2,] 1 0 
[3,] 1 0 
[4,] 1 1 
[5,] 0 0 
[6,] 1 0

and create a 2X2 table.

I stacked the data

head(StackedCoin)
Side Coins    HT
1 0 flips1 tails
2 1 flips1 heads
3 1 flips1 heads
4 1 flips1 heads
5 0 flips1 tails
6 1 flips1 heads

then created a simple 2×2 table (see Note 6 below for a simple explanation for why “.Table” as a named object).

.Table <- xtabs(~Coins+HT, data=StackedCoin)

Table 13. coins as 2X2 table.

We can calculate the phi coefficient for Table 13.

phi(.Table, digits = 3)
[1] -0.167

and would conclude no correlation between the two sets of coin tosses.

Now, agreement statistics. Both agree() and kappa2() expect an unstacked (wide) data format. It’s not necessary, but we’ll call a value of 1 as “heads up” and a value of 0 as “tails up” of the tossed coin.

Next, simple agreement.

agree(coins, tolerance=0)

R returns

Subjects = 24 
Raters = 2 
%-agree = 41.7

Now, a proper agreement statistic.

kappa2(coins) 

and R returns

Subjects = 24 
Kappa = -0.135
z = -0.7 
p-value = 0.484

Results with psych:cohen.kappa, which provides the 95% confidence interval as default output.

Cohen Kappa and Weighted Kappa correlation coefficients and confidence boundaries 
                 lower estimate upper
unweighted kappa -0.51    -0.14  0.24
weighted kappa   -0.51    -0.14  0.24
Number of subjects = 24

By chance, the two independent toss of 24 coins had 10 rows of “agreement” (H:H or T:T), simply due to chance. However, results of kappa find no agreement between the two sets of coin tosses.

In summary, we see that simple agreement and the kappa statistic don’t return similar results. Percent agreement reported 41%, whereas kappa was zero. We know percent agreement can’t reflect anything in the coin toss case because each toss is independent from the next. We’ll have you explore repeated trials of coin tosses to help convince you that simple agreement is not a reliable measure of agreement (see question 6 below).

Interpreting agreement values is not straightforward, and some argue agreement statistics should be avoided in favor of disagreement measures (for example, see discussion in Pontius and Millones 2011). Cohen suggested (Table 14)

Table 14. Landis and Koch’s (1977) interpretation of Cohen’s kappa.

Even a cursory look at Table 14 should lead to questions about scenarios in which a range of agreement “substantial” would provide little comfort about the agreement between the judges or the reliability of the instrument used, for example forensic cases or histology of tissue biopsies.

Contingency table analyses in Rcmdr

Assuming you have already summarized the data, you can enter the data directly in the Rcmdr contingency table form. If your data are not summarized, then you would use Rcmdr's Two-way table… for this. We will proceed under the assumption that you have already obtained the frequencies for each cell in the table.

Rcmdr: Statistics → Contingency tables → Enter and analyze two-way table…

Here you can tell R how many rows and how many columns (Fig 1).

Figure 1. Screenshot R Commander menu for 2X2 data entry with counts.

The default is a 2X2 table. For larger tables, use the sliders to increase the number of rows, columns, or both.

Next, enter the counts. You can edit every cell in this table, including the headers. In the next panel, I will show you the data entry and options. The actual calculation of the test statistic is done when you select the “Chi-square test of independence” check-box (Fig 2).

Figure 2. Display of Xiang et al data entered into R Commander menu.

After entering the data, click on the Statistics tab (Fig 3).

Figure 3. Screenshot Statistics options for contingency table.

If you also select “Components of chi-square statistic” option, then R will show you the contributions of each cell (O-E) towards the chi-square test statistic value. This is helpful to determine if rejection of the null is due to a subset of the categories, and it also forms the basis of the heterogeneity tests, a subject we will pick up in the next section.

Here’s the R output from R Commander. Note the “2, 2, byrow=TRUE” instructions (check out R help to confirm what

.Table <- matrix(c(103,41,95,123), 2, 2, byrow=TRUE)
dimnames(.Table) <- list("rows"=c("HG+", "HG-"), "columns"=c("Lived", "Died"))

.Table # Counts
     columns
rows Lived Died
 HG+   103   41
 HG-   95   123

.Test <- chisq.test(.Table, correct=FALSE)
.Test

Pearson's Chi-squared test

data: .Table
X-squared = 27.339, df = 1, p-value = 0.0000001708

round(.Test$residuals^2, 2) # Chi-square Components
    columns
rows Lived Died
HG+   7.46 9.00
HG-   4.93 5.95

remove(.Test)
remove(.Table)

end R output.

Lots of output, take a deep breath and remember… The minimum output we need to look at is… (Table 15)?

Table 15. Minimally expected outcome for report from a statistical test.

We can see that the p-value 1.7^-7 is much less than Type 1 α = 0.05, thus, by our decision criterion we reject the null hypothesis (provisionally of course, as science goes).

myNumber = 0.0000001708
format(myNumber, scientific = TRUE, digits=2)

[1] "1.7e-07"

Questions

1. Instead of R Commander, try the contingency table problem in R directly.

   myData <- matrix(c(103,41,95,123))
   chisq.test(myData)

   Is this the correct contingency table analysis? Why or why not?

2. For many years National Football League games that ended in ties went to “sudden death,” where the winner was determined by the first score in the extra period of play, regardless of whether or not the other team got an opportunity to possess the ball on offense. Thus, in more than 100 games (140), the team that won the coin toss and therefore got the ball first in overtime, won the game following either a kicked field goal or after a touchdown was scored. In 337 other games under this system, the outcome was not determined by who got the ball first. Many complained that the “sudden death” format was unfair and in 2013 the NFL changed its overtime rules. Beginning 2014 season, both teams got a chance to possess the ball in overtime, unless the team that won the coin toss also went on to score a touchdown at which time that team would be declared the winner. In this new era of over-time rules ten teams that won the coin toss went on to score a touchdown in their first possession and therefore win the game, whereas in 54 other overtime games, the outcome was decided after both teams had a chance on offense (data as of 1 December 2015). These data may be summarized in the table
   

   	First possession win?
   	Yes	No
   Coin flip years	140	337
   New era	10	54

   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?

3. Return to the Doll and Hill (1950) data: 2 men with lung cancer were nonsmokers, 647 men with lung cancer were cigarette smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were cigarette smokers.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
4. A more recent example from the study of the efficacy of St John’s Wort as a treatment for major depression (Shelton et al 2001). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
5. Return to the coin toss simulation and repeat comparisons between simple agreement and Cohen’s kappa for different values of N. Do you see any trend(s) for the two measures as sample size increases or decreases?
6. Repeat the coin toss simulation for N = 24 up to five times, each time calculate simple agreement and Cohen’s kappa. Provide a table of your five results plus the example, and generalize simple agreement vs Cohen’s kappa as a reliable measure of agreement. Hint: you may want to use Descriptive statistics to help make your case.

Quiz Chapter 9.2

Chi-square contingency tables

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• References and suggested readings

9.1 – Chi-square test: Goodness of fit

Introduction

We ask about the “fit” of our data against predictions from theory, or from rules that set our expectations for the frequency of particular outcomes drawn from outside the experiment. Three examples to illustrate goodness of fit, GOF, , A, B, and C follow.

A. For example, for a toss of a coin, we expect heads to show up 50% of the time. Out of 120 tosses of a fair coin, we expect 60 heads, 60 tails. Thus, our null hypothesis would be that heads would appear 50% of the time. If we observe 70 heads in an experiment of coin tossing, is this a significantly large enough discrepancy to reject the null hypothesis?

B. For example, simple Mendelian genetics makes predictions about how often we should expect particular combinations of phenotypes in the offspring when the phenotype is controlled by one gene, with 2 alleles and a particular kind of dominance.

For example, for a one locus, two allele system (one gene, two different copies like R and r) with complete dominance, we expect the phenotypic (what you see) ratio will be 3:1 (or round, wrinkle). Our null hypothesis would be that pea shape will obey Mendelian ratios (3:1). Mendel’s round versus wrinkled peas (RR or Rr genotypes give round peas, only rr results in wrinkled peas).

Thus, out of 100 individuals, we would expect 75 round and 25 wrinkled. If we observe 84 round and 16 wrinkled, is this a significantly large enough discrepancy to reject the null hypothesis?

C. For yet another example, in population genetics, we can ask whether genotypic frequencies (how often a particular copy of a gene appears in a population) follow expectations from Hardy-Weinberg model (the null hypothesis would be that they do).

This is a common test one might perform on DNA or protein data from electrophoresis analysis. Hardy-Weinberg is a simple quadratic expansion:

If p = allele frequency of the first copy, and q = allele frequency of the second copy, then p + q = 1,

Given the allele frequencies, then genotypic frequencies would be given by 1 = p² + 2pq + q².

Deviations from Hardy-Weinberg expectations may indicate a number of possible causes of allele change (including natural selection, genetic drift, migration).

Thus, if a gene has two alleles, and , with the frequency for , and for ,  (equivalently q = 1 – p) in the population, then we would expect 36 , 16 , and 48 individuals. (Nothing changes if we represent the alleles as A and a, or some other system, eg, dominance/recessive.)

Question. If we observe the following genotypes: 45  individuals, 34 individuals, and 21 individuals, is this a significantly large enough discrepancy to reject the null hypothesis?

Table 1. Summary of our Hardy Weinberg question

Recall from your genetics class that we can get the allele frequency values from the genotype values, eg,

We call these chi-square tests, tests of goodness of fit. Because we have some theory, in this case Mendelian genetics, or guidance, separate from the study itself, to help us calculate expected values in a chi-square test.

goodness of fit

For k groups, the equation for the chi-square test may be written as

where f_i is the frequency (count) observed (in class i) and f_i<hat> is the frequency (count) expected if the null hypothesis is true, sum over all k groups. Alternatively, here is a format for the same equation that may be more familiar to you… ?

where O_i is the frequency (count) observed (in class i) and E_i is the frequency (count) expected if the null hypothesis is true.

The degrees of freedom, df, for the GOF are simply the number of categories minus one, k – 1.

Explaining GOF

Why am I using the phrase “goodness of fit?” This concept has broad use in statistics, but in general it applies when we ask how well a statistical model fits the observed data.

At least for the chi-square test it is simple to see how the test statistic increases from zero as the agreement between observed data and expected data depart, where zero would be the case in which all observed values for the categories exactly match the expected values.

The goodness of fit test is designed to evaluate whether or not your data agree with a theoretical expectation (there are additional ways to think about this test, but this is a good place to start). Let’s take our time here and work with an example. The other type of chi-square problem or experiment is one for the many types of experiments in which the response variable is discrete, just like in the GOF case, but we have no theory to guide us in deciding how to obtain the expected values. We can use the data themselves to calculate expected values, and we say that the test is “contingent” upon the data, hence these types of chi-square tests are called contingency tables.

You may be a little concerned at this point that there are two kinds of chi-square problems, goodness of fit and contingency tables. We’ll deal directly with contingency tables in the next section, but for now, I wanted to make a few generalizations.

1. Both goodness of fit and contingency tables use the same chi-square equation and analysis. They differ in how the degrees of freedom are calculated.
2. Thus, what all chi-square problems have in common, whether goodness of fit or contingency table problems
   1. You must identify what types of data are appropriate for this statistical procedure? Categorical (nominal data type).
   2. As always, a clear description of the hypotheses being examined.

For goodness of fit chi-square test, the most important type of hypothesis is called a Null Hypothesis: In most cases the Null Hypothesis (H_O) is “no difference” “no effect”…. If H_O is concluded to be false (rejected), then an alternative hypothesis (H_A) will be assumed to be true. Both are specified before tests are conducted. All possible outcomes are accounted for by the two hypotheses.

From above, we have

A. H_O: Fifty out of 100 tosses will result in heads.

H_A: Heads will not appear 50 times out of 100.

B. H_O: Pea shape will equal Mendelian ratios (3:1).

H_A: Pea shape will not equal Mendelian ratios (3:1).

C. H_O: Genotypic frequencies will equal Hardy-Weinberg expectations.

H_A: Genotypic frequencies will not equal Hardy-Weinberg expectations

Assumptions: In order to use the chi-square, there must be two or more categories. Each observation must be in one and only one category. If some of the observations are truly halfway between two categories then you must make a new category (eg, low, middle, high) or use another statistical procedure. Additionally, your expected values are required to be integers, not ratio. The number of observed and the number of expected must sum to the same total.

The chi-square test is a good example of such tests, and we will encounter other examples too. Another common goodness of fit is the coefficient of determination, which will be introduced in linear regression sections (see Chapter 17.7 – Regression model fit). Still other examples are the likelihood ratio test (LRT), Akaike Information Criterion (AIC), and Bayesian Information Criterion (BIC), which are all used to assess fit of models to data. (See Graffelman and Weir [2018] for how to use AIC in the context of testing for Hardy Weinberg equilibrium.)

The likelihood ratio test is used to compare the goodness of fit of two hierarchically nested models to determine if a more complex model provides a significantly better fit to the data. It does this by forming a ratio of the likelihood of the data under the two models (the simpler “null” model and the more complex “alternative” model). A small ratio suggests the simpler model is a poor fit, while a value close to one indicates the added complexity is not statistically significant. AIC (and BIC) is a measure of relative fit — it balances goodness of fit and model complexity: A lower AIC suggests a better model by finding the optimal trade-off between fitting the data well and keeping the model simple. A model that is too complex will have a higher AIC than a simpler model with a nearly as good fit.

We use LRT, AIC and BIC when we talk about selecting best regression models (see Chapter 17.7 – Regression model fit). However, as we discuss the concept of comparing models, we need to distinguish between the concept of model choice — methods used to select a best model from competing options and estimating the “goodness of fit” of a single model.

How well does data fit the prediction?

Frequentist approach interprets the test as, how well does the data fit the null hypothesis, ? When you compare data against a theoretical distribution (eg, Mendel’s hypothesis predicts the distribution of progeny phenotypes for a particular genetic system), you test the fit of the data against the model’s predictions (expectations). Recall that the Bayesian approach asks how well does the model fit the data?

Table 2. A. 120 tosses of a coin, we count heads 70/120 tosses.

Table 3. B. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

Table 4. C. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

For completeness, instead of a goodness of fit test we can treat this problem as a test of independence, a contingency table problem. We’ll discuss contingency tables more in the next section, but or now, we can rearrange our table of observed genotypes for problem C, as a 2X2 table

Table 5. Problem C reported in 2X2 table format.

The contingency table is calculated the same way as the GOF version, but the degrees of freedom are calculated differently: df = number of rows – 1 multiplied by the number of columns – 1.

Thus, for a 2X2 table the df are always equal to 1.

Note that the chi-square value itself says nothing about how any discrepancy between expectation and observed genotype frequencies come about. Therefore, one can rearrange the  equation to make clear where deviance from equilibrium, D, occur for the heterozygote (het). We have

where D² is equal to

φ coefficient

The chi-square test statistic and its inference tells you about the significance of the association, but not the strength or effect size of the association. Not surprisingly, Pearson (1904) came up with a statistic to quantify the strength of association between two binary variables, now called the φ (phi) coefficient. Like the Pearson product moment correlation, the φ (phi) coefficient takes values from -1 to +1.

The formula for the absolute value of φ coefficient is

Thus, for A, B, and C examples, φ coefficient was 0.167, 0.190, and 0.771. Thus, only weak associations in examples A and B, but strong association in C. We’ll provide a formula to directly calculate φ coefficient from the cells of the table in 9.2 – Chi-square contingency tables.

Carry out the test and interpret results

What was just calculated? The chi-square, , test statistic.

Just like t-tests, we now want to compare our test statistic against a critical value — calculate degrees of freedom (df = k – 1, k equals the numbers of categories), and set a rejection level, Type I error rate. We typically set the Type I error rate at 5%. A table of critical values for the chi-square test is available in Appendix Table Chi-square critical values.

Obtaining Probability Values for the goodness-of-fit test of the null hypothesis:

As you can see from the equation of the chi-square, a perfect fit between the observed and the expected would be a chi-square of zero. Thus, asking about statistical significance in the chi-square test is the same as asking if your test statistic is significantly greater than zero.

The chi-square distribution is used and the critical values depend on the degrees of freedom. Fortunately for  and other statistical procedures we have Tables that will tell us what the probability is of obtaining our results when the null hypothesis is true (in the population).

Here is a portion of the chi-square critical values for probability that your chi-square test statistic is less than the critical value (Fig 1).

Figure 1. A portion of critical values of the chi-square at alpha 5% for degrees of freedom between 1 and 10. A more inclusive table is provided in the Appendix, Table of Chi-square critical values.

For the first example (A), we have df = 2 – 1 = 1 and we look up the critical value corresponding to the probability in which Type I = 5% are likely to be smaller iff (“if and only if”) the null hypothesis is true. That value is 3.841; our test statistic was 3.330, and therefore smaller than the critical value: so we do not reject the null hypothesis.

Interpolating p-values

How likely is our test statistic value of 3.333 and the null hypothesis was true? (Remember, “true” in this case is a shorthand for our data was sampled from a population in which the HW expectations hold). When I check the table of critical values of the chi-square test for the “exact” p-value, I find that our test statistic value falls between a p-value 0.10 and 0.05 (represented in the table below). We can interpolate

Table 6. Interpolated p-value for critical value not reported in chi-square table.

If we assume the change in probability between 2.706 and 3.841 for the chi-square distribution is linear (it’s not, but it’s close), then we can do so simple interpolation.

We set up what we know on the right hand side equal to what we don’t know on the left hand side of the equation,

and solve for x. Then, x is equal to 0.0724

R function pchisq() gives a value of p = 0.0679. Close, but not the same. Of course, you should go with the result from R over interpolation; we mention how to get the approximate p-value by interpolation for completeness, and, in some rare instances, you might need to make the calculation. Interpolating is also a skill used to provide estimates where the researcher needs to estimate (impute) a missing value.

Interpreting p-values

This is a pretty important topic, so much so that we devote an entire section to this very problem — see 8.2 – The controversy over proper hypothesis testing. If you skipped the chapter, but find yourself unsure how to interpret the p-value, then please go back to Ch 8.2. OK, that commercial message, what does it mean to “reject the chi-square null hypothesis?” These types of tests are called goodness of fit in the following sense — if your data agree with the theoretical distribution, then the difference between observed and expected should be very close to zero. If it is exactly zero, then you have a perfect fit. In our coin toss case, if we say that the ratio of heads:tails do not differ significantly from the 50:50 expectation, then we accept the null hypothesis.

You should try the other examples yourself! A hint, the degrees of freedom are one (1) for example B and two (2) for example C.

R code

Printed tables of the critical values from the chi-square distribution, or for any statistical test for that matter are fine, but with your statistical package R and Rcmdr, you have access to the critical value and the p-value of your test statistic simply by asking. Here’s how to get both.

First, let’s get the critical value.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared quantiles (Fig 2).

Figure 2. R Commander menu for Chi-squared quantiles.

I entered “0.05” for the probability because that’s my Type I error rate α. Enter “1” for Degrees of freedom, then click “upper tail” because we are interested in obtaining the critical value for α. Here’s R’s response when I clicked “OK.”

qchisq(c(0.05), df=1, lower.tail=FALSE)
[1] 3.841459

Next, let’s get the exact P-value of our test statistic. We had three from three different tests:  for the coin-tossing example, for the pea example, and for the Hardy-Weinberg example.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared probabilities… (Fig 3).

Figure 3. R Commander menu for Chi-squared probabilities.

I entered “3.333” because that is one of the test statistics I want to calculate for probability and “1” for Degrees of freedom because I had k – 1 = 1 df for this problem. Here’s R’s response when I clicked “OK.”

pchisq(c(3.333), df=1, lower.tail=FALSE)
[1] 0.06790291

I repeated this exercise for  I got ; for I got .

Nice, right? Saves you from having to interpolate probability values from the chi-square table.

3. How to get the goodness of fit  in Rcmdr.

R provides the goodness of fit  (the command is chisq.text()), but Rcmdr thus far does not provide a menu option to link to the function. Instead, R Commander provides a menu for contingency tables, which also is a chi-square test, but is used where no theory is available to calculate the expected values (see Chapter 9.2). Thus, for the goodness of fit chi-square, we will need to bypass Rcmdr in favor of the script window. Honestly, other options are as quick or quicker: calculate by hand, use a different software (eg, Microsoft Excel), or many online sites provide JavaScript tools.

So how to get the goodness of fit chi-square while in R? Here’s one way. At the command line, type

chisq.test (c(O1, O2, ... On), p = c(E1, E2, ... En))

where O1, O2, … On are observed counts for category 1, category 2, up to category n, and E1, E2, … En are the expected proportions for each category. For example, consider our Heads/Tails example above (problem A).

In R, we write and submit

chisq.test(c(70,30),p=c(1/2,1/2))

R returns

Chi-squared test for given probabilities.
data: c(70, 30) 
X-squared = 16, df = 1, p-value = 0.00006334

Easy enough. But not much detail — details are available with some additions to the R script. I’ll just link you to a nice website that shows how to add to the output so that it looks like the one below.

mike.chi <- chisq.test(c(70,30),p=c(1/2,1/2))

Let’s explore one at time the contents of the results from the chi square function.

names(mike.chi)   #The names function 
[1] "statistic" "parameter" "p.value" "method" "data.name" "observed"
[7] "expected" "residuals" "stdres"

Now, call each name in turn.

mike.chi$residuals
[1] 2.828427 -2.828427
mike.chi$obs
[1] 70 30
mike.chi$exp
[1] 50 50

And finally, let’s get the summary output of our statistical test.

mike.chi
Chi-squared test for given probabilities.
data: c(70, 30)
X-squared = 16, df = 1, p-value = 6.334e-05

GOF and spreadsheet apps

Easy enough with R, but it may even easier with other tools. I’ll show you how to do this with spreadsheet apps and with and online at graphpad.com.

Let’s take the pea example above. We had 16 wrinkled, 84 round. We expect 25% wrinkled, 75% round.

Now, with R, we would enter

chisq.test(c(16,80),p=c(1/4,3/4))

and the R output

Chi-squared test for given probabilities
data: c(16, 80)
X-squared = 3.5556, df = 1, p-value = 0.05935

Microsoft Excel and the other spreadsheet programs (Apple Numbers, Google Sheets, LibreOffice Calc) can calculate the goodness of fit directly; they return a P-value only. If the observed data are in cells A1 and A2, and the expected values are in B1 and B2, then use the procedure =CHITEST(A1:A2,B1:B2).

Table 7. A spreadsheet with formula visible.

The P-value (but not the Chi-square test statistic) is returned. Here’s the output from Calc.

Table 8. Spreadsheet example from Table 7 with calculated P-value.

You can get the critical value from MS Excel (=CHIINV(alpha, df), returns the critical value), and the exact probability for the test statistic =CHIDIST(x,df), where x is your test statistic. Putting it all together, a general spreadsheet template for goodness of fit calculations calculations of test statistic and p-value might look like

Table 9. Spreadsheet template with formula to calculate chi-square goodness of fit on two groups.

	A	B	C	D	E
1	f1	0.75
2	f2	0.25
3	N	=SUM(A5,A6)
4	Obs	Exp	Chi.value	Chi.sqr
5	80	=B1*3	=((A5-B5)^2)/B5	=SUM(C5,C6)
6	16	=B2*3	=((A6-B6)^2)/B6		=CHIDIST(D5, COUNT(A5:A6-1)
7
8

Microsoft Excel can be improved by writing macros, or by including available add-in programs, such as the free PopTools, which is available for Microsoft Windows 32-bit operating systems only.

Another option is to take advantage of the internet — again, many folks have provided java or JavaScript-based statistical routines for educational purposes. Here’s an easy one to use www.graphpad.com.

In most cases, I find the chi-square goodness-of-fit is so simple to calculate by hand that the computer is redundant.

Questions

1. A variety of p-values were reported on this page with no attempt to reflect significant figures or numbers of digits (see Chapter 8.2). Provide proper significant figures and numbers of digits as if these p-values were reported in a science journal.

1. 0.0724
2. 0.0679
3. 0.03766692
4. 0.004955794
5. 0.00006334
6. 6.334e-05
7. 0.05935
8. 0.058714340077662

2. For a mini bag of M&M candies, you count 4 blue, 2 brown, 1 green, 3 orange, 4 red, and 2 yellow candies.

1. What are the expected values for each color?
2. Calculate using your favorite spreadsheet app (eg, Numbers, Excel, Google Sheets, LibreOffice Calc)
3. Calculate using R (note R will reply with a warning message that the “Chi-squared approximation may be incorrect”; see 9.2 Yates continuity correction)
4. Calculate using Quickcalcs at graphpad.com
5. Construct a table and compare p-values obtained from the different applications

3. CYP1A2 enzyme involved with metabolism of caffeine. Folks with C at SNP rs762551 have higher enzyme activity than folks with A. Populations differ for the frequency of C. Using R or your favorite spreadsheet application, compare the following populations against global frequency of C is 33% (frequency of A is 67%).

1. 286 persons from Northern Sweden: f(C) = 26%, f(A) = 73%
2. 4532 Native Hawaiian persons: f(C) = 22%, f(A) = 78%
3. 1260 Native American persons: f(C) = 30%, f(A) = 70%
4. 8316 Native American persons: f(C) = 36%, f(A) = 64%
5. Construct a table and compare p-values obtained for the four populations.

Quiz Chapter 9.1

Chi-square test: Goodness of fit

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• References and suggested readings

8.6 – Confidence limits for the estimate of population mean

Introduction

In Chapter 3.4 and Chapter 8.3, we introduced the concept of providing a confidence interval for estimates. We gave a calculation for an approximate confidence interval for proportions and for the Number Needed to Treat (Chapter 7.3). Even an approximate confidence interval gives the reader a range of possible values of a population parameter from a sample of observations.

In this chapter we review and expand how to calculate the confidence interval for a sample mean, . Because  is derived from a sample of observations, we use the t-distribution to calculate the confidence interval. Note that if the population was known (population standard deviation), then you would use normal distribution. This was the basis for our recommendation to adjust your very approximate estimate of a confidence interval for an estimate by replacing the “2” with “1.96” when you multiply the standard error of the estimate (SE) in the equation estimate . As you can imagine, the approximation works for large sample size, but is less useful as sample size decreases.

Consider ; it is a point estimate of , the population mean (a parameter). But our estimate of  is but one of an infinite number of possible estimates. The confidence interval, however, gives us a way to communicate how reliable our estimate is for the population parameter. A 95% confidence interval, for example, tells the reader that we are willing to say (95% confident) the true value of the parameter is between these two numbers (a lower limit and an upper limit). The point estimate (the sample mean) will of course be included between the two limits.

Instead of 95% confidence, we could calculate intervals for 99%. Since 99% is greater than 95%, we would communicate our certainty of our estimate.

Question 1: For 99% confidence interval, the lower limit would be smaller than the lower limit for a 95% confidence interval.

1. True
2. False

When we set the Type I error rate, (alpha) = 0.05 (5%), that means that 5% of all possible sample means from a population with mean, , will result in t values that are larger than OR smaller than .

Why the t-distribution?

We use the t-test because, technically, we have a limited sample size and the t-distribution is more accurate than the normal distribution for small samples. Note that as sample size increases, the t-distribution is not distinguishable from the normal distribution and we could use  (Fig 1).

Figure 1. Critical values at Type I rate of 5% of t-distribution . Blue dashed line is z = 1.96.

Here’s the equation for calculating the confidence interval based on the t-distribution. These set the limits around our estimate of the sample mean. Together, they’re called the 95% confidence interval, .

Here’s a simplified version of the same thing, but generalized to any Type I level…

This statistic allows us to say that we are 95% confident that the interval includes the true value for . For this confidence interval you need to identify the critical t value at 5%. Thus, you need to know the degrees of freedom for this problem, which is simply , the sample size minus one.

It is straightforward to calculate these by hand, but…

Set the Type I error rate, calculate the degrees of freedom (df):

• samples for one sample test
• pairs of samples for paired test
• samples for two independent sample test

and lookup the critical value from the t table (or from the t distribution in R). Of course, it is easier to use R.

In R, for the one tail critical value with seven degrees of freedom, type at the R prompt

qt(c(0.05), df=7, lower.tail=FALSE)
[1] 1.894579

For the two-tail critical value

qt(c(0.025), df=7, lower.tail=FALSE)
[1] 2.364624

Or, if you prefer to use R Commander, then follow the menu prompts to bring up the t quantiles function (Fig 2 and Fig 3).

Figure 2. Drop down menu to get t-distribution.

Figure 3. Menu for t quantiles, with values entered for the two-tail example.

You should confirm that what R calculates agrees with the critical values tabulated in the Table of Critical values for the t distribution provided in the Appendix.

A worked example

Let’s revisit our lizard example from last time (see Chapter 8.5). Prior to conducting any inference test, we decide acceptable Type I error rates (cf justify alpha discussion in Ch8.1); For this example, we set Type I error rate to be 1% for a 99% confidence interval.

The Rcmdr output was

t.test(lizz$bm, alternative='two.sided', mu=5, conf.level=.99)
data: lizz$bm
t = -1.5079, df = 7, p-value = 0.1753
alternative hypothesis: true mean is not equal to 5 
99 percent confidence interval:
1.984737 6.199263
sample estimates:
mean of x
4.092

Sort through the output and identify what you need to know.

Question 1: What was the sample mean?

1. 5
2. -1.5079
3. 7
4. 0.1753
5. 1.984737
6. 6.199263
7. 4.092

Question 2: What was the most likely population mean?

1. 5
2.  Answer
3. -1.5079
4. 7
5. 0.1753
6. 1.984737
7. 6.199263
8. 4.092

Question 3: This was a “one-tailed” test of the null hypothesis?

1. True
2. False

The output states “alternative hypothesis: true mean is not equal to 5” — so it was a two-tailed test.

Question 4: What was the lower limit of the confidence interval?

1. 5
2. -1.5079
3. 7
4. 0.1753
5. 1.984737
6. 6.199263
7. 4.092

The 99% confidence interval, , is , which means we are 99% certain that the population mean is between (lower limit) and (upper limit). In Chapter 8.5 we calculated the , is .

Confidence intervals by nonparametric bootstrap sampling.

Bootstrapping is a general approach to estimation or statistical inference that utilizes random sampling with replacement (Kulesa et al. 2015). In classic frequentist approach, a sample is drawn at random from the population and assumptions about the population distribution are made in order to conduct statistical inference. By resampling with replacement from the sample many times, the bootstrap samples can be viewed as if we drew from the population many times without invoking a theoretical distribution. A clear advantage of the bootstrap is that it allows estimation of confidence intervals without assuming a particular theoretical distribution and thus avoids the burden of repeating the experiment. Which method to prefer? For cases where assumption of a particular distribution is unwarranted (eg, what is the appropriate distribution when we compare medians among samples?), bootstrap may be preferred (and for small data sets, percentile bootstrap may be better). We cover bootstrap sampling of confidence intervals in Chapter 19.2 Bootstrap sampling.

Conclusions

The take home message is simple.

• All estimates should — must? — be accompanied by a Confidence Interval
• The more confident we wish to be, the wider the confidence interval will be

Note that the confidence interval concept combines DESCRIPTION (the population mean is between these limits) and INFERENCE (and we are 95% certain about the values of these limits). It is good statistical practice to include estimates of confidence intervals for any estimate you share with readers. Any statistic that can be estimated should be accompanied by a confidence interval and, as you can imagine, formulas are available to do just this. For example, earlier this semester we calculated NNT.

Questions

1. To gain practice with calculations of confidence intervals, calculate the approximate confidence interval, the 95% and the 99% confidence intervals based on the t distribution, for each of the following.

• • = 13, = 1.3, = 10
  • = 13, = 1.3, = 30
  • = 13, = 2.6, = 10
  • = 13, = 2.6, = 30

Quiz Chapter 8.6

Confidence limits for the estimate of population mean

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8.5 – One sample t-test

Introduction.

We’re now talking about the traditional, classical two group comparison involving continuous data types. Thus begins your introduction to parametric statistics. One sample tests involve questions like, how many — what proportion of — people would we expect are shorter or taller than two standard deviations from the mean? This type of question assumes a population and we use properties of the normal distribution and, hence, these are called parametric tests because the assumption is that the data has been sampled from a particular probability distribution.

However, when we start asking questions about a sample statistic (eg, the sample mean), we cannot use the normal distribution directly, ie, we cannot use Z and the normal table as we did before (Chapter 6.7). This is because we do not know the population standard deviation and therefore must use an estimate of the variation (s) to calculate the standard error of the mean.

With the introduction of the t-statistic, we’re now into full inferential statistics-mode. What we do have are estimates of these parameters. The t-test — aka Student’s t-test — was developed for the purpose of testing sample means when the true population parameters are not known.

The equation of the one sample t-test. Note the resemblance in form with the Z-score!

where is the sample standard error of the sample mean (SEM).

For example, weight change of mice given a hormone (leptin) or placebo. The , but under the null hypothesis, the mean change is “really” zero (). How unlikely is our value of 5 g?

Speaking of null hypotheses, can you say (or write) the null and alternative hypotheses in this example? How about in symbolic form?

We want to know if our sample mean could have been obtained by chance alone from a population where the true change in weight was zero.

$s = 3, n = 20$

and

$s_{\overline{X}} = \frac{s}{\sqrt{n}} = \frac{3}{\sqrt{20}} = 0.6708$

and we take these values and plug them into our equation of the t-test

$t = \frac{\left(5 – 0\right)}{0.67} = 7.45$

Then recall that Degrees of Freedom are DF = n – 1 so we have DF = 20 – 1 = 19 for the one sample t-test. And the Critical Value is found in the appropriate table of critical values for the t distribution (Fig 1)

Figure 1. Table of a portion of the Critical values of the t distribution. Red selections highlight critical value for t-test at α = 5% and df = 19.

Or, and better, use R

qt(c(0.025), df=19, lower.tail=FALSE)

where qt() is function call to find t-score of the p^th percentile (cf 3.3 – Measures of dispersion) of the Student t distribution. For a two tailed test, we recall that 0.025 is lower tail and 0.025 is upper tail.

In this example we would be willing to reject the Null Hypothesis if there was a positive OR a negative change in weight.

This was an example of a “two-tailed test” which is “2-tail” or α₍₂₎ in Table of critical values of the t distribution.

Critical Value for α₍₂₎ = 0.05, df = 19, = 2.093

Do we accept or reject the Null Hypothesis?

A typical inference workflow.

Note the general form of how the statistical test is processed, a form which actually applies to any statistical inference test.

1. Identify the type of data
2. State the null hypothesis (2 tailed? 1 tailed?)
3. Select the test statistic (t-test) and determine its properties
4. Calculate the test statistic (the value of the result of the t-test)
5. Find degrees of freedom
6. For the DF, get the critical value
7. Compare critical value to test statistic
8. Do we accept or reject the null hypothesis?

And then we ask, given the results of the test of inference, What is the biological interpretation? Statistical significance is not necessarily evidence of biological importance. In addition to statistical significance, the magnitude of the difference — the effect size — is important as part of interpreting results from an experiment. Statistical significance is at least in part because of sample size — the large the sample size, the smaller the standard error of the mean, therefore even small differences may be statistically significant, yet biologically unimportant. Effect size is discussed in Ch9.1 – Chi-square test: Goodness of fit, Ch11.4 – Two sample effect size and Ch12.5 – Effect size for ANOVA.

R Code.

Let’s try a one-sample t-test. Consider the following data set: body mass of four geckos and four Anoles lizards (Dohm unpublished data).

For starters, let’s say that you have reason to believe that the true mean for all small lizards is 5 grams (g).

Geckos: 3.186, 2.427, 4.031, 1.995
Anoles: 5.515, 5.659, 6.739, 3.184

Get the data into R (Rcmdr)

By now you should be able to load this data in one of several ways. If you haven’t already entered the data, check out Part 07. Working with your own data in Mike’s Workbook for Biostatistics.

Once we have our data.frame, proceed to carry out the statistical test.

To get the one-sample t-test in Rcmdr, click on Statistics → Means → Single-sample t-test… Because there is only one numerical variable, Body.mass, that is the only one that shows up in the Variable (pick one) window (Fig 2).

Figure 2. Screenshot Rcmdr single-sample t-test menu.

Type in the value 5.0 in the Null hypothesis: m = u box.

Question 1: Quick! Can you write, in plain old English, the statistical null hypothesis???

Answer 1: For example: No difference between gecko and Anolis lizard mean body mass.

Click OK

The results go to the Output Window.

t.test(lizards$Body.mass, alternative='two.sided', mu=5.0, conf.level=.95)
One Sample t-test
data: lizards$Body.mass
t = -1.5079, df = 7, p-value = 0.1753
alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
2.668108 5.515892
sample estimates:
mean of x 4.092

end of R output

Let’s identify the parts of the R output from the one sample t-test. R reports the name of the test and identifies

1. The dataset$variable used (lizards$Body.mass). The data set was called “lizards” and the variable was “Body.mass”. R uses the dollar sign ($) to denote the dataset and variable within the data set.
2. The value of the t test statistic was (t = -1.5079). It is negative because the sample mean was less than the population mean — you should be able to verify this!
3. The degrees of freedom, df = 7
4. The p-value = 0.1753
5. 95% confidence interval of the population mean; lower limit = 2.668108, upper limit = 5.515892
6. The sample mean = 4.092

Take a step back and review.

Let’s make sure we “get” the logic of the hypothesis testing we have just completed.

Consider the one-sample t-test.

Step 1. Define H_O and H_A. The null hypothesis might be that a sample mean, , is equal to μ = 5.

$H_0 : \mu = 5$

The alternate is that the sample mean is not equal to 20.

$H_A : \mu \ne 5$

Where did the value 5 come from? It could be a value from the literature (does the new sample differ from values obtained in another lab?). The point is that the value is known in advance, before the experiment is conducted, and that makes it a one-sample t-test.

One tailed hypothesis or two?

We introduced you to the idea of “tails of a test” (Ch08.4). As you should recall, a null/alternative hypothesis for a two-tailed test may be written as

Null hypothesis

versus the alternative hypothesis

where  is the sample mean and is the population mean.

Alternatively, we can write one-tailed tests of null/alternative hypothesis

for the null hypothesis versus the alternative hypothesis

Question 2: Are all possible outcomes of the one-tailed test covered by these two hypotheses?

Answer 2: Yes

Question 3: What was the SEM for this problem?

Answer 3: It would be the sample standard deviation divided by the square root of the sample size.

Step 2. Decide how certain you wish to be (with what probability) that the sample mean is different from μ. As stated previously, in biology, we say that we are willing to be incorrect 5% of the time (Cowles and Davis 1982; Cohen 1994). This means we are likely to correctly reject the null hypothesis 100% – 5% = 95% of the time, which is the definition of statistical power. We do this by setting the Type I error to be 5% (alpha, α = 0.05). The Type I error is the chance that we will reject a null hypothesis, but the true condition in the population we sampled was actually “no difference.”

Step 3. Carry out the calculation of the test statistic. In other words, get the value of t from the equation above by hand, or, if using R (yes!) simply identify the test statistic value from the R output after conducting the one sample t test.

Step 4. Evaluate the result of the test. If the value of the test statistic is greater than the critical value for the test, then you conclude that the chance (the P-value) that the result could be from that population is not likely and you therefore reject the null hypothesis.

Question 4: What is the critical value for a one-sample t-test with df = 7?

Answer 4: From R, we get + 2.365 for the two-tailed test. R code was qt(c(.025), df=7, lower.tail=FALSE)

Hint; you need the table or better, use R

Rcmdr: Distributions → Continuous distributions → t distributions → t quantiles

You also need to know three additional things to answer this question.

1. You need to know alpha (α), which we have said generally is set at 5%.
2. You also need to know the degrees of freedom (DF) for the test. For a one sample t-test, DF = n – 1, where n is the sample size.
3. You also must know whether your test is one or two-tailed.
4. You then use the t-distribution (the tables of the t-distribution at the back of your book) to obtain the critical value. Note that if you use R, the actual p-value is returned.

Why learn the equations when I can just do this in R?

Rcmdr does this for you as soon as you click OK. Rcmdr returns the value of the test statistic and the p-value. R does not show you the critical value, but instead returns the probability that your test statistic is as large as it is AND the null hypothesis is true. From our one-sample t-test example, the Rcmdr output. The simple answer is that in order to understand the R output properly you need to know where each item of the output for a particual test comes from and how to interpret it. Thus, the best way is to have the equations available and to understand the algorithmic approach to statistical inference.

And, this is as good of time as any to show you how to skip the RCmdr GUI and go straight to R.

First, create your variables. At the R prompt enter the first variable

liz <- c("G","G","G","G","A","A","A","A")

and then create the second variable

bm <- c(3.186,2.427,4.031,1.995,5.515,5.659,6.739,3.184)

Next, create a data frame. Think of a data frame as another word for worksheet.

lizz <- data.frame(liz,bm)

Verify that entries are correct. At the R prompt type “lizz” wthout the quotes and you should see

lizz
 liz bm
1 G 3.186
2 G 2.427
3 G 4.031
4 G 1.995
5 A 5.515
6 A 5.659
7 A 6.739
8 A 3.184

End of R output

Carry out the t-test by typing at the R prompt the following

t.test(lizz$bm, alternative='two.sided', mu=5, conf.level=.95)

And, like the Rcmdr output we have for the one-sample t-test the following R output

One Sample t-test
data: lizards$Body.mass
t = -1.5079, df = 7, p-value = 0.1753 alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
2.668108 5.515892
sample estimates:
mean of x
4.092

End of R output

which, as you probably guessed, is the same as what we got from RCmdr.

Question 5: From the R output of the one sample t-test, what was the value of the test statistic?

1. -1.5079
2. 7
3. 0.1753
4. 2.668108
5. 5.515892
6. 4.092

Answer 5: -1.5079

Why fall back on statistical tables? Think of this as a basic skill. In statistics and for some statistical tests, Rcmdr and other software may not provide the information needed to decide that your test statistic is large, and a table in a statistics book is the best way to evaluate the test.

For now, double check Rcmdr by looking up the critical value from the t-table.

Check critical value against our test statistic

Df = 8 – 1 = 7

The test is two-tailed, therefore α(2)

α = 0.05 (note that two-tailed critical value is 2.365. T was equal to 1.51 (since t-distribution is symmetrical, we can ignore the negative sign), which is smaller than 2.365 and so we would agree with Rcmdr — we cannot reject the null hypothesis.

Question 6: From the R output of the one sample t-test, what was the P-value?

1. -1.5079
2. 7
3. 0.1753
4. 2.668108
5. 5.515892

Answer 6: 0.1753

Question 7: We would reject the null hypothesis

1. False
2. True

Answer 7: False — p-value, 17.5%, is greater than Type I error of 5%.

Questions

Seven questions, with answers, were provided for you within the text in this chapter. Here’s one more, but without answers.

8. Here’s a small data set for you to try your hand at the one-sample t-test and Rcmdr. The dataset contains cell counts, five counts of the numbers of beads in a liquid with an automated cell counter (Scepter, Millipore USA). The true value is 200,000 beads per milliliter fluid; the manufacturer claims that the Scepter is accurate within 15%. Does the data conform to the expectations of the manufacturer? Write a hypothesis then test your hypothesis with the one-sample t-test. Here’s the data.

Quiz Chapter 8.5

One sample t-test

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8.4 – Tails of a test

Introduction

The basics of statistical inference is to establish the null and alternative hypotheses. Starting with the simplest cases, where there is one sample of observations and the comparison is against a population (theory) mean, how many possible comparisons can be made? The next simplest is the two-sample case, where we have two sets of observations and the comparison is against the two groups. Again, how many total comparisons may be made?

Let , “X bar”, equal the sample mean and , “mu”, represent the population mean. For sample means, designate groups by a subscript, 1 or 2. We then have Table 1.

Table 1. Possible hypothesis involving two groups

Comparison	One-same	Two-sample
1.
2.
3.
4.
5.
6.

Classical statistics classifies inference into null hypothesis, H_O, vs. alternative hypotheses, H_A, and specifies that we test null hypotheses based on the value of the estimated test statistic (see discussion about critical value and p-value, Chapter 8.2). From the list of six possible comparisons we can divide them into one-tailed and two-tailed differences (Table 1). By “tail” we are referring to the ends or tails of a distribution (Figure 1, Figure 2); where do our results fall on the distribution?

Two-tailed hypotheses: Comparison 1 and comparison 2 in the table above are two-tailed hypotheses. We don’t ask about the direction of any difference (less than or greater than).

Figure 1 shows the “two-tailed” distribution — if our results fall to the left , , or to the right we reject the null hypothesis (blue regions in the curve). We divide the type I error into two equal halves.

Figure 1. Two-tailed distribution.

RcmdrMisc::plotDistr(x = seq(-4, 4, length.out = 100),
  p = dnorm(seq(-4, 4, length.out = 100)),
  regions = list(c(-Inf, -1.96), c(1.96, Inf)),
  xlab="X", ylab="Density",
  col=c("blue"),
  legend=FALSE)

Figure 2 shows the “one-tailed” distribution — if our alternative hypothesis was that the sample mean was less than the population mean, then our fall to the left, , for the “lower tail” of the distribution. If, however, our alternative hypothesis was that the sample mean was greater than the population mean, then our region of interest falls to the right, . Again, we reject the null hypothesis (blue regions in the curve). Note for one-tailed hypothesis, all Type I error occurs in the one area, not both, so (alpha) remains 0.05 over the entire rejection region (Fig 2).

Figure 2. One-tailed distribution, lower tail (left) and upper tail (right).

library(tigerstats)
pnormGC(1.645, region="above", mean=0, sd=1,graph=TRUE)
pnormGC(-1.645, region="below", mean=0, sd=1,graph=TRUE)

One-tailed hypotheses: Comparison 3 through comparison 6 in the table are one-tailed hypotheses. The direction of the difference matters.

Examples

Let’s consider some examples. We learn best by working through cases.

Chemotherapy as an approach to treat cancers owes its origins to the work of Dr Sidney Farber among others in the 1930s and ’40s (DeVita and Chu 2008; Mukherjee 2011). Following up on the observations of others that folic acid (vitamin B₉) improved anemia, Dr Farber believed that folic acid might reverse the course of leukemia (Mukherjee 2011). In 1946 he recruited several children with acute lymphoblastic leukemia and injected them with folic acid. Instead of ameliorating their symptoms (eg, white blood cell counts and percentage of abnormal immature white blood cells, called blast cells), treatments accelerated progression of the disease. That’s a scientific euphemism for the reality — the children died sooner in Dr. Faber’s trial than patients not enrolled in his study. He stopped the trials. Clearly, adding folic acid was not a treatment against this leukemia.

Question 1. Do you think these experiments are one sample or two sample? Hint: Is there mention of a control group?

Answer: There’s no mention of a control group, but instead, Dr. Faber would have had plenty of information about the progression of this disease in children. This was a one sample test.

Question 2. What would be a reasonable interpretation of Dr Faber’s alternative hypothesis with respect to percentage of blast cells in patients given folic acid treatment? Your options are

1. Folic acid supplementation has an effect on blast counts.
2. Folic acid supplementation reduces blast counts.
3. Folic acid supplementation increases blast counts.
4. Folic acid supplementation has no effect on blast counts.

Answer: At the start of the trials, it is pretty clear that the alternative hypothesis was intended to be a one-tailed test (option 2). Dr. Faber’s alternative hypothesis clearly was that he believed that addition of folic acid would reduce blast cell counts. However, that they stopped the trials shows that they recognized that the converse had occurred, that blast counts increased; this means that, from a statistician’s point of view, Dr Faber’s team was testing a two-sided hypothesis (option 1).

Another example.

Dr Farber reasoned that if folic acid accelerated leukemia progression, perhaps anti-folic compounds might inhibit leukemia progression. Dr Farber’s team recruited patients with acute lymphoblastic leukemia and injected them with a folic acid agonist called aminopterin. Again, he predicted that blast counts would reduce following administration of the chemical. This time, and for the first time in recorded medicine, blast counts of many patients drastically reduced to normal levels and the patients experienced remissions. The remissions were not long lasting and all patients eventually succumbed to leukemia. Nevertheless, these were landmark findings — for the first time a chemical treatment was shown to significantly reduce blast cell counts, even leading to remission, if however brief (Mukherjee 2011).

Try Question 3 and Question 4 yourself.

Question 3. Do you think these experiments are one sample or two sample? Hint: Is there mention of a control group?

Question 4. What would be a reasonable interpretation of Dr Faber’s alternative hypothesis with respect to percentage of blast cells in patients given aminopterin treatment? Your options are

1. Aminopterin supplementation has an effect on blast counts.
2. Aminopterin supplementation reduces blast counts.
3. Aminopterin supplementation increases blast counts.
4. Aminopterin supplementation has no effect on blast counts.

Pros and Cons to One-sided testing

Here’s something to consider: why not restrict yourself to one-tailed hypothesis?

Here’s the pro-argument for one-tailed tests. Strictly speaking you gain statistical power to test the null hypothesis. For example, look up the t-test distribution for degrees of freedom equal to 20 and compare (one tail) vs. (two-tail). You will find that for the one-tailed test, the critical value of the t-distribution with 20 df is 1.725, whereas for the two-tailed test, the critical value of the t-distribution with the same numbers of df is 2.086. Thus, the difference between means can be much smaller in the one-tailed test and prove to be “statistically significant.” Put simply, with the same data, we will reject the Null Hypothesis more often with one-tailed tests.

Or better yet, if during exploratory data analysis you see a clear difference between the groups and it is in the direction your scientific intuition suggests it should be, shouldn’t you switch to a one-tailed hypothesis? That’s a hard no. You would be “guilty” of p-hacking — the inappropriate manipulation of data analysis to get a more favored, statistically significant result.

The con-argument. If you use a one-tailed test you MUST CLEARLY justify its use and be aware that a deviation in the opposite direction MUST be ignored! More specifically, you interpret a one-tailed result in the opposite direction as acceptance of the null — you cannot, after the fact, change your mind and start speaking about “statistically significant differences” if you had specified a one-tailed hypothesis and the results showed differences in the opposite direction.

Questions

1. For a Type I error rate of 5% and the following degrees of freedom, compare the critical values for one tail test and a two tailed test of the null hypothesis.

• 5 df
• 10 df
• 15 df
• 20 df
• 25 df
• 30 df

2. Using your findings from Additional Question 1, make a scatterplot with degrees of freedom on the horizontal axis and critical values on the vertical axis. What trend do you see for the difference between one and two tailed tests as degrees of freedom increase?
3. A clinical nutrition researcher wishes to test the hypothesis that a vegan diet lowers total serum cholesterol levels compared to an omnivorous diet. What kind of hypothesis should he use, one-tailed or two-tailed? Justify your choice.
4. Spironolactone, introduced in 1953, is used to block aldosterone in hypertensive patients. A newer drug eplerenone, approved by the FDA in 2002, is reported to have the same benefits as spironolactone (reduced mortality, fewer hospitalization events), but with fewer side effects compared with spironolactone. Does this sentence suggest a one-tailed test or a two-tailed test?
5. Write out the appropriate null and alternative hypothesis statements for the spironolactone and eplerenone scenario.
6. You open up a bag of Original Skittles and count the number of green, orange, purple, red, and yellow candies in the bag. What kind of hypothesis should be used, one-tailed or two-tailed? Justify your choice.
7. Verify the probability values from the table of standard normal distribution for Z equal to -1.96, -1.645, 1.645, and 1.96.

Quiz Chapter 8.4

Tails of a test

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8.2 – The controversy over proper hypothesis testing

Introduction

Over the next several chapters we will introduce and develop an approach to statistical inference, which has been given the title “Null Hypothesis Significance Testing” or NHST.

In outline, NHST proceeds with

• statements of two hypotheses, a null hypothesis, H_O, and an alternative hypothesis, H_A
• calculate a test statistic comparison of the null hypothesis (assuming some characteristic of data).
• The value of the test statistic is to be compared to a critical value for the test, identified for the assumed probability distribution at associated degrees of freedom for the statistical function, and assigned Type I error rate.

We will expand on these statements later in this chapter, so stay with me here. Basically, the null hypothesis is often a statement like the responses of subjects from the treatment and control groups are the same, eg, no treatment effect. Note that the alternative hypothesis, eg, hypertensive patients receiving hydalazine for six weeks have lower systolic blood pressure than patients receiving a placebo (Campbell et al 2011), would be the scientific hypothesis we are most interested in. But in the Frequentist NHST approach we test the null hypothesis, not the alternative hypothesis. This framework over proper hypothesis testing is the basis of the Bayesian vs Frequentist controversy.

Consider the independent sample t-test (see Chapter 8.5 and 8.6), our first example of a parametric test.

After plugging in the sample means and the standard error for the difference between the means, we calculate t, the test statistic of the t-test. The critical value is treated as a cut-off value in the NHST approach. We have to set our Type I error rate before we start the experiment, and we have available the degrees of freedom for the test, which follows from the sample size. With these in hand, the critical value is found by looking in the t-table of probabilities (or better, use R).

For example, what is the critical value of t-test with 10 degrees of freedom and Type I error of 5%?

In Rcmdr, choose Distributions → Continuous distributions → t distribution → t quantiles… (Fig 1).

Figure 1. Screenshot t-quantiles Rcmdr menu.

Note we want Type I equal to 5%. Since there are two tails for our test, we divide 5% by two and enter 0.025 and select the Upper tail.

R output

> qt(c(.025), df=10, lower.tail=FALSE)
[1] 2.228139

which is the same thing we would get if we look up on the t-distribution table (Fig 2).

Figure 2. Screenshot of portion of t-table with highlighted (red) critical value for 10 degrees of freedom.

If the test statistic is greater than the critical value, then the conclusion is that the null hypothesis is to be provisionally rejected. We would like to conclude that the alternative hypothesis should favored as best description of the results. However, we cannot — the p-value simply tells us how likely our results would be obtained and if the null hypothesis was true. Confusingly, however, you cannot interpret the p-value as telling you the probability (how likely) that the null hypothesis is true. If however the test statistic is less than the critical value, then the conclusion is that the null hypothesis is to be provisionally accepted.

The test statistic can be assigned a probability or p-value. This p-value is judged to be large or small relative to an a priori error probability level cut off called the Type I error rate. Thus, NHST as presented in this way may be thought of as a decision path — if the test statistic is greater than the critical value, which will necessarily mean that the p value is less than the Type I error rate, then we make one type of conclusion (reject H_O). In contrast, if the test statistic is less than the critical value, which will mean that the p-value associated with the test statistic will be greater than the Type I error rate, then we conclude something else about the null hypothesis. The various terms used in this description of NHST were defined in Chapter 8.3.

Sounds confusing, but, you say, OK, what exactly is the controversy? The controversy has to do whether the probability or p-value can be interpreted as evidence for a hypothesis. In one sense, the smaller the p-value, the stronger the case to reject the null hypothesis, right? However, just because the p-value is small — the event is rare — how much evidence do we have that the null hypothesis is true? Not necessarily, and so we can only conclude that the p-value is one part of what we may need for evidence for or against a hypothesis (hint: part of the solution is to consider effect size — introduced in Chapter 9.2 — and the statistical power of the test, see Ch 11). What follows was covered by Goodman (1988) and others. Here’s the problem. Consider tossing a fair coin ten times, with the resulting trial yielding nine out of ten heads (eg, a value of one, with tails equal to zero).

R code

set.seed(938291156)
rbinom(10,1,0.5)
[1] 1 1 1 1 1 1 0 1 1 1

While rare (binomial probability 0.0098), do we take this as evidence that the coin is not fair? By itself, the p-value provides no information about the alternative hypothesis. More about p-value follows below in sections What’s wrong with the p-value from NHST? and The real meaning and interpretation of P-values.

Statisticians have been aware of limitations of the NHST approach for years (see editorial by Wasserstein et al 2019), but only now is the message getting attention of researchers in the biosciences and other fields. In fact, the New York Times recently had a nice piece by F.D. Flam (“The Odds, Continually Updated,” 29 Sep 2014) on the controversy and the Bayesian alternative. Like most controversies there are strong voices on either side, and it can be difficult as an outsider to know which position to side with (Fig 3).

Figure 3. “Frequentists vs. Bayesians,” https://xkcd.com/1132/.

The short answer is — as you go forward do realize that there is a limitation to the frequentist approach and to be on the correct side of the controversy, you need to understand what you can conclude from statistical results. NHST is by far the most commonly used approached in biosciences (eg, out of 49 research articles I checked from four randomly selected issues of 2015 PLoS Biology, 43 used NHST, 2 used a likelihood approach, none used Bayesian statistics). The NHST is also the overwhelming manner in which we teach introductory statistics courses (eg, checking out the various MOOC courses at www.coursera.org, all of the courses related to Basic Statistics or Inferential Statistics are taught primarily from the NHST perspective). However, right from the start I want to emphasize the limits of the NHST approach.

If the purpose of science is to increase knowledge, then the NHST approach by itself is an inadequate framework at best, and in the eyes of some, worthless! Now, I think this latter sentiment is way over the top, but there is a need for us to stop before we begin, in effect, to set the ground rules for what can be interpreted from the NHST approach. The critics of NHST have a very important point, and that needs to be emphasized, but we will also defend use and teaching of this approach so that you are not left with the feeling that somehow this is a waste of time or that you are being cheated from learning the latest knowledge on the subject of statistical inference. The controversy hinges on what probability means.

P-values, statistical power, and replicability of research findings

Science, as a way of knowing how the world works, is the only approach that humans have developed that has been empirically demonstrated to work. Note how I narrowed what science is good for — if we are asking questions about the material world, then science should be your toolkit. Some (eg, Platt 1964), may further argue that there are disciplines in science that have been more successful (eg, molecular biology) than others (eg, evolutionary psychology, cf discussion in Ryle 2006) at advancing our knowledge about the material world. However, to the extent research findings are based solely on statistical results there is reason to believe that many studies in fact have not recovered truth (Ioannidis 2005).

In a review of genomics, it was reported that findings of gene expression differences by many microarray studies were not reproducible (Allison et al 2006). The consensus is that confidence in the findings should hold only for the most abundant gene transcripts of many microarray gene expression profiling studies, a conclusion that undercuts the perceived power of the technology to discover new causes of disease and the basis for individual differences for complex phenotypes. Note that when we write about failure of research reproducibility we are not including cases of alleged fraud (Carlson 2012 on Duke University oncogenomics case), we are instead highlighting that these kinds of studies often lack statistical power; hence, when repeated, the experiments yield different results.

Frequentist and Bayesian Probabilities

Turns out there is a lot of philosophical problems around the idea of “probability,” and three schools of thought. In the Fisherian approach to testing, the researcher devises a null hypothesis, H_O, collects the data, then computes a probability (p-value) of the result or outcome of the experiment. If the p-value is small, then this is inferred as little evidence in support of the null hypothesis.

In the Frequentists’ approach, the one we are calling NHST, the researcher devises two hypotheses, the null hypothesis, H_O, and an alternative hypothesis, H_A. The results are collected from the experiment and, prior to testing, a Type I error rate (α, chance) is defined. The Type I error rate is set to some probability and refers to the chance of rejecting the null hypothesis purely due to random chance. The Frequentist then computes a p-value of result of the experiment and applies a decision criterion: If P-value greater than Type I error rate, then provisionally accept null hypothesis. In both the Fisherian and Frequentist approaches, the probability, again defined at the relative frequency of an event over time, is viewed as a physical, objective and well-defined set of values.

Bayesian approach: based on Bayes conditional probability, one identifies the prior (subjective) probability of an hypothesis, then, adjusts the prior probability (down or up) as new results come in. The adjusted probability is known as posterior probability and it is equal to the likelihood function for the problem. The posterior probability is related to the prior probability and this function can be summarized by the Bayes factor as evidence the evidence against the null hypothesis. And that’s what we want, a metric of our evidence for or against the null hypothesis.

We can calibrate the Bayesian probability to the frequentist p-value (Selke et al 2001; Goodman 2008; Held 2010; Greenland and Poole 2012). Methods to achieve this calibration vary, but the Fagan nomogram Held (2010) proposed is a good tool for us as we go forward. We can calculate our NHST p-value, but then convert the p-value to a Bayes factor by looking at the nomogram. I mention this here not as part of your to do list, but rather as a way past the controversy: the NHST p-value can be interpreted as a Bayesian conditional probability, but they do not test the same hypothesis.

Likelihood

Before we move on there is one more concept to introduce, that of likelihood. We describe a model (an equation) we believe can generate the data we observe. By constructing different models with different parameters (hypotheses), you generate a statistic that yields a likelihood value. If the model fits the data, then the likelihood function has a small value. The basic idea then is to compare related, but different models to see which fits the data better. We will use this approach when comparing linear models when we introduce multiple regression models in Chapter 18.

What’s wrong with the p-value from NHST?

Well, really nothing is “wrong” with the p-value.

Where we tend to get into trouble with the p-value concept is when we try and interpret it. See below, Why is this important to me as a beginning student? The p-value is not evidence for a position, it is a statement about error rates. The p-value from NHST can be viewed as the culmination of a process that is intended to minimize the chance that the statistician makes an error.

In Bayesian terms, the p-value from NHST is the probability that we observe the data (eg, the differences between two sample means), assuming the null hypothesis is true. If we want to interpret the p-value in terms of evidence for a proposition, then we want the conditional error probability.

Sellke et al (2001) provided a calibration of p-values and, assuming that the prior probabilities of the null hypothesis and the alternative hypothesis are equal (that is, that each have a prior probability of 0.5), by using a formula provided by them (equation 3), we can correct our NHST p-value into a probability that can be interpreted as evidence in favor of the interpretation that the null hypothesis is true. In Bayesian terms this is called the posterior probability of the null hypothesis. The formula is

where e is Euler’s number, the base of the natural logarithm (ln), and p is the p-value from the NHST. This calibration works as long as   (Sellke et al 2001).

By convention we set the Type I error at 5% (cf Cohen 1994). How strong of evidence is a p-value near 5% against the null hypothesis being true, again, under the assumption that the prior probability of the null hypothesis being true is 50%? Using the above formula I constructed a plot of the calculated conditional error probability values against p-values (Fig 4).

Figure 4. Conditional error probability values plotted against p-values.

As you can see, a p-value of 5% is not strong evidence at just 0.289. Not until p-values are smaller than 0.004 does the conditional error probability value dip below 0.05, suggesting strong evidence against the null hypothesis being true.

R note: For those of you keeping up with the R work, here’s the code for generating this plot. Text after “#” are comments and are not interpreted by R.

At the R prompt type each line

NHSTp = seq(0.00001,0.37,by=0.01)      #create a sequence of numbers between 0.0001 and 0.37 with a step of 0.01
CEP = (1+(-1*exp(1)*NHSTp*log(NHSTp))^-1)^-1    #equation 3 from Sellke et al 2001
plot(NHSTp,CEP,xlab="P-value", ylab="Conditional error probability",type="l",col="blue")

Why is this important to me as a beginning student?

As we go forward I will be making statements about p-values and Type I error rates and null hypotheses and even such things as false positives and false negative. We need to start to grapple with what exactly can be said by p-values in the context of statistical inference, and to recognize that we will sometimes state conclusions that cut some corners when it comes to interpreting p-values. And yet, you (and all consumers of statistics!) are expected to recognize what p-values mean. Always.

The real meaning and interpretation of P-values

This is as good of a time as any to make some clarification about the meaning of p-value and the whole inference concept. Fisher indeed came up with the concept of the p-value, but its use as a decision criterion owes to others and Fisher disagreed strongly with use of the p-value in this way (Fisher 1955; Lehmann 1993).

Here are some common p-value corner-cutting statements to avoid using (after Goodman 2008; Held 2010). P-values are sometimes interpreted, incorrectly, as

Table. Incorrect interpretations of NHST p-values

1. the probability of obtaining the observed data under the assumption of no real effect
2. an observed type-I error rate
3. the false discovery rate, ie, the probability that a significant finding is a “false positive”
4. the (posterior) probability of the null hypothesis.

So, if p-values don’t mean any of these things, what does a p-value mean? It means that we begin by assuming that there is no effect of our treatments — the p-value is then the chance we will get as large of a result (our test statistic) and the null hypothesis is true. Note that this definition does not include a statement about evidence of the null hypothesis being true. To get evidence of “truth” we need additional tools, like the Bayes Factor and the correction of the p-value to the conditional error probability (see above). Why not dump all of the NHST and go directly to a Bayesian perspective, as some advise? The single best explanation was embedded in the assumption we made about the prior probability in order to calculate the conditional error probability. We assumed the prior probability was 50%. For many, many experiments, that is simply a guess. The truth is we generally don’t know what the prior probability is. Thus, if this assumption is incorrect, then the justification for the formula by Sellke et al (2001) is weakened, and we are no closer to establishing evidence than before.

The take home message is that it is unlikely that a single experiment will provide strong evidence for the truth. Thus the message is repeat your experiments — and you already knew that!  And the Bayesians can tell us that with the addition of more and more data reduces the effect of the particular value of the prior probability on our calculation of the conditional error probability. So, that’s the key to this controversy over the p-value.

Reporting p-values

Estimated p-values can never be zero. Students may come to use software that may return p-values like “0” — I’m looking at you Google Sheets re: default results from CHISQ.TEST() — but again, this does not mean the probability of the result is zero. The software simply reports values to two significant figures and failed to round. Some journals may recommend that 0 should be replaced by p < 0.01 or even < 0.05 inequalities, but the former lacks precision and the latter over-emphasizes the 5% Type I error rate threshold, the “statistical significance” of the result. In general, report p-value to three significant figures and four digits. If a p-value is small, use scientific notation and maintain significant digits. Thus, a p-value of 0.004955794 should be reported as 0.00496 and a p-value of 0.0679 should be reported as 0.0679. Use R’s signif() function, for example p-value reported as 6.334e-05, then

signif(6.334e-05,3)
[1] 6.33e-05

Rounding and significant figures were discussed in Chapter 3.5. See Land and Altman (2015) for guidelines on reporting p-values and other statistical results.

Questions

1. Revisit Figure 1 again and consider the following hypothesis — the sun will rise tomorrow.
   • If we take the Frequentist position, what would the null hypothesis be?
   • If we take the Bayesian approach, identify the prior probability.
   • Which approach, Bayesian or Frequentist, is a better approach for testing this hypothesis?
2. Consider the pediatrician who, upon receiving a chest X-ray for a child notes the left lung has a large irregular opaque area in the lower quadrant. Based on the X-ray and other patient symptoms, the doctor diagnoses pneumonia and prescribes a broad-spectrum antibiotic. Is the doctor behaving as a Frequentist or a Bayesian?
3. With the p-value interpretations listed in the table above in hand, select an article from PLoS Biology, or any of your other favorite research journals, and read how the authors report results of significance testing. Compare the precise wording in the results section against the interpretative phrasing in the discussion section. Do the authors fall into any of the p-value corner-cutting traps?

Quiz Chapter 8.2

The controversy over proper hypothesis testing

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing<.span>
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8.1 – The null and alternative hypotheses

Introduction

Classical statistical parametric tests — t-test (one sample t-test, independent sample-t-test), analysis of variance (ANOVA), correlation, and linear regression— and nonparametric tests like  (chi-square: goodness of fit and contingency table),  share several features that we need to understand. It’s natural to see all the details as if they are specific to each test, but there’s a theme that binds all of the classical statistical inference in order to make  claim of “statistical significance.”

• a calculated test statistic
• degrees of freedom associated with the calculation of the test statistic
• a probability value or p-value which is associated with the test statistic, assuming a null hypothesis is “true” in the population from which we sample.
  • Recall from our previous discussion (Chapter 8.2) that this is not strictly the interpretation of p-value, but a short-hand for how likely the data fit the null hypothesis. P-value alone can’t tell us about “truth.”
• in the event we reject the null hypothesis, we provisionally accept the alternative hypothesis.

Statistical Inference in the NHST Framework

By inference we mean to imply some formal process by which a conclusion is reached from data analysis of outcomes of an experiment. The process at its best leads to conclusions based on evidence. In statistics, evidence comes about from the careful and reasoned application of statistical procedures and the evaluation of probability (Abelson 1995).

Formally, statistics is rich in inference process (Bzdok et al, 2018). We begin by defining the classical frequentist, aka Neyman-Pearson approach, to inference, which involves the pairing of two kinds of statistical hypotheses: the null hypothesis (H_O) and the alternative hypothesis (H_A). The null hypothesis assumes that any observed difference between the groups are the results of random chance (introduced in Ch 2.3 – A brief history of (bio)statistics ), where “random chance” implies that outcomes occur without order or pattern (Eagle 2021). Whether we reject or accept the null hypothesis — a short-hand for the more appropriate “provisionally reject” in light of the humility that our results may reflect a rare sampling event — based on the outcome of our statistical inference, the inference is evaluated against a decision criterion (cf discussion in Christiansen 2005), a fixed statistical significance level (Lehmann 1992). Significance level refers to the setting of a p-value threshold before testing is done. The threshold is often set to Type I error of 5% (Cowles & Davis 1982), but researchers should always consider whether this threshold is appropriate for their work (Benjamin et al 2017).

This inference process is referred to as Null Hypothesis Significance Testing, NHST. Additionally, a probability value will be obtained for the test outcome or test statistic value. In the Fisherian likelihood tradition, the magnitude of this statistic value can be associated with a probability value, the p-value, of how likely the result is given the null hypothesis is “true.” (Again, keep in mind that this is not strictly the interpretation of p-value, it’s a short-hand for how likely the data fit the null hypothesis. P-value alone can’t tell us about “truth”, per our discussion, Chapter 8.2.)

-1*log(0.000134,10)
[1] 3.872895

> 10^(-1*3.872895)
[1] 0.0001340001

NHST Workflow

We presented in the introduction to Chapter 8 without discussion a simple flow chart to illustrate the process of decision (Fig 1, Chapter 8). Here, we repeat the flow chart diagram and follow with descriptions of the elements.

Figure 1. Flow chart of inductive statistical reasoning.

What’s missing from the flow chart is the very necessary caveat that interpretation of the null hypothesis is associated with two kinds of error, Type I error and Type II error. These points and others are discussed in the following sections.

We start with the hypothesis statements. For illustration we discuss hypotheses in terms of comparisons involving just two groups, also called two sample tests. One sample tests in contrast refer to scenarios where you compare a sample statistic to a population value. Extending these concepts to more than two samples is straightforward, but we leave that discussion to Chapters 12 – 18.

Null hypothesis

By far the most common application of the null hypothesis testing paradigm involves the comparisons of different treatment groups on some outcome variable. These kinds of null hypotheses are the subject of Chapters 8 through 12.

The Null hypothesis (H_O) is a statement about the comparisons, eg, between a sample statistic and the population, or between two treatment groups. The former is referred to as a one tailed test whereas the latter is called a two-tailed test. The null hypothesis is typically “no statistical difference” between the comparisons.

For example, a one sample, two tailed null hypothesis.

and we read it as “there is no statistical difference between our sample mean and the population mean.” For the more likely case in which no population mean is available, we provide another example, a two sample, two tailed null hypothesis.

Here, we read the statement as “there is no difference between our two sample means.” Equivalently, we interpret the statement as both sample means estimate the same population mean.

Under the Neyman-Pearson approach to inference we have two hypotheses: the null hypothesis and the alternative hypothesis. The hull hypothesis was defined above.

Alternative hypothesis

Alternative hypothesis (H_A): If we conclude that the null hypothesis is false, or rather and more precisely, we find that we provisionally fail to reject the null hypothesis, then we provisionally accept the alternative hypothesis. The view then is that something other than random chance has influenced the sample observations. Note that the pairing of null and alternative hypotheses covers all possible outcomes. We do not, however, say that we have evidence for the alternative hypothesis under this statistical regimen (Abelson 1995). We tested the null hypothesis, not the alternative hypothesis. Thus, it is incorrect to write that, having found a statistical difference between two drug treatments, say aspirin and acetaminophen for relief of migraine symptoms, it is not correct to conclude that we have proven the case that acetaminophen improves improves symptoms of migraine sufferers.

For the one sample, two tailed null hypothesis, the alternative hypothesis is

and we read it as “there is a statistical difference between our sample mean and the population mean.” For the two sample, two tailed null hypothesis, the alternative hypothesis would be

and we read it as “there is a statistical difference between our two sample means.”

Alternative hypothesis often may be the research hypothesis.

It may be helpful to distinguish between technical hypotheses, scientific hypothesis, or the equality of different kinds of treatments. Tests of technical hypotheses include the testing of statistical assumptions like normality assumption (see Chapter 13.3) and homogeneity of variances (Chapter 13.4). The results of inferences about technical hypotheses are used by the statistician to justify selection of parametric statistical tests (Chapter 13). The testing of some scientific hypothesis like whether or not there is a positive link between lifespan and insulin-like growth factor levels in humans (Fontana et al 2008), like the link between lifespan and IGFs in other organisms (Holtzenberger et al 2003), can be further advanced by considering multiple hypotheses and a test of nested hypotheses and evaluated either in Bayesian or likelihood approaches (Chapter 16 and Chapter 17). The advantage of the Bayesian approach is that it returns evidence for or against this hypothesis: if the posterior probability is greater than the prior probability, we can we say that we have evidence for the hypothesis. This is very appealing — the challenge is that the prior probability is inherently subjective (Chapter 6.4), limited by the amount and quality of existing information pertaining to the hypothesis (Prelec 2004).

How to interpret the results of a statistical test.

Any number of statistical tests may be used to calculate the value of the test statistic. For example, a one sample t-test may be used to evaluate the difference between the sample mean and the population mean (Chapter 8.5) or the independent sample t-test may be used to evaluate the difference between means of the control group and the treatment group (Chapter 10). The test statistic is the particular value of the outcome of our evaluation of the hypothesis and it is associated with the p-value. In other words, given the assumption of a particular probability distribution, in this case the t-distribution, we can associate a probability, the p-value, that we observed the particular value of the test statistic and the null hypothesis is true in the reference population.

By convention, we determine statistical significance (Cox 1982; Whitley & Ball 2002) by assigning ahead of time a decision probability called the Type I error rate, often given the symbol α (alpha). The practice is to look up the critical value that corresponds to the outcome of the test with degrees of freedom like your experiment and at the Type I error rate that you selected. The Degrees of Freedom (DF, df, or sometimes noted by the symbol v), are the number of independent pieces of information available to you. Knowing the degrees of freedom is a crucial piece of information for making the correct tests. Each statistical test has a specific formula for obtaining the independent information available for the statistical test. We first were introduced to DF when we calculated the sample variance with the Bessel correction, n – 1, instead of dividing through by n. With the df in hand, the value of the test statistic is compared to the critical value for our null hypothesis. If the test statistic is smaller than the critical value, we fail to reject the null hypothesis. If, however, the test statistic is greater than the critical value, then we provisionally reject the null hypothesis. This critical value comes from a probability distribution appropriate for the kind of sampling and properties of the measurement we are using. In other words, the rejection criterion for the null hypothesis is set to a critical value, which corresponds to a known probability, the Type I error rate.

Before proceeding with yet another interpretation, and hopefully less technical discussion about test statistics and critical values, we need to discuss the two types of statistical errors. The Type I error rate is the statistical error assigned to the probability that we may reject a null hypothesis as a result of our evaluation of our data when in fact in the reference population, the null hypothesis is, in fact, true. In Biology we generally use Type I error α = 0.05 level of significance. We say that the probability of obtaining the observed value AND H_O is true is 1 in 20 (5%) if α = 0.05. Put another way, we are willing to reject the Null Hypothesis when there is only a 5% chance that the observations could occur and the Null hypothesis is still true. Our test statistic is associated with the p-value; the critical value is associated with the Type I error rate. If and only if the test statistic value equals the critical value will the p-value equal the Type I error rate.

The second error type associated with hypothesis testing is, β, the Type II statistical error rate. This is the case where we accept or fail to reject a null hypothesis based on our data, but in the reference population, the situation is that indeed, the null hypothesis is actually false. We must also keep in mind that, as we evaluate the p-value relative to the magnitude of the Type I error rate, and if we reject the null hypothesis, we still do not say that we have evidence for the alternative hypothesis.

Thus, we end with a concept that may take you a while to come to terms with — there are four, not two possible outcomes of an experiment.

Outcomes of an experiment

What are the possible outcomes of a comparative experiment\? We have two treatments, one in which subjects are given a treatment and the other, subjects receive a placebo. Subjects are followed and an outcome is measured. We calculate the descriptive statistics aka summary statistics, means, standard deviations, and perhaps other statistics, and then ask whether there is a difference between the statistics for the groups. So, two possible outcomes of the experiment, correct\? If the treatment has no effect, then we would expect the two groups to have roughly the same values for means, etc., in other words, any difference between the groups is due to chance fluctuations in the measurements and not because of any systematic effect due to the treatment received. Conversely, then if there is a difference due to the treatment, we expect to see a large enough difference in the statistics so that we would notice the systematic effect due to the treatment.

Actually, there are four, not two, possible outcomes of an experiment, just as there were four and not two conclusions about the results of a clinical assay. The four possible outcomes of a test of a statistical null hypothesis are illustrated in Table 1.

Table 1. When conducting hypothesis testing, four outcomes are possible.

In the actual population, a thing happens or it doesn’t. The null hypothesis is either true or it is not. But we don’t have access to the reference population, we don’t have a census. In other words, there is truth, but we don’t have access to the truth. We can weight, assigned as a probability or p-value,  our decisions by how likely our results are given the assumption that the truth is indeed “no difference.”

If you recall, we’ve seen a table like Table 1 before in our discussion of conditional probability and risk analysis (Chapter 7.3). We made the point that statistical inference and the interpretation of clinical tests are similar (Browner and Newman 1987). From the perspective of ordering a diagnostic test, the proper null hypothesis would be the patient does not have the disease. For your review, here’s that table (Table 2).

Table 2. Interpretations of results of a diagnostic or clinical test.

Thus, a positive diagnostic test result is interpreted as rejecting the null hypothesis. If the person actually does not have the disease, then the positive diagnostic test is a false positive.

Questions

1. Match the corresponding entries in the two tables. For example, which outcome from the inference/hypothesis table matches specificity of the test \?
2. Find three sources on the web for definitions of the p-value. Write out these definitions in your notes and compare them.
3. In your own words distinguish between the test statistic and the critical value.
4. All of our discussions have been about testing the null hypothesis, about accepting or rejecting, provisionally, the null hypothesis.
5. What are the p-values for -logP of 5, 6, 7, 8, 9, and 10 \? Complete the p-value -logP transform table.
6. Instead of log₁₀ transform, create a similar table but for negative natural log transform. Which is more convenient? Hint: log(x, base=exp(1))

Quiz Chapter 8.1

The null and alternate hypothesis

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8 – Inferential statistics

Introduction

Statistical methods are important in biology because results of experiments are usually not clear-cut and therefore tests to support decisions between competing hypotheses are needed.

We will limit ourselves to a general discussion with examples, but beginning in this chapter, we start our introductions of specific types of statistical tests. As a reminder, our statistical philosophy is frequentist and follows the Null Hypothesis Significance Testing or NHST approach. Discussion of Bayesian statistical approaches are included as appropriate.

Thus, all statistical tests we will talk about share the following requirements or properties.

1. The type of data we have dictates which test or tests are appropriate.
2. We start with a clear description of the null hypotheses.
   • (Fig 1, Test H₀). See Chapter 8.1.
3. Set the Type I error rate, alpha, α (Fig 1, Select Type I error rate).
   • By convention, 5% is often used (Cowles and David 1982).
   • See also our discussion in Chapter 8.1 and Chapter 8.2.
4. We must be aware of the assumptions our statistical tests make and what, if any, modifications to them we can make.
   • Assumptions of parametric tests are discussed in Chapter 13.
5. Correct computation of the test statistic and degrees of freedom.
   • Fig 1, Calculate test statistics and get df.
   • Each statistical test results in a test statistic with associated degrees of freedom. These are described in chapters and chapter sections devoted to the different kinds of statistical tests.
6. Comparison of the critical value and the test statistic value, with interpretation and significance testing (p-value, Bayesian Factor, cf. discussion in Goodman 2008).
   • Fig 1, Test statistic greater than critical value?
7. Report result: Fig 1, test statistic value, df, and p-value.
   • Interpretation of p-value is done throughout the rest of this book, but in earnest, discussion is presented in Chapter 8.2.

We can provide a flow-chart of these steps (Fig 1).

Figure 1. NHST decision flow chart.

While we want to avoid the impression that statistical analysis is simply a matter of following a step-by-step protocol (Fig 1), it nevertheless may be helpful to think of it as such, understanding all the while that there are caveats and assumptions that accompany the choices we make while following the protocol.

Suggested homework

Homework 5

Quizzes in this chapter

A total of 38 questions among the several subchapters, a mix of true or false and multiple choice question format.

Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

6.11 – F distribution

Introduction

The F distribution is the probability distribution associated with the F statistic and named in honor of R. A. Fisher. The F distribution is used as the null distribution of the ANOVA test statistic. The F distribution is the ratio of two chi-square distributions, with degrees of freedom v₁ and v₂ for numerator and denominator, respectively.

We can for illustration purposes define the F statistic as a ratio of two variances,

The F statistic has two sets of degrees of freedom, one for the numerator and one for the denominator. The actual formula for the F distribution is quite complicated and in general we don’t use the F distribution in a way that involves parameter estimation. Rather, it is used in evaluating the statistical significance of the F statistic. Therefore, we produce but a few graphs and a table of critical values to illustrate the distribution.

We call the result of this calculation the F test statistic. We evaluate how often that value or greater of a test statistic will occur by applying the F distribution function. A few graphs to get a sense of what the distribution looks like for varying v₁, v₂ held to ten degrees of freedom (Fig 1).

Figure 1. Animated GIF plot of F distribution value for range of degrees of freedom.

By convention in the Null Hypothesis Significance Testing protocol (NHST), we compare the test statistic to a critical value. The critical value is defined as the value of the test statistic that occurs at the Type I error rate, which is typically set to 5%., per our presentations in Chapter 6.7, 6.9, and 6.10. The justification for NHST approach to testing of statistical significance is developed in Chapter 8.

Table of Critical values of the F distribution, one tail (upper)

Degrees of freedom v1 = 1 – 4, v2 = 10

For the complete F table see Appendix 20.4

χ², t and F distributions are related

χ², t and F distributions are all distributions indexed by their degrees of freedom. With some algebra, these three distributions can be shown to be related to each other. The probabilities tabled in the chi-squared are part of the F-distribution.

Some interesting relationships between the F distribution and other distributions can be shown. By definition we claimed that the F distribution is built on ratio of chi-square distributions, so that should indicate to you the relationship between the two kinds of continuous probability distributions. However, one can also show relationships to other distributions for the F distribution. For example, for the case of v₁ = 1 and v₂ = any value, then F_1,v2 = t², where t refers to the t distribution.

Questions

1. What happens to the shape of the F distribution as df1 is held constant, but df2 degrees of freedom are increased from 1 to 5 to 20 to 100?
2. What happens to the shape of the F distribution as df1 degrees of freedom are increased from 1 to 5 to 20 to 100 while df2 is held constant?

Be able to answer these questions using the F table, Appendix 20.4, or using R/Rcmdr

3. For probability α = 5%, and numerator degrees of freedom equal to 1, what is the critical value of the F distribution (upper tail) for 5 degree of freedom? For 20 df? For 30 df?

Quiz Chapter 6.11

F distribution

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Chapter 6 contents

• Introduction
• Some preliminaries
• Ratios and proportions
• Combinations and permutations
• Types of probability
• Discrete probability distributions
• Continuous distributions
• Normal distribution and the normal deviate (Z)
• Moments
• Chi-square distribution
• t distribution
• F distribution
• References and suggested readings

6.10 – t distribution

Introduction

Student’s t distribution is a sampling distribution where values are sampled from a normal distributed population, but σ the standard deviation and μ the mean of the population are not known. When sample size is large and we know σ the standard deviation, we would use the Z-score to evaluate probabilities of the sample mean. The t-distribution applies when σ is not known and sample size is small (eg, less than 30, per rule of thirty).

Because of its thicker tails compared to the normal distribution, the t-distribution accounts for the uncertainty in estimating the population variability. This is important because we would require a more extreme value to be considered “statistically significant” compared to the normal distribution, making your conclusions more conservative.

The equation of the t-test is

where the difference between “X bar,” the sample mean, and μ, the population mean, is divided by the standard error of the mean, , defined in Chapter 3.2 and again in Chapter 3.3. This formulation of the t-test is called the one sample t-test (Chapter 8.5).  We call the result of this calculation the test statistic for t. We evaluate how often that value or greater of a test statistic will occur by applying the t distribution function.

There are many t-distributions, actually, one for every degree of freedom. Like the normal distribution, the t distribution is always symmetrical about the mean. But it is stacked up (leptokurtic) around the middle at low degrees of freedom. As degrees of freedom increase, the t distribution spreads and becomes increasingly like the normal distribution.

Relationship between t distribution and standard normal curve.

First, here is our standard normal plot, mean = 0, standard deviation = 1 (Fig 1).

Figure 1. Density plot of standard normal distribution

Next, here’s the t-distribution for five degrees of freedom (Fig 2).

Figure 2. Density plot of t-distribution for five degrees of freedom.

Lets see what happens to the shape of the t-distribution as we increase the degrees of freedom from df = 5, 10, 20, 50, 1000, 10000 (Fig 3). The last graphic in the series is the standard normal curve again (Fig 3).

Figure 3. Animated GIF of density plot t distribution, from df = 5 to 10,000 plus standard normal curve.

By convention in the Null Hypothesis Significance Testing protocol (NHST), we compare the test statistic to a critical value. The critical value is defined as the value of the test statistic that occurs at the Type I error rate, which is typically set to 5%. We introduced logic of NHST approach in Chapter 6.9 with the chi-square distribution. Again ,this is just an introduction; we teach it now as a sort of mechanical understanding to develop. The justification for this approach to testing of statistical significance is developed in Chapter 8.

Table of Critical values of the t distribution for df  1 – 5, one tail (upper)

See Appendix 20.3 for a complete table of t-distribution.

Questions

Be able to answer these questions using the t table, Appendix 20.3, or using Rcmdr

1. For probability α = 5%, what is the critical value of the t distribution (upper tail) for 1 degree of freedom? For 5 df? For 20 df? For 30 df?
2. The value of the t test statistic is given as 12. With 3 degrees of freedom, what is the approximate probability of this value, or greater from the t distribution?

Quiz Chapter 6.10

t distribution

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Chapter 6 contents

• Introduction
• Some preliminaries
• Ratios and proportions
• Combinations and permutations
• Types of probability
• Discrete probability distributions
• Continuous distributions
• Normal distribution and the normal deviate (Z)
• Moments
• Chi-square distribution
• t distribution
• F distribution
• References and suggested readings