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