10.1 – Compare two independent sample means

Introduction

We introduced the concept of comparing a sample statistic (mean) against a population parameter (Chapter 6.7, Normal deviate) or one-sample t-test against a specified mean (eg, from published data or from theory, Chapter 8.5).

Consider now a basic experimental design, the randomized control trial, or RCT (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Control Trial.

Subjects randomly selected from population of interest, then again — random assignment — once recruited into one of two treatment groups. Importantly, subjects belong to one treatment arm only: no subject simultaneously receives the treatment and the control. This is in contrast to the paired design, in which subjects receive both treatments (see Chapter 10.3).

In inferential statistics about an experiment, we are more likely trying to test if sample means are different. For example

• two species grown in common garden, do they differ in growth rate?
• human subjects given a new treatment have better outcomes compare to those receiving a control treatment (eg, placebo).

The equivalent null hypothesis is that two samples are pulled from the same population. We write the null hypothesis as

and the corresponding alternative hypothesis, H_A, then must be

Question: Is this a one tailed or two-tailed hypothesis?

Answer: Two-tailed (review Chapter 8.4)

Worked example

We introduce the two-sample t-test, or better, the independent sample t-test.

where the numerator is the difference between the two sample means and the denominator is the standard error of the differences between the two groups standard errors. The formula for

The choice of independent sample over two-sample is best because it emphasizes that the two groups (the two samples), must be comprise of independent sampling units. This is a pretty straight-forward requirement; you have randomly assigned twenty individuals to two groups, a control group (n = 10) and a treatment group (n = 10). Individuals are either in the control group or they are in the treatment group — they cannot simultaneously appear in both groups.

We will work our way through this test by example. For starters, let’s use the same lizard dataset (see Example data set, below), four body mass recordings (grams) each for house geckos (Hemidactylus frenatus, Fig 2) and the Carolina anole (Anolis carolensis, Fig 3), two of many lizard species introduced to Hawaiʻi.

Figure 2. Male Hemidactylus frenatus, central Oahu, M. Dohm.

Figure 3. Male Anolis carolinensis, ʻAkaka Falls, Big Island of Hawaiʻi, M. Dohm.

Example data set

Geckos: 3.186, 2.427, 4.031, 1.995

Anoles: 5.515, 5.659, 6.739, 3.184

Question: How would you go about creating a data frame with the values in long form (stacked worksheet), including a label variable and the body mass?

Answer: At the R prompt, type

Geckos <- c(3.186, 2.427, 4.031, 1.995); Anoles = c(5.515, 5.659, 6.739, 3.184) #create two vectors
bmass <- c(Geckos, Anoles)    #combine the two vectors into a single vector holding all of the body mass records
species <- c("gecko", "gecko", "gecko", "gecko", "anole", "anole", "anole", "anole")   #create your label variable
lizards <- data.frame(species, bmass)  #create your data frame
lizards   #print your data frame  species bmass
1   gecko 3.186
2   gecko 2.427
3   gecko 4.031
4   gecko 1.995
5   anole 5.515
6   anole 5.659
7   anole 6.739
8   anole 3.184

Note also that you can enter data into the Data editor by creating the data frame first then adding values. To edit the data frame “lizards” type fix(lizards) at the R prompt, then close the data frame when you have added or changed values as needed.

As always, begin with an exploration of the data, including a graph (Fig 4).

Figure 4. Box plot of lizard body mass.

We can see already that there’s greater spread of data for the Anoles compared to the Geckos, but the median values differ. Small sample sizes can be a problem for analyses as we can only have reduced confidence in our conclusions. However, we press on for the sake of demonstration.

Let’s test the null hypothesis, H₀, ie, the two species of lizards have the same mean body mass.

Rcmdr: Statistics → Means → Independent-samples t-test…

In this next image I posted the Rcmdr menu popup for the Independent Samples t-test (Fig 5). Later versions of Rcmdr split the settings for this command into two tabs; the first tab allows for the selection of the variables and setting the hypotheses whereas the second tab, labeled Options, permits additional choices. The default selections need your attention: to actually conduct the t-test you need to answer “No” to the question, “Assume equal variance?” (Fig 6).

Figure 5. Rcmdr menu for Independent sample t-test.

Select the Options tab (Fig 6) to select null hypothesis and to select the t-test and not the Welch-test (which is the default, ie, No to the prompt “Assume equal variances?”).

Figure 6. Rcmdr Options menu for Independent sample t-test.

Let’s look at the results and break down the parts of the test.

t.test(Body.mass~Lizard, alternative='two.sided', conf.level=.95, 
+ var.equal=TRUE, data=lizards)

Two Sample t-test

data: Body.mass by Lizard 
t = 2.7117, df = 6, p-value = 0.03503
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
0.2308685 4.4981315 
sample estimates:
mean in group Anolis mean in group Gecko 
5.27425 2.90975

Consider the R session output above and answer the following questions.

Questions for the worked example

1. Which lizard group had the greater mean value, Anolis or Gecko?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

Try another example

DNA damage, changes in the chemical structure of nucleotide bases or breakage of the DNA chains, occurs in cells under many circumstances. The comet assay, or single-cell gel electrophoresis, is one method for visualizing and measuring DNA strand breaks in cells. Exposed cells are mixed with a low-melting temperature agarose and placed onto a microscope slide. The cells are then lysed with an alkaline detergent and high salts. When current is applied across the slide, undamaged DNA remains in the nucleus, whereas damaged DNA extends towards the anode to form a comet-like tail, with imaging assisted by including a fluorescent dye like Sybr-Green. Examples of comets are shown below (Fig 7).

Figure 7. Comet examples. A Intact cell, no DNA damage, B Cell with some DNA damage, a slight tail to the right is evident, C  Cell with significant DNA damage, a large tail is evident. M. Dohm.

In an experiment, immortalized lung epithelial cells were exposed to dilute copper solutions for 30 minutes then washed with PBS. The comet assay was applied to these cells and for comparison, to cells without copper exposure but otherwise treated the same way (controls). The data are available at the bottom of this page (scroll down or click here).

Again, you should begin all analyses with an exploration of the data, including a graph (Fig 8).

Figure 8. Boxplot of comet tail lengths for cells with and without (control) exposure to copper in the cell medium for 30 minutes.

Let’s look at the R output for the t-test analysis.

    Two Sample t-test

data:  CometTail by Treatment
t = -5.8502, df = 38, p-value = 9.139e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -22.39865 -10.88213
sample estimates:
mean in group Control  mean in group Copper 
             11.14533              27.78571

Consider the R session output above and answer the following questions.

Questions for Comet assay data set

1. Which cell group had the greater mean value, Copper-exposed or Control-exposed cells?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

T test from summary statistics

In some cases you may only have to summary statistics for data, eg, the means and the standard deviations. We can use the equations of the t test to write a simple formula, where the user provides the known means, standard deviations, and sample size. For example, create a simple function with readline for user input.

myTtest <- function() {
   mnx <- as.numeric(readline(prompt="Enter mean of x: "))
   stdevx <- as.numeric(readline(prompt="Enter sd of x: "))
   nx <- as.numeric(readline(prompt="Enter n of x: "))
   mny <- as.numeric(readline(prompt="Enter mean of y: "))
   stdevy <- as.numeric(readline("Enter sd of y: "))
   ny <- as.numeric(readline(prompt="Enter n of y: "))
   myTvalue <- abs(((mnx-mny)-0)/sqrt(((stdevx^2)/nx)+(stdevy^2)/ny))
   myDF <- as.integer(nx+ny-2)
   myPvalue <- pt(myTvalue,myDF,lower.tail=FALSE)*2
   myResults <- c(myTvalue, myDF, myPvalue)
   report <- c("T-test: ", "df: ", "two-tailed p-value: ")
   cat(sprintf("%s %3.3f, ", report, myResults))
}

then run the function by typing myTest() at the R prompt and entering the means, standard deviations, and sample size when prompted.

myTtest()
Enter mean of x: 2.91
Enter sd of x: .895
Enter n of x: 4
Enter mean of y: 5.27
Enter sd of y: 1.497
Enter n of y: 4
T-test: 2.706, df: 6.000, two-tailed p-value: 0.035

P-values from confidence intervals

While we expect certain reporting criteria for published, it is not uncommon that one or more elements are missing. For example, while means plus or minus standard deviations for two groups may be reported, the confidence interval of the difference may not be reported, or, an inexact p-value is reported like “< 0.05,” but we “need” an exact p-value for our meta-analysis.  A little math, and these missing statistics can be calculated.

We can use the lizard study as an example; while all of the expected elements were reported, including the p-value (0.03503), let’s say the p value was reported as “statistically significant at p < 0.05.”

We need

lower and upper confidence intervals: 0.2309 and 4.498, respectively.

critical value from t distribution, df = 6 (2 tailed, therefore 0.05/2). A little R code we have:

the standard error, which can be calculated as

and then the p-value,

which is pretty close to the result from R (p = 0.03503). The difference is likely due to the small sample size.

Questions

1. Don’t forget to work through the Questions for the Comet tail data set (scroll up or click here).
2. Microsoft Excel, LibreOffice Calc, and Google sheets spreadsheet software all include t-test functions and return the p-value. Consider two variables big (100, 110, 120, 100, 110, 210, 200) and small (0,1,1,2,0,1,0). (Note — these two groups are obviously very different, calculating a t-test on their difference is silly, just for this question.)  If formatting is set to the default two decimal places for Number cell category, the p-value will return as “0.00.” How should you report the p-value in this case?

Quiz Chapter 10.1

Compare two independent sample means

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Comet assay data set

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

10 – Inferences, Quantitative: Two Sample tests

Testing differences in averages: From one group to two

A one sample parametric test compares the mean against a population value. The population value may come literally from census information, or, more likely, it comes from some applicable theory. The one sample t-test was presented along with how to calculate the confidence interval in the previous chapter.

In this chapter, we also extend to consider two samples, tests about hypotheses for two groups. The two groups may consist of observations on different subjects, and thus the two groups are independent of each other — an independent sample t-test may be used to test null hypothesis. This is called a crossed design or between-subjects design –where each group has its own unique set of observations, and no relationship exists between the observations in the two groups. We discuss crossed experimental designs in Chapter 14.1.

A common experimental design is to measure individuals two or more times, eg, observations like body mass index, BMI, recorded on individuals at the start of an exercise program, and again on the same individuals some time after a treatment — a repeated measures design. In this case, the measures are paired and are, thus, not independent, and a paired-sample t-test would be advised. If the repeated measures occur at different times, then the design is also longitudinal design or within-subjects design. Paired data are examples of nested design — the data points are clustered within individual units. We discuss nested experimental designs in Chapter 14.5.

Two sample parametric tests are used to answer questions about the mean where the data are collected from two random samples of independent observations, each from an underlying normal distribution. The samples may be independent or paired, in which different hypotheses are tested.

Suggested homework

Homework 7: t-tests and ANOVA

Quizzes in this chapter

A total of 27 questions among the several subchapters, a mix of true or false and multiple choice question format.

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

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