10.2 – Digging deeper into t-test Plus the Welch test

Introduction

We need to spend some more time with the independent sample t-test; by tearing it apart, we can learn about how parametric tests work in general.

Our assumptions for the independent sample t-test are like the one-sample t-test the data must be continuous and normally distributed (one of our standard assumptions for parametric tests, see Chapter 13). The formula is very similar to the one-sample t-test, except that now we have two sample means and the formula for the standard error (SE) has also changed.

    \begin{align*} t = \frac{\bar{X}_{1}-\bar{X}_{2}}{s_{\bar{X}_{1}-\bar{X}_{2}}} \end{align*}

We see that the test statistic t is large if the numerator is large compared to the denominator. Large values of t will be evidence in favor of rejecting the null hypothesis.

The numerator is straightforward: we subtract one sample mean from the other — if there is no difference between the samples, then this difference will be close to zero.

The denominator requires additional discussion. What is it called? It is the pooled standard error of the mean (pooled SEM). Provided the assumption of equal variances holds — an additional standard assumptions for parametric tests, see Chapter 13 — then sample variances estimate the population variance and we can use this information to our advantage to best test the hypothesis about the sample means. In other words, we don’t have to lose a degree of freedom to account for differences in variability for the two groups (see Welch test). More degrees of freedom means more statistical power to test the null hypothesis and at the same time, more confidence that the test is performing to its best.

Lets break down the pooled standard error of the mean in order to see how the assumption of equal variances affects the t-test. We assume that s_{1}^2 = s_{2}^2

Recall that

    \begin{align*} H_{0}: \bar{X}_{1} = \bar{X}_{2} \end{align*}

Now we want a “pooled SE” that is the pooled standard error for both samples. The variance of the difference between the means can be written as

    \begin{align*} \sigma^2_{\bar{X}_{1}-\bar{X}_{2}} = \frac{\sigma_{1}^2}{n_{1}} + \frac{\sigma_{2}^2}{n_{2}} \end{align*}

First we need to calculate the pooled variance. where v1 and v2 are degrees of freedom for sample one and sample two, respectively.

Note 1This is just simply a combined formula of the sample variance.

    \begin{align*} s^2_{p} = \frac{SS_{1} + SS_{2}}{v_{1} + v_{2}} \end{align*}

What is SS? The sum of squares , where SS1 and SS2 refer to sum of squares for each of the two groups.

    \begin{align*} SS = \sum_{i=1}^{n}\left ( X_{i} - \bar{X}\right )^2 \end{align*}

You should recognize SS from your definition of the sample variance, Chapter 3.3.

And the standard error for the difference between two means can now be written as So the 2-sample t-test can be written to reflect the pooled sample standard error of the difference between two sample means. We can see how unequal sample size is accommodated by the t-test.

Conducting the test by hand follows the same form as the one-sample t-test. Find the degrees of freedom (DF), but now for each sample.

Finally we evaluate with the critical value in Table (Appendix, Table of Student’s t distribution) and compare the t test statistic against the critical value with the appropriate degrees of freedom.

Because this is the t-test, again, we are assuming that the variances are the same between the two populations (homoscedasticity) and this allows us to pool the variance. As it turns out, the T-test is not overly sensitive to other deviations from the assumptions, but if the variances are in fact different, then the standard formula may yield incorrect Type I error rates compared to stated probability level (α).

However, it would be poor statistical choice to use a test where there are alternatives. This is why in part that R (Rcmdr) sets as the default for the t-test that variances are unequal! In fact, R does not do a t-test unless you change the default to assume equal variances, which, as we now know, is the t-test.

Welch test

What to do when assumptions to t-test are not met? Many options have been proposed, and Welch’s approximate t is a good alternative to the two-sample t-test — it would be appropriate if the normal assumption still held.

    \begin{align*} t = \frac{\left ( \bar{X}_{1}-\bar{X}_{2} \right )}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} \end{align*}

The degrees of freedom for the Welch’s test are now

    \begin{align*} df \approx \frac{\left ( \frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}} \right )^2}{\frac{s_{1}^4}{n_{1}^2 \cdot v_{1}} + \frac{s_{2}^4}{n_{2}^2\cdot v_{2}}} \end{align*}

Note 2: The Welch test removes the pooled estimate of the standard error, replaces with both variance estimates directly.

As a default option, R and Rcmdr uses a variation of Welch’s test when you select to do the t-test without making the assumption of equal variance (Fig 1).

Screenshot Rcmdr t-test, Welch test option

Figure 1. Screenshot Rcmdr t-test options. Default is “No” for Assume equal variances, ie, the Welch test.

The Welch test is not a nonparametric test, it is a different formulation of the t-test.

Justification for beginning with t-test

It’s unlikely that you will need the t-test in today’s research climate. Data sets are large, experiments are complex with multiple variables and samples. Why do we have to consider the t-test, and then a separate test in the case for unequal sample size? I view it as a teaching moment. It makes the general point that ALL statistical tests make assumptions about how the calculations are done and as to the nature of the data set.

This is our first experience with what to do if there is a violation of an assumption of parametric tests (see Chapter 13). Here, the assumption is that the two groups have equal sample size. When they do not, the standard t-test tends to biased estimates. On the other hand, if the assumptions are met, the standard test is the best test because it has more power to do what we intended — that is, it is best at the actual test of the null hypothesis! Take heart — this point is not always appreciated even by scientists (cf. Fagerland 2012).

Note 3: Expert advice will differ here, but you can always calculate the test statistic, even if the assumptions do not hold. The problem, the concern is with the interpretation.

Let’s us approach the problem of violation of assumptions in a couple of ways as an introduction to how, in general, to approach choice of statistical tests.

  1. Power of a test. Much current statistical research focuses on learning about how a particular statistical test works when assumptions are violated. Thus, in addition to learning what tests are designed to do, we need to consider the effects of violations of assumptions on the performance of the test (namely, is the Type I error at the stated alpha level?). This is a matter of statistical power; power of a test reflects how able the test is to get you the correct result even if assumptions are violated. Often tests perform well if sample sizes are large despite violation of assumptions. We mention without proving that the 2-sample T-test is robust to violations of normality assumption, to requirement for equal sample size, and even to unequal variances. But good experimental design attempts to meet the assumptions because the test does better!
  2. Alternate forms of some tests are available to handle some aspects of test violations. For example, the simple 2-sample T-test can be modified to accommodate different variances (Welch’s formula). Or you must find a different test (eg, nonparametric tests).

In conclusion, all tests begin with consideration of the assumptions. In some cases we can test our assumptions. For example, we learned about testing the assumption of normal distributions of sample data. We can also test the assumption of equal variances.

Questions

  1. Consider a clinical trial in which resting blood pressure is recorded on hypertensive subjects at the start of the trial, then 6 weeks after subjects have received daily supplements of flaxseed. Thus, for each subject there are two measures of blood pressure, BEFORE and AFTER.
    • Write the null hypothesis
    • Write the alternative hypothesis
    • Justify why or why not the hypothesis should be two tailed. Explain why or why not an independent sample t-test may be used to compare groups.

Quiz Chapter 10.2

Digging deeper into t-test Plus the Welch test

Chapter 10 contents

6.6 – Continuous distributions

Large numbers and the Central Limit Theorem

Imagine we’ve collected (sampled) data from a population and now want to summarize the data sample. How do we proceed? A good starting point is to plot the data in a histogram and note the shape of the sample distribution. Not to get too far ahead of ourselves here, but much of the classical inferential statistics demands that we are able to assume that the sampled values come from a certain kind of distribution called the normal, or Gaussian distribution.

Consider a random sample drawn from a normally distributed population of the following series of graphs, Figures 1 – 4:

n = 20

Figure 1. Sample size = 20, drawn from population with known μ = 0 and σ = 1.

n = 100

Figure 2. Sample size = 100, also drawn from population with known μ = 0 and σ = 1.

n = 1000

Figure 3. Sample size n = 1000, once again drawn from population with known μ = 0 and σ = 1.

n = one million

Figure 4. And lastly, sample size n = 1 million also drawn from population with known μ = 0 and σ = 1.

These graphs illustrate a fundamental point in statistics: for many kinds of measurements in biology, the more data you sample, the more likely the data will approach a normal distribution. This series of simulations was a quick and dirty “proof” of the Central Limit Theorem, which is one of the two fundamental theorems of probability, the other being that of Law of Large Numbers (ie, large-sample statistics). Basically the CLT says that for a large number of random samples, the sample mean will approach the population mean,  μ, and the sample variance will approach the population variance σ2; the distribution of the large sample will converge on the normal distribution.

As the sample size gets bigger and bigger, the resulting sample means and standard deviations get closer and closer to the true value (remember — I TOLD the program to grab numbers from the Z distribution with a mean of zero and standard deviation = 1), obeying the Law of Large Numbers.

Simulation

I used the computer to generate sample data from a population. This process is called a simulation. R can make new data sets by sampling from known populations with specified distribution properties that we determine in advance — a very powerful tool — a technique used for many kinds of statistics (eg, Monte Carlo methods, bootstrapping, etc., see Chapter 19).

Note 1: How I got the data. All of these data are from a simulation where I asked told R, “grab random numbers from an infinitely large population, with mean = 0 and standard deviation = 1.”

  1. The first graph is for a sample of 20 points;
  2. the second for 100;
  3. the third for 1,000;
  4. and lastly, 1 million points.

To generate a sample from a normal population, the Rcmdr call the menu by selecting:

Rcmdr: Distributions → Continuous distributions → Normal distribution → Sample from normal distribution…

Rcmdr menu: sample from normal distribution

Figure 5. Screenshot Rcmdr menu, sample from a normal distribution.

The menu pops up. I entered Mean (μ) = 0 and Standard deviation (σ) = 1 number of samples = 10, unchecked all boxes under the “add to data set.” I left the object name as “NormalSamples,” you can, of course, change it as needed. R code derived from these requests were

normalityTest(~obs, test="shapiro.test", data=NormalSamples)
NormalSamples <- as.data.frame(matrix(rnorm(10*1, mean=0, sd=1), ncol=1))
rownames(NormalSamples) <- paste("sample", 1:10, sep="")
colnames(NormalSamples) <- "obs"

This results in a new data.frame called NormalSamples with a single variable called obs.

Note 2: About pseudorandom number generator, PRNG. An algorithm is used for creating a sequence of numbers that are like random numbers. We say “like” or “pseudo” random numbers because the algorithm requires a starting number called the seed, rather than a truly random process, ie, a source of entropy outside of the computer. The default PRNG algorithm in base R is Mersenne Twister (Wikipedia), though there are many others included in base R (bring up the help menu by typing ?RNGkind at the prompt), as well as other packages, like random, which can be used to generate truly random numbers (source of entropy is “atmospheric noise,” per citation in the package, see also random.org).

rnorm() was the function used to sample from a normal distribution. If you run the function over and over again (eg, use a for loop), each time you will get different samples. For example, results from three runs

for (i in 1:3){
print(rnorm(5))
}
[1] -0.4221672 -1.4317800 -1.8310352 0.4181184 -1.1596058
[1] -0.2034944 1.1809083 1.5925296 -2.0763677 1.6982357
[1] -1.0967218 -0.3205041 -1.7513838 -0.3335311 -1.8808454

However, if you set the seed to the same number before calling rnorm, you’ll get the same sampled numbers.

for (i in 1:3){
set.seed(1) 
print(rnorm(5))
}
[1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078
[1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078
[1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078

The seed number can be any number, it does not have to equal 1.

After creating a sample of numbers drawn from the normal distribution, make a histogram, Rcmdr: Graphs → Histogram… (see Chapter 4.2).

The normal distribution

More on the importance of normal curves in a moment. Sometimes people call these “bell-shaped” curves. You may also see such distributions referred to as Gaussian Distributions, but the normal curve is but one of many Gaussian -type distributions. Moreover, not all “bell-shaped” curves are NORMAL. For a distribution to be “normally distributed” it must follow a specific formula.

    \begin{align*} Y_{i} = \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-\left (X_{i}-\mu \right )^2}{2\sigma^2}} \end{align*}

This formula has a lot of parts, but once we break down the parts, we’ll see that there is just two things to know. First, let’s identify the parts of the equation:

Y_{i} is the height of the curve (normal density)
\pi (pie) is a constant = 3.14159… (R, use pi)
µ is the population mean
σ2 is the population variance
σ is the square-root of the variance or the population standard deviation
e is the natural logarithm (R, use exp())
X_{i} is the individual’s value

Why the Normal distribution is so important in classical statistics

With these distinctions out of the way, the first important message about the normal curve is that it permits us to say how likely (ie, how probable) is a particular value if the observation comes from a population with mean µ and standard deviation σ and the population from which the sample was drawn came from a normal distribution.

The second message: all we need to know to recreate the normal distribution for a set of data is the mean and the variance (or the standard deviation) for the population!! With just these two parameters, we can then determine the expected proportion of observations expected for each value of X. Note — we generally do not know these two because they are population parameters: we must estimate them from samples, using our sample statistics, and that’s where the first big assumption in conducting statistical analyses comes into play!!

An example for calculating the normal distribution when knowing the mean and variance

µ = 5, σ2 = 10 (thus, the standard deviation, σ = 3.16)

The formula becomes

    \begin{align*} Y_{i} = \frac{0.398947}{3.16} e^{\frac{-\left (X_{i}-5 \right )^2}{2\cdot 10}} \end{align*}

Now, plug in different values of X (for example, what’s the probability that a value of X could be 0, 1, 2, …, 10? If we really do have a normal curve with mean = 5, and variance = 10).

The normal equation returns the proportion of observations in a normal population for each X value:

When i = 5 Y5 = 0.12616 This is the proportion of all data points that have an X = 5 When i = 1 Y1 = 0.019995 This is the proportion of all data points that have an X = 1

We can keep going, getting the proportion expected for each value of X, then make the plot (Fig 6).

some normal points

Figure 6. Frequency expected for a few points (X: 0 – 10) drawn from a normal distribution, calculated using the formula and example values. 

Here’s the R code for the plot

X = seq(0,10, by=1)
Y = (0.398947/3.16)*exp((-1*(X-5)^2)/20)
plot(Y~X, ylab="Frequency (Y)", cex=1.5, pch=19,col="blue")
lines(X,Y, col="blue", lwd=2)

Next up, more about the normal distribution, Chapter 6.7.

Questions

  1. For a mean of 0 and standard deviation of 1, apply the equation for the normal curve for X = (-4, -3.5, -3, -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4). Plot your results.
  2. Sample from a normal distribution with different sample size, means, and standard deviations, each time make a histogram and compare the shape of the histograms.
  3. For  a subset of

Quiz Chapter 6.6

Continuous distributions

Chapter 6 contents