t-Test

STUDENT'S T-TEST OF A SINGLE MEAN

A simple example of such an application is the test of a mean. When σ is unknown and sample size is large, we can be confident that the estimated standard error of the mean ^σx¯ (standard deviation of the sampling distribution of means under H0) will be equal to the true σx¯, and the ratio we will evaluate and that will be referred to as a normal sampling distribution is the standardized score z = [X¯ − E(X¯)]/σx¯. However, when sample size is small, we cannot have this confidence in our estimate of the standard error; then, the ratio we use is not a z-score, but t = [X¯ − E(X¯)]/^σx¯, in which the denominator is not a constant but a random variable, which varies along with the sample size, that is, with the DEGREES OF FREEDOM.

The t-distribution has a mean of 0 and a symmetric and unimodal shape, but it is flatter in the middle and denser at the extremes in comparison with the NORMAL DISTRIBUTION. As sample size n grows large, the distribution of t approaches the standardized normal distribution.

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Now, suppose we hypothesize that the mean income of a particular group, say, the chemists, is larger than 2,000 euros per month (Ha : μ > 2,000) as against the null hypothesis H0 :μ = 2,000. The standard deviation σ is unknown. We draw a small RANDOM SAMPLE of n = 26. We obtain X¯ = 2,040 and s = 100. Our estimate of ^σx¯ will be

and consequently, the t-ratio is

In the t-table, we find for a one-tailed test and α = 0.05 a critical t-value of t* = 1.706. Our sample value is higher, hence, the mean income of chemists is significantly higher than 2,000. Even in such a small sample, we obtain a significant result. However, had we performed a two-tailed test (t* = 2.056), then the result would have been nonsignificant.

STUDENT'S T-TEST OF DIFFERENCE OF MEANS

Another example of application of Student's t is the t-test of DIFFERENCE OF MEANS. We can think of two groups in an experimental design, such as a group of men and a group of women, for which the scores of a QUANTITATIVE VARIABLE are considered, let us say their smoking behavior measured as the number of cigarettes per day. In this example, the dependent variable Y (smoking) is CONTINUOUS, and the independent variable X (representing the two groups) is DICHOTOMOUS. We could give the experimental group and the control group the codes X = 1 and X = 2, respectively. We want to investigate whether the mean number of cigarettes for men (X = 1) is significantly different from (two-tailed test) or significantly lower than (one-tailed test) the mean number of cigarettes for women (X = 2). The null hypothesis states that the difference of means is equal to zero. Take two groups of five individuals, that is, n1 = n2 = 5. Let the series of scores for the total sample be2234456789 with mean 5 and variation 54. The group of men with scores 22344hasa mean of 3 and a variation of 4. The group of women with scores56789hasa mean of 7 and a variation of 10. The dispersions are unequal: variations 4 and 10. In order to perform a t-test legitimately, the distributions should be approximately normal (in the case of small groups), and the variances must not be significantly different. For both conditions, there exist a couple of test procedures, the test of normality and the test of homoskedasticity (LEVENE's TEST), respectively. Assuming that both conditions are fulfilled, the calculations are as follows.

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