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Consider the following hypothetical data on pass‐fail grades (“0” is fail, “1” is pass) for a seminar course with 10 attendees:

grade studytime 0 30 0 25 0 59 0 42 0 31 1 140 1 90 1 95 1 170 1 120

To conduct the two‐sample t‐test, we generate the relevant vectors in R then carry out the test:

> grade.0 <- c(30, 25, 59, 42, 31) > grade.1 <- c(140, 90, 95, 170, 120) > t.test(grade.0, grade.1) Welch Two Sample t-test data: grade.0 and grade.1 t = -5.3515, df = 5.309, p-value = 0.002549 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -126.00773 -45.19227 sample estimates: mean of x mean of y 37.4 123.0

Using a Welch adjustment for unequal variances (Welch, 1947) automatically generated by R, we conclude a statistically significant difference between means (p = 0.003). With 95% confidence, we can say the true mean difference lies between the lower limit of approximately −126.0 and the upper limit of approximately −45.2. As a quick test to verify the assumption of equal variances (and to confirm in a sense whether the Welch adjustment was necessary), we can use var.test which will produce a ratio of variances and evaluate the null hypothesis that this ratio is equal to 1 (i.e., if the variances are equal, the numerator of the ratio will be the same as the denominator):

> var.test(grade.0, grade.1) F test to compare two variances data: grade.0 and grade.1 F = 0.1683, num df = 4, denom df = 4, p-value = 0.1126 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.01752408 1.61654325 sample estimates: ratio of variances 0.1683105

The var.test yields a p‐value of 0.11, which under most circumstances would be considered insufficient reason to doubt the null hypothesis of equal variances. Hence, the Welch adjustment on the variances was probably not needed in this case as there was no evidence of an inequality of variances to begin with.

Carrying out the same test in SPSS is straightforward by requesting (output not shown):

t-test groups = grade(0 1) /variables = studytime.

A classic nonparametric equivalent to the independent‐samples t‐test is the Wilcoxon rank‐sum test. It is a useful test to run when either distributional assumptions are known to be violated or when they are unknown and sample size too small for the central limit theorem to come to the “rescue.” The test compares rankings across the two samples instead of actual scores. For a brief overview of how the test works, see Kirk (2008, Chapter 18) and Howell (2002, pp. 707–717), and for a more thorough introduction to nonparametric tests in general, see the following chapter on ANOVA in this book, or consult Denis (2020) for a succinct chapter and demonstrations using R. We can request the test quite easily in R:

> wilcox.test(grade.0, grade.1) Wilcoxon rank sum test data: grade.0 and grade.1 W = 0, p-value = 0.007937 alternative hypothesis: true location shift is not equal to 0

We see that the obtained p‐value still suggests we reject the null hypothesis, though the p‐value is slightly larger than for the Welch‐corrected parametric test.