Applied Univariate, Bivariate, and Multivariate Statistics. Daniel J. Denis

      Notice that the pooled estimate of the variance is nothing more than an averaged weighted sum, each variance being weighted by its respective sample size. This idea of weighting variances as to arrive at a pooled value is not unique to t‐tests. Such a concept forms the very fabric of how MS error is computed in the analysis of variance as we shall see further in Chapter 3 when we discuss the ANOVA procedure in some depth.

      2.20.3 Two‐Sample t‐Tests in R

      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.

2.21 STATISTICAL POWER

      Power, first and foremost, is a probability. Power is the probability of rejecting a null hypothesis given that the null hypothesis is false. It is equal to 1 − β (i.e., 1 minus the type II error rate). If the null hypothesis were true, then regardless of how much power one has, one would still not be able to reject the null. We may think of it somewhat in terms of the sensitivity of a statistical test for detecting the falsity of the null hypothesis. If the test is not very sensitive to departures from the null (i.e., in terms of a particular alternative hypothesis), we will not detect such departures. If the test is very sensitive to such departures, then we will correctly detect these departures and be able to infer the statistical alternative hypothesis in question.

      A useful analogy for understanding power is to think of a sign on a billboard that reads “H₀ is false.” Are you able to detect such a sign with your current glasses or contact lenses that you are wearing? If not, you lack sufficient power. That is, you lack the sensitivity in your instrument (your reading glasses) to correctly detect the falsity of the null hypothesis, and in doing, be in a position to reject it. Alternatively, if you have 20/20 vision, you will be able to detect the false null with ease, and reject it with confidence. A key point to note here is that if H₀ is false, it is false regardless of your ability to detect it, analogous to a virus strain being present but biomedical engineers lacking a powerful enough microscope to see it. If the null is false, the only question that remains is whether or not you will have a powerful enough test to detect its falsity. If the null were not false on the other hand, then regardless of your degree of power, you will not be able to detect its falsity (because it is not false to begin with).

      Power is a function of four elements, all of which will be featured in our discussion of the p‐value toward the conclusion of this chapter: