This exercise requires ideas from Section 2.6. In a two-sample experiment, when each item in one sample is paired with an item in the other,

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This exercise requires ideas from Section 2.6. In a two-sample experiment, when each item in one sample is paired with an item in the other, the paired t test (Section 6.8) can be used to test hypotheses regarding the difference between two population means. If one ignores the fact that the data are paired, one can use the two-sample t test (Section 6.7) as well. The question arises as to which test has the greater power. The following simulation experiment is designed to address this question. 

Let (X₁, Y₁), . . . , (X₈, Y₈) be a random sample of eight pairs, with X₁, . . . , X₈ drawn from an N(0, 1) population and Y₁, . . . , Y₈ drawn from an N(1, 1) population. It is desired to test H₀ :μ_X − μ_Y = 0 versus H₁ :μ_X −μ_Y = 0. Note that μ_X = 0 and μ_Y = 1, so the true difference between the means is 1. Also note that the population variances are equal. If a test is to be made at the 5% significance level, which test has the greater power?

Let D_i = X_i − Y_i for i = 1, . . . , 10. The test statistic for the paired t test is D̅/(s_D/√8), where s_D is the standard deviation of the D_i (see Section 6.8). Its null distribution is Student’s t with seven degrees of freedom. Therefore the paired t test will reject H₀ if | D̅/(s_D/√8)| > t_7,.025 = 2.365, so the power is √(|D̅/(s_D/√8)| > 2.365). 

For the two-sample t test when the population variances are equal, the test statistic is D̅/(s_p√1/8 + 1/8) = D̅/(s_p/2), where s_p is the pooled standard deviation, which is equal in this case to p(s²_X+ s²_Y )/2. (See page 443. Note that D̅ =X –Y̅.) The null distribution is Student’s t with 14 degrees of freedom. Therefore the two-sample t test will reject H₀ if | D̅/(s_p√1/8 + 1/8)| > t_14,.025 = 2.145, and the power is P(|D̅/(s_p√1/8 + 1/8)| > 2.145). The power of these tests depends on the correlation between X_i and Y_i .

a. Generate 10,000 samples X^∗_1i, . . . , X^∗_8i froman N(0, 1) population and 10,000 samples Y^∗_1i, . . . , Y^∗_8i from an N(1, 1) population. The random variables X^∗_ki and Y^∗_ki are independent in this experiment, so their correlation is 0. For each sample, compute the test statistics D̅^∗/(s^∗_D/√8)and D̅^∗/(s^∗_p/2). Estimate the power of each test by computing the proportion of samples for which the test statistics exceeds its critical point (2.365 for the paired test, 2.145 for the two sample test). Which test has greater power?

b. As in part (a), generate 10,000 samples X^∗_1i, . . . , X^∗_8i from an N(0, 1) population. This time, instead of generating the values Y^∗ independently, generate them so the correlation between X^∗_ki and Y^∗_ki is 0.8. This can be done as follows: Generate 10,000 samples Z^∗_1i, . . . , Z_8i from an N(0, 1) population, independent of the X^∗ values. Then compute Y_ki = 1 + 0.8 X^∗_ki + 0.6 Z^∗_ki. The sample Y^∗_1i, . . . , Y^∗_8i will come from an N(1, 1) population, and the correlation between X^∗_ki and Y^∗_ki will be 0.8, which means that large values of X^∗_ki will tend to be paired with large values of Y^∗_ki, and vice versa. Compute the test statistics and estimate the power of both tests, as in part (a). Which test has greater power?