This exercise requires ideas from Section 2.6. In a two-

Chapter 6, Problem 8E

(choose chapter or problem)

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 \(\left(X_{1}, Y_{1}\right)\), . . . , \(\left(X_{8}, Y_{8}\right)\) be a random sample of eight pairs, with \(X_{1}\), . . . , \(X_{8}\) drawn from an N(0, 1) population and \(Y_{1}\), . . . , \(Y_{8}\) drawn from an N(1, 1) population. It is desired to test \(H_{0}: \mu_{X}-\mu_{Y}=0\) versus \(H_{1}: \mu_{X}-\mu_{Y} \neq 0\). Note that \(\mu_{X}=0\) and \(\mu_{Y}=1\), 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 \(\overline{D} /\left(s_{D} / \sqrt{8}\right)\), 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 \(H_{0}\) if \(\left|\overline{D} /\left(s_{D} / \sqrt{8}\right)\right|>t_{7, .025}=2.635\), so the power is \(P\left(\left|\overline{D} /\left(s_{D} / \sqrt{8}\right)\right|>2.365\right)\).

For the two-sample t test when the population variances are equal, the test statistic is \(\overline{D} /\left(s_{p} \sqrt{1 / 8+1 / 8}\right)=\overline{D} /\left(s_{p} / 2\right)\), where \(s_{p}\) is the pooled standard deviation, which is equal in this case to \(\sqrt{\left(s_{X}^{2}+s_{Y}^{2}\right) / 2}\). (See page 443. Note that \(\overline{D}=\overline{X}-\overline{Y}\). The null distribution is Student’st with 14 degrees of freedom. Therefore the two-sample t test will reject \(H_{0}\) if \(\left|\overline{D} /\left(s_{p} \sqrt{1 / 8+1 / 8}\right)\right|>t_{14, .025}=2.145\), and the power is \(P\left(\left|\overline{D} /\left(s_{p} \sqrt{1 / 8+1 / 8}\right)\right|>2.145\right)\). The power of these tests depends on the correlation between \(X_{i}\) and \(Y_{i}\).

a. Generate 10,000 samples \(X_{1 i}^{*}\), . . . , \(X_{8 i}^{*}\) from an N(0, 1) population and 10,000 samples \(Y_{1 i}^{*}\), . . . , \(Y_{8 i}^{*}\) from an N(1, 1) population. The random variables \(X_{k i}^{*}\) and \(Y_{k i}^{*}\) are independent in this experiment, so their correlation is 0. For each sample, compute the test statistics \(\overline{D}^{*} /\left(s_{D}^{*} / \sqrt{8}\right)\) and \(\overline{D^{*}} /\left(s_{p}^{*} / 2\right)\). 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?

As in part (a), generate 10,000 samples \(X_{1 i}^{*}\)i, . . . , \(X_{8 i}^{*}\) from an N(0, 1) population. This time, instead of generating the values \(Y^{*}\) independent, generate them so the correlation between \(X_{k i}^{*}\) and \(Y_{k i}^{*}\) is 0.8. This can be done as follows: Generate 10,000 samples \(Z_{1 i}^{*}\), . . . , \(Z_{8 i}\) from an N(0, 1) population, independent of the \(X^{*}\) values. Then compute \(Y_{k i}^{*}=1+0.8 X_{k i}^{*}+0.6 Z_{k i}^{*}\). The sample \(Y_{1 i}^{*}\), . . . , \(Y_{8 i}^{*}\) will come from an N(1, 1) population, and the correlation between \(X_{k i}^{*}\) and \(Y_{k i}^{*}\) will be 0.8m which means that large values of \(X_{k i}^{*}\) will tend to be paired with large values of \(Y_{k i}^{*}\), and vice versa. Compute the test statistics and estimate the power of both tests, as in part (a). Which test has greater power?

Equation Transcription:

Text Transcription:

(X_1,Y_1)

(X_8,Y_8)

X_1

X_8

Y_1

Y_8

H_0:{mu}_X-Y=0

H_1:{mu}_X-Y{not=}0

{mu}_X=0

{mu}_Y=1

D_i=X_i-Y_i

i=1

overline{D}/(s_D/sqrt{8})

s_D

D_i

H_0

|overline{D}/(s_D/sqrt{8})|>t_{7,.025}=2.635

P(|overline{D}/(sD/sqrt{8})|>2.365)

overline{D}/(s_{p}sqrt{1/8+1/8})=overline{D}/(s_p/2)

s_p

sqrt{(sX2+sY2)/2}

overline{D}=overline{X}-overline{Y}

H_0

|overline{D}/(s_{p}sqrt{1/8+1/8})|>t_{14,.025}=2.145

P(|overline{D}/(s_{p}sqrt{1/8+1/8})|>2.145)

X_i

Y_i

X_1i^*

X_8i^*

Y_1i^*

Y_8i^*

X_ki^*

Y_ki^*

overline{D}^*/(s_D^*/8)

overline{D*}/(s_p^*/2)

X_1i^

X_8i^*

Y^*

X_ki^*

Y_ki^*

Z_1i^*

Z_8i

X^*

Y_ki=1+0.8X_ki^*+0.6Z_ki^*

Y_1i^*

Y_8i^*

X_ki^*

Y_ki^*

X_ki^*

Y_ki^*