<p> Homework 3 Solution</p><p>Question 1 SAS Code: DATA Rats; INPUT Group BP; DATALINES; 1 384 1 369 1 354 1 375 1 366 1 423 2 152 2 157 2 179 2 182 2 176 2 149 ; PROC UNIVARIATE DATA = Rats NORMAL; CLASS Group; VAR BP; RUN; PROC TTEST DATA = Rats; CLASS Group; VAR BP; RUN;</p><p>Selected Output: The UNIVARIATE Procedure Variable: BP Group = 1</p><p>Tests for Normality Test Statistic p Value Shapiro-Wilk W 0.869398 Pr < W 0.2238 Kolmogorov-Smirnov D 0.24254 Pr > D >0.1500 Cramer-von Mises W-Sq 0.0744 Pr > W-Sq 0.2112 Anderson-Darling A-Sq 0.444501 Pr > A-Sq 0.1864</p><p>The UNIVARIATE Procedure Variable: BP Group = 2</p><p>Tests for Normality Test Statistic p Value Shapiro-Wilk W 0.85134 Pr < W 0.1614 Kolmogorov-Smirnov D 0.254371 Pr > D >0.1500</p><p>1 Tests for Normality Test Statistic p Value Cramer-von Mises W-Sq 0.077782 Pr > W-Sq 0.1896 Anderson-Darling A-Sq 0.454293 Pr > A-Sq 0.1737</p><p>The TTEST Procedure Method Variances DF t Value Pr > |t| Pooled Equal 10 18.51 <.0001 Satterthwait Unequal 8.3215 18.51 <.0001 e</p><p>Equality of Variances Method Num DF Den DF F Value Pr > F Folded 5 5 2.63 0.3121 F</p><p>Interpretation: Normality Test: In order to choose appropriate test, we check the normality for both groups. Both group have the same Null Hypothesis and Alternative Hypothesis. </p><p>The p-value for the group at is , and the p-value for the group at is . Since both p-values are greater than , we fail to reject the null hypothesis and conclude that the normality assumption is met and both group follow normal distribution. Variance Test: Under normality assumption, we then test if two population variances are equal. The null hypothesis and the alternative hypothesis are </p><p>The p-value of the F-test is . Since the p-value is greater than , we fail to reject the null hypothesis at the significance level and conclude that . Pooled-Variance T-Test: The Null Hypothesis and Alternative Hypothesis: </p><p>The p-value of the pooled-variance t-test is much less than . Since the p-value is less than , we reject the null hypothesis at the significance level and conclude that the rats have higher blood pressure at than .</p><p>2 Question 2 SAS Code:</p><p>DATA Sales; INPUT Before After; Diff = After - Before; DATALINES; 12 18 18 24 25 24 9 14 14 19 16 20 ; PROC UNIVARIATE DATA = Sales NORMAL ALPHA = 0.01; VAR Diff; RUN;</p><p>Selected Output: The UNIVARIATE Procedure Variable: Diff Moments</p><p>Tests for Location: Mu0=0 Test Statistic p Value Student's t t 3.86680 Pr > |t| 0.0118 1 Sign M 2 Pr >= |M| 0.2188 Signed S 9.5 Pr >= |S| 0.0625 Rank</p><p>Tests for Normality Test Statistic p Value Shapiro-Wilk W 0.735678 Pr < W 0.0144 Kolmogorov-Smirnov D 0.308159 Pr > D 0.0739 Cramer-von Mises W-Sq 0.138153 Pr > W-Sq 0.0242 Anderson-Darling A-Sq 0.77049 Pr > A-Sq 0.0211</p><p>Interpretation: Normality Test: In order to choose appropriate test, we check the normality for the paired data. The Null Hypothesis and Alternative Hypothesis. </p><p>3 The p-value for Shapiro-Wilk test is . Since , we reject the null hypothesis and conclude that the paired difference does NOT follow a normal distribution. Wilcoxon Signed Rank Test: We then use Wilcoxon signed rank test since the normality assumption is not met. The Null Hypothesis and Alternative Hypothesis are </p><p>The p-value of the Wilcoxon signed rank test is . Since the p-value is greater than , we fail to reject the null hypothesis and cannot conclude that the mean weekly sales for all salespersons increase after the course at the significance level .</p><p>4</p>