Homework Solutions – 10.2 and 10.3

Homework Solutions – 10.2 and 10.3 -...

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Homework Solutions – 10.2 and 10.3 10.25 Paired T-Test and CI: Whole Foods, Fairway Paired T for Whole Foods - Fairway N Mean StDev SE Mean Whole Foods 10 2.92000 2.01993 0.63876 Fairway 10 2.24000 1.51291 0.47843 Difference 10 0.680000 0.656540 0.207616 99% lower bound for mean difference: 0.094224 T-Test of mean difference = 0 (vs > 0): T-Value = 3.28 P-Value = 0.005 Since critical value t = 2.8214 < 3.28, reject null. There is enough evidence to conclude that prices are higher at Whole Foods. 10.31 Test and CI for Two Proportions Sample X N Sample p 1 136 240 0.566667 2 224 260 0.861538 Difference = p (1) - p (2) Estimate for difference: -0.294872 95% CI for difference: (-0.370323, -0.219421) Test for difference = 0 (vs not = 0): Z = -7.34 P-Value = 0.00 Since critical value Z = +/- 2.58, reject H 0 . A significant difference exists in the
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Unformatted text preview: proportion of males and females who enjoy shopping for clothes. b. Since p-value =0.0 < .01, reject null hypothesis. 10.37 a. Test and CI for Two Proportions Sample X N Sample p 1 82 200 0.410000 2 104 200 0.520000 Difference = p (1) - p (2) Estimate for difference: -0.11 95% CI for difference: (-0.207162, -0.0128384) Test for difference = 0 (vs not = 0): Z = -2.21 P-Value = 0.027 Critical value: Z = +/- 1.96. Since the test statistic of -2.21 < -1.96, reject the null hypothesis. There is evidence of a significant difference in the proportion in the proportion who get their news primarily from newspapers between those respondents 36 to 50 years old and those above 50 years old. b. Since the p- value of 0.027 < .05, you should reject the null hypothesis....
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This note was uploaded on 02/24/2010 for the course ACCT 302 taught by Professor Staff during the Spring '08 term at Old Dominion.

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