Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 9 - Inferences from Two Samples - 9-3 Two Means: Independent Samples - Basic Skills and Concepts - Page 465: 19


a) Fail to reject the null hypothesis. b)$\mu_1-\mu_2$ is between -23.62787 and 4.44929.

Work Step by Step

a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1$ is less than $\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(13.125-22.71429)-(0)}{\sqrt{8.96036^2/24+18.60285^2/14}}=-1.81.$ The degree of freedom: $min(n_1-1,n_2-1)=min(24-1,14-1)=13.$ The corresponding P-value by using the table: p is between 0.025 and 0.05. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.01$, because it is more than 0.0.025, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.01}=2.65.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.65\sqrt{\frac{8.96036^2}{24}+\frac{18.60285^2}{14}}=14.03858.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(13.125-22.71429)-14.03858=-23.62787 and $\overline{x_1}-\overline{x_2}+E$=(13.125-22.71429)+14.03858=4.44929.
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