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: 7

Answer

a) We fail to reject the null hypothesis. b) $\mu_1-\mu_2$ is between -0.26 and 0.36.

Work Step by Step

a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1>\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(0.49-0.44)-(0)}{\sqrt{0.96^2/20+1.4^2/20}}=0.132.$ The degree of freedom: $min(n_1-1,n_2-1)=min(20-1,20-1)=19.$ The corresponding P-value by using the table: p>0.1. 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 $\alpha=0.05$, because it is more than 0.05, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.025}=2.093.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.093\sqrt{\frac{0.49^2}{20}+\frac{0.44^2}{20}}=0.31.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(0.49-0.44)-0.31=-0.26 and$\overline{x_1}-\overline{x_2}+E$=(0.49-0.44)+0.31=0.36.
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