## Elementary Statistics (12th Edition)

Published by Pearson

# Chapter 9 - Inferences from Two Samples - 9-3 Two Means: Independent Samples - Basic Skills and Concepts: 20

#### Answer

a) Reject the null hypothesis. b)$\mu_1-\mu_2$ is between 5.02452 and 17.3088.

#### Work Step by Step

a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1$ is more than $\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(147.58333-136.41667)-(0)}{\sqrt{10.63834^2/12+5.21289^2/12}}=3.265.$ The degree of freedom: $min(n_1-1,n_2-1)=min(12-1,12-1)=11.$ The corresponding P-value by using the table: p is less than 0.005. 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 less than $\alpha=0.05$, because it is less than 0.005, hence we reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.05}=1.796.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=1.796\sqrt{\frac{10.68384^2}{12}+\frac{5.21289^2}{12}}=6.14214.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(147.58333-136.41667)-6.14214=5.02452 and $\overline{x_1}-\overline{x_2}+E$=(147.58333-136.41667)+6.14214=17.3088.

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