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: 12

Answer

a) Fail to reject the null hypothesis. b)$\mu_1-\mu_2$ is between -18.08353 and 7.25019.

Work Step by Step

a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1\ne\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(-20.5+15.08333)-(0)}{\sqrt{12.38401^2/12+15.62317^2/12}}=-0.941.$ 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 more than 0.2. 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 0.2, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.025}=2.201.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.201\sqrt{\frac{12.38401^2}{12}+\frac{15.62317^2}{12}}=12.66686.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(-20.5+15.08333)-12.66686=-18.08353 and$\overline{x_1}-\overline{x_2}+E$=(-20.5+15.08333)-12.66686=7.25019.
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