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: 15


a) Fail to reject the null hypothesis. b)$\mu_1-\mu_2$ is between -1.0808 and 4.7613.

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: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(28.44075-26.6005)-(0)}{\sqrt{7.394076^2/40+5.359442^2/40}}=1.274.$ The degree of freedom: $min(n_1-1,n_2-1)=min(40-1,40-1)=39.$ 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: $t_{\alpha/2}=t_{0.025}=2.023.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.023\sqrt{\frac{7.394076^2}{40}+\frac{5.359442^2}{40}}=2.92105.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(28.44075-26.6005)-2.92105=-1.0808 and$\overline{x_1}-\overline{x_2}+E$=(28.44075-26.6005)+2.92105=4.7613.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.