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: 18

Answer

a) Fail to reject the null hypothesis. b) $\mu_1-\mu_2$ is between 4.6832 and 7.4038.

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: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(69.825-63.7815)-(0)}{\sqrt{0.79974^2/10+2.59665^2/40}}=12.533.$ The degree of freedom: $min(n_1-1,n_2-1)=min(10-1,40-1)=9.$ The corresponding P-value by using the table: p is more than 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 $\alpha=0.01$, because it is more than 0.1, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.01}=2.821.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.821\sqrt{\frac{0.7994^2}{10}+\frac{2.59665^2}{40}}=1.3603.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(69.825-63.7815)-1.3603=4.6832 and $\overline{x_1}-\overline{x_2}+E$=(69.825-63.7815)+1.3603=7.4038.
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