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: 16

Answer

a) Reject the null hypothesis. b)$\mu_1-\mu_2$ is between 1.4599 and 10.49982.

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: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(92.88462-86.90476)-(0)}{\sqrt{15.34451^2/78+8.988352^2/21}}=2.282.$ The degree of freedom: $min(n_1-1,n_2-1)=min(78-1,21-1)=20.$ The corresponding P-value by using the table: p is between 0.01 and 0.025. 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 0.025, hence we reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.05}=1.725.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=1.725\sqrt{\frac{15.34451^2}{78}+\frac{8.988352^2}{21}}=4.51996.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(92.88462-86.90476)-4.51996=1.4599 and$\overline{x_1}-\overline{x_2}+E$=(92.88462-86.90476)+4.51996=10.49982.
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