Answer
a) We reject the null hypothesis. b) $\mu_1-\mu_2$ is between 0.83 and 6.33.
Work Step by Step
a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1 \gt \mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(15.89-12.31)-(0)}{\sqrt{5.9^2/35+5.48^2/36}}=2.647.$ The degree of freedom: $min(n_1-1,n_2-1)=min(35-1,36-1)=34.$ The corresponding P-value by using the table: $0.01 \lt P \lt 0.02$. 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 0.02, hence we reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.025}=2.032.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.032\sqrt{\frac{5.9^2}{35}+\frac{5.48^2}{36}}=2.75.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(15.89-12.31)-2.75=0.83 and$\overline{x_1}-\overline{x_2}+E$=(15.89-12.31)+2.75=6.33.
