Answer:
a) Test of hypothesis Null hypothesis: $H_0: \mu_1
= \mu_2$Alternative hypothesis: $H_1: \mu_1 \\eq \mu_2$Level of significance
= $\alpha = 0.01$The test statistic is given as $t
= \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$We need to calculate the value of the test statistic:$t
= \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$t
= \frac{28.2271 - 25.9703}{\sqrt{\frac{(7.785511)^2}{49} + \frac{(4.807094)^2}{49}}}$$t = 2.526$
The degrees of freedom are calculated as: $df
= \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}}$Substituting the values, we get $df = 94.243$The P-value for a two-tailed test at a significance level of 0.01 is calculated as P-value
= $2P(t_{94.243} > 2.526)$Using a t-distribution table or a calculator, the P-value is found to be P-value $
= 0.013$The conclusion of the hypothesis test is: Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
(Option B is correct) b) Confidence interval 99% confidence interval for the difference of the means is given by $\overline{x}_1 - \overline{x}_2 \pm t_{\alpha/2, df}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$Substituting the values, we get the confidence interval:$(28.2271 - 25.9703) \pm t_{0.005, 94.243}\sqrt{\frac{(7.785511)^2}{49} + \frac{(4.807094)^2}{49}}$$2.2568 \pm 2.6737$$-0.4169 < \mu_1 - \mu_2 < 5.9305$
Since the confidence interval includes the value 0, it supports the conclusion of the test that there is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI. Therefore, the correct answer is D: Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.