Step 1 :Given the following data for male and female BMI: \n\n\(\mu_1\) (Male BMI), \(\mu_2\) (Female BMI), \(n_1 = 46\) (Male sample size), \(n_2 = 46\) (Female sample size), \(\bar{x}_1 = 27.2376\) (Male sample mean), \(\bar{x}_2 = 24.9608\) (Female sample mean), \(s_1 = 8.797257\) (Male sample standard deviation), \(s_2 = 5.702369\) (Female sample standard deviation), and \(\alpha = 0.05\) (Significance level).
Step 2 :First, we need to set up the null and alternative hypotheses for the test. The null hypothesis is that the mean BMI for males and females is the same, and the alternative hypothesis is that the mean BMI for males and females is not the same. So, \n\n\(H_0: \mu_1 = \mu_2\) \n\n\(H_1: \mu_1 \neq \mu_2\)
Step 3 :Next, we calculate the standard error (SE) using the formula: \n\n\(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = 1.5457427012971017\)
Step 4 :Then, we calculate the test statistic (t) using the formula: \n\n\(t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = 1.4729488925223047\)
Step 5 :We also calculate the degrees of freedom (df) using the formula: \n\n\(df = n_1 + n_2 - 2 = 90\)
Step 6 :Then, we find the P-value associated with the calculated t-value and the degrees of freedom. The P-value is 0.14425360399554354.
Step 7 :Since the P-value is greater than the significance level of 0.05, we 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.
Step 8 :Finally, we construct a 95% confidence interval for the difference in mean BMI between males and females. The confidence interval is \([-0.794, 5.348]\).