1-9.3)
Part 5 of 6
points
Points: 0 of 1
Save
Given in the table are the BMI statistics for random samples of men and women. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts.
\begin{tabular}{|c|c|c|}
\hline & Male BMI & Female BMI \\
\hline $\boldsymbol{\mu}$ & $\mu_{1}$ & $\mu_{2}$ \\
\hline $\mathbf{n}$ & 46 & 46 \\
\hline $\bar{x}$ & 27.2376 & 24.9608 \\
\hline $\mathbf{s}$ & 8.797257 & 5.702369 \\
\hline
\end{tabular}
a. Test the claim that males and females have the same mean body mass index (BMI).
What are the null and alternative hypotheses?
A. $\mathrm{H}_{0}: \mu_{1}=\mu_{2}$
\[
H_{1}: \mu_{1}> \mu_{2}
\]
C. $H_{0}: \mu_{1} \neq \mu_{2}$
\[
H_{1}: \mu_{1}< \mu_{2}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu_{1} \geq \mu_{2} \\
H_{1}: \mu_{1}< \mu_{2}
\end{array}
\]
The test statistic, $\mathrm{t}$, is 1.47 . (Round to two decimal places as needed.)
The $\mathrm{P}$-value is 0.146 . (Round to three decimal places as needed.)
State the conclusion for the test.
A. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
B. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
C. 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.
D. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
b. Construct a confidence interval suitable for testing the claim that males and females have the same mean BMI.
$< \mu_{1}-\mu_{2}< \square$
(Round to three decimal places as needed.)
View an example
Get more help -
Clear all
Check answer
Answer
Finally, we construct a 95% confidence interval for the difference in mean BMI between males and females. The confidence interval is \([-0.794, 5.348]\).
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 \([-0.794, 5.348]\).
