Question

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.
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 not 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.
$\square<\mu_1 - \mu_2<\square$
(round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

We use the formula \(df=\min(n_1 - 1,n_2 - 1)\). Here \(n_1=n_2 = 45\), so \(df=45- 1=44\). For a two - tailed test with \(\alpha = 0.05\), the critical value \(t_{\alpha/2}\) from the t - distribution table (or using a calculator) is approximately \(t_{0.025}\approx 2.015\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error \(E\) for two independent samples (variances not assumed equal) is \(E=t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\)
Substitute \(s_1 = 8.600843\), \(n_1 = 45\), \(s_2=5.119136\), \(n_2 = 45\), \(t_{\alpha/2}=2.015\)

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Step3: Calculate the confidence interval

The formula for the confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E\)
Substitute \(\bar{x}_1 = 27.3776\), \(\bar{x}_2=24.8991\), \(E = 2.996\)
\((27.3776 - 24.8991)-2.996<\mu_1-\mu_2<(27.3776 - 24.8991)+2.996\)
\(2.4785-2.996<\mu_1-\mu_2<2.4785 + 2.996\)
\(- 0.5175<\mu_1-\mu_2<5.4745\)

Answer:

\(-0.518<\mu_1-\mu_2<5.475\)