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 0.01 significance level for both parts.
a. test the claim that males and females have the same mean body mass index (bmi).
what are the null and alternative hypotheses?
a. ( h_{0}: mu_{1} geq mu_{2} )
( h_{1}: mu_{1}<mu_{2} )
b. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}>mu_{2} )
c. ( h_{0}: mu_{1}
eq mu_{2} )
( h_{1}: mu_{1}<mu_{2} )
d. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}
eq mu_{2} )
the test statistic, ( t ), is 2.22. (round to two decimal places as needed.)
the p - value is (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The formula for degrees of freedom \(df\) when \(n_1 = n_2=n\) (here \(n = 40\)) and variances are not assumed equal is \(df=\frac{(s_1^{2}/n_1 + s_2^{2}/n_2)^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}\)

Substitute \(s_1 = 7.101911\), \(n_1=40\), \(s_2 = 4.141761\), \(n_2 = 40\)

\(s_1^{2}/n_1=\frac{7.101911^{2}}{40}\approx1.260\)

\(s_2^{2}/n_2=\frac{4.141761^{2}}{40}\approx0.429\)

\((s_1^{2}/n_1 + s_2^{2}/n_2)^{2}=(1.260 + 0.429)^{2}\approx2.853\)

\(\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{1.260^{2}}{39}+\frac{0.429^{2}}{39}\approx\frac{1.588}{39}+\frac{0.184}{39}\approx0.0407 + 0.0047=0.0454\)

\(df=\frac{2.853}{0.0454}\approx62.84\approx63\)

Step2: Calculate the P - value

Since \(t = 2.22\) and it is a two - tailed test (\(H_1:\mu_1
eq\mu_2\)), we use the t - distribution table or a calculator.

Using a calculator (e.g., in R: \(2*(1 - pt(2.22,63))\)), or using the formula for the P - value of a two - tailed t - test \(P=2P(T >|t|)\) where \(T\) follows a t - distribution with \(df = 63\)

\(P\approx0.029\)

Answer:

\(0.029\)