In Exercises 5–20, 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. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

BMI We know that the mean weight of men is greater than the mean weight of women, and the mean height of men is greater than the mean height of women. A person’s body mass index (BMI) is computed by dividing weight (kg) by the square of height (m). Given below are the BMI statistics for random samples of females and males taken from Data Set 1 “Body Data” in Appendix B.

a. Use a 0.05 significance level to test the claim that females and males have the same mean BMI.

b. Construct the confidence interval that is appropriate for testing the claim in part (a).

c. Do females and males appear to have the same mean BMI?

Female BMI: n = 70, \(\bar x\) = 29.10, s = 7.39

Male BMI: n = 80, \(\bar x\) = 28.38, s = 5.37

For a sample of 70 females, the mean BMI is equal to 29.10 kg/\({{\rm{m}}^2}\),and the standard deviation of the BMI is equal to 7.39 kg/\({{\rm{m}}^2}\). For another sample of 80 males, the mean BMI is equal to 28.38 kg/\({{\rm{m}}^2}\),and the standard deviation of the BMI is equal to 5.37 kg/\({{\rm{m}}^2}\).

It is claimed that the mean BMI of females is equal to the mean BMI of males.

Let\({\mu _1}\)and\({\mu _2}\)be the population mean BMIs of females and males, respectively.

Null hypothesis: The population mean BMI of females is equal to the population mean BMI of males.

Symbolically,

\({H_0}:{\mu _1} = {\mu _2}\).

Alternate hypothesis: The population mean BMI of females is not equal to the population mean BMI of males.

Symbolically,

\({H_1}:{\mu _1} \ne {\mu _2}\).

The sample mean BMI for females\(\left( {{{\bar x}_1}} \right)\)is equal to 29.10.

The sample mean BMI for males\(\left( {{{\bar x}_2}} \right)\)is equal to 28.38.

The sample standard deviation\(\left( {{s_1}} \right)\)of BMI for females is equal to 7.39.

The sample standard deviation\(\left( {{s_2}} \right)\)of BMI for males is equal to 5.37.

The sample size\(\left( {{n_1}} \right)\)is equal to 70, and the sample size\(\left( {{n_2}} \right)\)is equal to 80.

Substitute the respective values in the above formula and simplify.

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}}} }}\\ = \frac{{\left( {29.10 - 28.38} \right) - 0}}{{\sqrt {\frac{{{{\left( {7.39} \right)}^2}}}{{70}} + \frac{{{{\left( {5.37} \right)}^2}}}{{80}}} }}\\ = 0.674\end{array}\).

The degrees of freedom is the smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below.

\(\begin{array}{c}\left( {{n_1} - 1} \right) = 70 - 1\\ = 69\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = 80 - 1\\ = 79\end{array}\)

Thus, the value of the degrees of freedom is equal to the smaller one of the values 69 and 79, which is 69.

The critical values can be obtained using the t-distribution table with degrees of freedom equal to 69 and the significance level equal to 0.05 for a two-tailed test.

Thus, the critical values are -1.9949 and 1.9949.

The p-value for t equal to 0.674 is equal to 0.5026.

a.

As the test statistic value lies between the critical values and the p-value is greater than 0.05, the null hypothesis is failed to reject.

Therefore,there is not sufficient evidence to reject the claim thatfemales and males have the same mean BMI.

b.

The confidence interval has the following expression:

\(CI = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\).

The margin of error is given by the following formula:
\(E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \).

Here,\(\alpha = 0.05\). Thus, the value of\({t_{\frac{\alpha }{2}}}\)for 69 degrees of freedom is equal to 1.9949.

Substitute the respective values in the above formula to compute the margin of error.

\(\begin{array}{c}E = 1.9949 \times \sqrt {\frac{{{{7.39}^2}}}{{70}} + \frac{{{{5.37}^2}}}{{80}}} \\ = 2.13056\end{array}\).

Substituting the value of E and the sample means, the following confidence interval is obtained:

\(\begin{array}{c}\left( {29.10 - 28.38} \right) - 2.13056 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {29.10 - 28.38} \right) + 2.13056\\ - 1.41 < \left( {{\mu _1} - {\mu _2}} \right) < 2.85\end{array}\)

Thus, the 95% confidence interval of the difference in the two population means is equal to (-1.41, 2.85).

It can be observed that 0 lies within the interval. This implies that there is no significant difference between the two population means.

Therefore,there is not sufficient evidence to reject the claim thatfemales and males have the same mean BMI.

c.

The sample mean BMI for females is equal to 29.10 kg/\({{\rm{m}}^2}\),and the sample mean BMI for males is equal to 28.38 kg/\({{\rm{m}}^2}\).

Thus, it appears that females and males have the same mean BMI.