Question

A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 15 men had a mean height of 69.4 inches with a standard deviation of 3.09 inches. A random sample of 7 women had a mean height of 64.9 inches with a standard deviation of 2.58 inches. Determine the 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women. Assume that the population variances are equal and that the two populations are normally distributed.

Answer 1

Answer:

(0.6613, 8.3387)

Step-by-step explanation:

Let the heights of men be the first population and the heights of women be the second population. Then

We have $n_{1} = 15$, $\bar{x}_{1} = 69.4$, $s_{1} = 3.09$ and

$n_{2} = 7$, $\bar{x}_{2} = 64.9$, $s_{2} = 2.58$. The pooled

estimate is given by  

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(3.09)^{2}+(7-1)(2.58)^{2}}{15+7-2} = 8.68$

The 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{7}}$, i.e.,

$(69.4-64.9)\pm t_{0.005}2.946\sqrt{\frac{1}{15}+\frac{1}{7}}$

where $t_{0.005}$ is the 0.5th quantile of the t distribution with (15+7-2) = 20 degrees of freedom. So  

$4.5\pm(-2.845)(2.946)(0.458)$, i.e.,

(0.6613, 8.3387)