Question

A study to compare the means of the age x at first marriage of men and women obtained the information shown, where the men and women were sampled independently of one another. We may assume that the populations of ages are normally distributed with approximately equal standard deviations.

	n	$LaTeX: \bar{x}$ x ¯	s
male	50	26.8	0.21
female	50	25.1	0.19

The correct formula to employ to use these data to construct a 90% confidence interval for the difference in mean age at first marriage between men and women is:

A.   $LaTeX: (\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2} \sqrt{s^2_p ( \frac{1}{n_1}+\frac{1}{n_2} )}$

B. $LaTeX: b\pm t_{\alpha/2} \frac{s_e}{SS_xx}$

C. $LaTeX: \bar{d} \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

D. $LaTeX: \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

E.   $LaTeX: \hat{p} \pm z_{\alpha/2} \sqrt{\frac{ \hat{p} \hat{q}}{n}}$

Answer 1

We may assume that the populations of ages are normally distributed with approximately equal standard deviations. So, we calculate the pooled standard deviation to compute the 90% confidence interval for the difference in mean age at first marriage between men and women.

Therefore, the correct formula as follows :

ni n2

Answer : A)