Question

In a typical month, men spend $178 and women spend $96 on leisure activities according to results from an International Communications Research (ICR) for American Express poll, as reported in USAToday June 25, 2015. Suppose random samples of 100 men and 100 women were taken from the population of male and female college students. Each student was asked to determine his or her expenditures for leisure activities in the prior month. The sample data results had a standard deviation of $75 for the men and $50 for the women.

A. Assuming normality in leisure activity expenditures, is the difference found in the ICR poll significant at alpha = .05?
B. Construct a 95% confidence interval for the difference in the expenditures. Interpret this interval.

Answer 1

A. Assuming normality in leisure activity expenditures, is the difference found in the ICR poll significant at alpha = .05?

The hypothesis being tested is:

H0: µ1 = µ2

H1: µ1 ≠ µ2

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that there is a difference found in the ICR poll.

B. Construct a 95% confidence interval for the difference in the expenditures. Interpret this interval.

The 95% confidence interval for the difference in the expenditures is between 64.224 and 99.776.

We are 95% confident that the true mean difference in the expenditures is between 64.224 and 99.776.

Men	Women
178	96	mean
75	50	std. dev.
100	100	n
	198	df
	82.000	difference (Men - Women)
	4,062.500	pooled variance
	63.738	pooled std. dev.
	9.014	standard error of difference
	0	hypothesized difference
	9.097	t
	9.96E-17	p-value (two-tailed)
	64.224	confidence interval 95.% lower
	99.776	confidence interval 95.% upper
	17.776	margin of error