Question

The management of a supermarket wanted to investigate if the male customers spend less money, on average, than the female customers. A sample of 25 male customers who shopped at this supermarket showed that they spent an average of $75 with a standard deviation of $17.50. Another sample of 20 female customers who shopped at the same supermarket showed that they spent an average of $89 with a standard deviation of $14.40. Assume that the amounts spent at this supermarket by all male and all female customers are normally distributed with equal but unknown population standard deviations.

a. Construct a 99% confidence interval for the difference between the mean amounts spent by all male and all female customers at this supermarket.

b. Using the 5% significance level, can you conclude that the mean amount spent by all male customers at this supermarket is less than all female customers?

Answer 1

Solution:

Given

n1= 25 sample size of male customers

n2= 20 sample size of female customers.

sample mean of male customers.

. Sample mean of female customers.

s1 = $ 17.50 sample standard deviations of male customers.

s2 = $ 14.40 sample standard deviations of female customers.

a) The 99% confidence interval for the difference between the mean amount spent by all male and female customers is

Where

At

from t table

( - 27.10055, -0.899446)

The 99% confidence interval for difference between the mean amount spent by all male and female customers is

( -27.10055, -0.899446)

b) To test the hypothesis

. Vs.

Test statistic

Where

t = - 2.880030

Test statistic t = -2.88

At

from t table.

The t critical value is = 1.681

Decision:

Reject Ho

Conclusion : Reject Ho, There is sufficient evidence to conclude that the mean amount spent by all male customers at this supermarket is less than all female customers.