Question

1) You wish to extend your study of the monthly sales of this particular type of men’s shirt so as to be able to compare the mean sales of the item at two types of store that make up the chain, superstores and regular stores. The population standard deviation of the monthly sales of the shirt at superstores is thought to equal 65 shirts, and the population standard deviation of the monthly sales of the shirt at regular stores is thought to equal 85 shirts. A random sample of 35 superstores is selected. The mean sales in a recent month for these stores are found to equal 487.6 shirts. A random sample of 40 regular stores is selected. The mean sales in the same recent month for these stores are found to equal 444.3 shirts. At each of the 5% and 2% levels of significance, are mean sales for the population of superstores greater than mean sales for the population of regular stores? If at either significance level you do observe that mean sales at the superstores are greater than mean sales at the regular stores, are mean sales greater by more than 25 shirts? If the software allows for it, construct 95% and 98% confidence intervals for the differences in mean sales between the two types of stores. Explain the meanings of these intervals.

Answer 1

H0 : mu1 = mu2
Ha: mu1 > mu2

test statistics;

t = (x1-x2)/sqrt(s1^2/n2+s2^2/n2)
= (487.6 - 444.3)/sqrt(65^2/35 + 85^2/40)
= 2.4944

p value = .0088

Reject H0 at 5% and 2%

H0 : mu1 - mu2 =25
Ha: mu1 - mu2 .25

test statistics;

t = (x1-x2)-d/sqrt(s1^2/n2+s2^2/n2)
= ((487.6 - 444.3)-25)/sqrt(65^2/35 + 85^2/40)
= 1.0542

p value = 0.1496

fail to Reject H0 at 5% and 2%

t value at 95% = 2.0322

CI = (x1-x2)+/- t * sqrt(s1^2/n2+s2^2/n2)
= (487.6 - 444.3) +/- 2.0322 * sqrt(65^2/35 + 85^2/40)
= (8.0228,78.5772)

t value at 98% = 2.4411

CI = (x1-x2)+/- t * sqrt(s1^2/n2+s2^2/n2)
= (487.6 - 444.3) +/- 2.4411 * sqrt(65^2/35 + 85^2/40)
= (0.9246,85.6754)