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

Answer:

a)

Given,

Sp^2 = ((n1-1)s1^2 + (n2-1)s2^2)/(n1+n2-2)

substitute values & then we get

= 16.2034

alpha = 0.01

degree of freedom = n1 + n2 - 2

= 25 + 20 - 2

= 43

t(alpha/2 , df) = 2.695

99% CI = (x1 - x2) +/- t*Sp*sqrt(1/n1 + 1/n2)

substitute values

= (75 - 89) +/- 2.695*16.2034*sqrt(1/25 + 1/20)

= - 14 +/- 13.1004

= (-27.1004 , - 0.8996)

b)

Ho : u1 - u2 = 0

Ha : u1 - u2 < 0

test statistic t = (x1 - x2)/Sp*sqrt(1/n1 + 1/n2)

substitute values

= (75 - 89)/16.2034*sqrt(1/25 + 1/20)

= - 2.8801

P value = 0.0031

Here we observe that p value < alpha, so we reject Ho.

So there is sufficient evidence to support the claim.