Question

A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the female in the household rather than the male in the household. A coupon book was sent at random to either the male or female in a random sample of 50 male-female households. A month later, a coupon book was sent to the other member of the pair. The manufacturer found that the mean difference (female − male) in the number of coupons redeemed was 1.5 with a standard deviation of 4.75.

Do these data provide convincing evidence that the mean number of coupons redeemed is greater when the coupons were addressed to a female? Use ? = 0.01. (4-step process)

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 0
Alternative Hypothesis, Ha: μ > 0

Rejection Region
This is right tailed test, for α = 0.01 and df = 49
Critical value of t is 2.405.
Hence reject H0 if t > 2.405

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (1.5 - 0)/(4.75/sqrt(50))
t = 2.233

P-value Approach
P-value = 0.0151
As P-value >= 0.01, fail to reject null hypothesis.

There is not sufficient evidence to conclude that he mean number of coupons redeemed is greater when the coupons were addressed to a female