Question

Businesses such as General Mills, Kellogg's, and Betty Crocker regularly use coupons to build brand allegiance and stimulate sales. Marketers believe that the users of paper coupons are different from the users of e-coupons accessed through the Internet. One survey recorded the age of each person who redeemed a coupon along with the type of coupon (either paper or electronic). The sample of 24 traditional paper-coupon clippers had a mean age of 35.8 with a standard deviation of 11.2. The sample of 35 e-coupon users had a mean age of 41.8 years with a standard deviation of 4.8. Assume the population standard deviations are not the same.

Using a significance level of 0.02, test the hypothesis of no difference in the mean ages of the two groups of coupon clients.

Hint: For the calculations, assume e-coupon as the first sample.

Find the degrees of freedom for unequal variance test. (Round down answer to nearest whole number.)

State the decision rule for 0.02 significance level: H0: μe-coupon = μtraditional ; H1: μe-coupon ≠ μtraditional. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

Test the hypothesis of no difference in the mean ages of the two groups of coupon clients.

Answer 1

Given that,
mean(x)=35.8
standard deviation , s.d1=11.2
number(n1)=24
y(mean)=41.8
standard deviation, s.d2 =4.8
number(n2)=35
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.02
from standard normal table, two tailed t α/2 =2.5
since our test is two-tailed
reject Ho, if to < -2.5 OR if to > 2.5
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =35.8-41.8/sqrt((125.44/24)+(23.04/35))
to =-2.473
| to | =2.473
critical value
the value of |t α| with min (n1-1, n2-1) i.e 23 d.f is 2.5
we got |to| = 2.47332 & | t α | = 2.5
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.4733 ) = 0.021
hence value of p0.02 < 0.021,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.473
critical value: -2.5 , 2.5
decision: do not reject Ho
p-value: 0.021
we do not have enough evidence to support the claim that no difference in the mean ages of the two groups of coupon clients.