Question

You would like to determine if a prospective customer is more likely to purchase a product after viewing a promotional advertisement for that product. You have the following data from the focus group.

Customer	Likely to Purchase	Likely to Purchase After Promotion
A	54	61
B	39	40
C	62	57
D	78	80
E	90	93
F	25	44
G	35	40

Conduct a hypothesis test using a 0.05 level of significance. Remember to show your work and indicate your conclusion.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.943
since our test is left-tailed
reject Ho, if to < -1.943
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -4.571
We have d = -4.571
pooled variance = calculate value of Sd= √S^2 = sqrt [ 474-(-32^2/7 ] / 6 = 7.39
to = d/ (S/√n) = -1.636
critical Value
the value of |t α| with n-1 = 6 d.f is 1.943
we got |t o| = 1.636 & |t α| =1.943
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.6364 ) = 0.07644
hence value of p0.05 < 0.07644,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -1.636
critical value: reject Ho, if to < -1.943
decision: Do not Reject Ho
p-value: 0.07644
we do not have enough evidence to support the claim that if a prospective customer is more likely to purchase a product after viewing a promotional advertisement for that product.