Question

A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use

α = 0.05

and the following data to test the hypothesis and comment on the value of the commercial.

Individual	Purchase Rating
	After	Before
1	6	5
2	6	5
3	7	8
4	4	3
5	3	5
6	9	8
7	7	5
8	6	6

State the null and alternative hypotheses. (Use μ_d = mean rating after − mean rating before.)

H₀:μ_d = 0

H_a:μ_d ≠ 0

Calculate the value of the test statistic. (Round your answer to three decimal places.)

Calculate the p-value. (Round your answer to four decimal places.)

p-value =

Answer 1

After	Before	Difference
6	5	1
6	5	1
7	8	-1
4	3	1
3	5	-2
9	8	1
7	5	2
6	6	0

Sample mean of the difference using excel function AVERAGE(), x̅_d = 0.3750

Sample standard deviation of the difference using excel function STDEV.S(), s_d = 1.3025

Sample size, n = 8

Null and Alternative hypothesis:

Ho : µ_d = 0

H1 : µ_d ≠ 0

Test statistic:

t = (x̅_d)/(s_d/√n) = (0.375)/(1.3025/√8) = 0.814

df = n-1 = 7

p-value = T.DIST.2T(ABS(0.8143), 7) = 0.4423

Decision:

p-value > α, Do not reject the null hypothesis