Question

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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	4
3	7	7
4	4	3
5	3	6
6	9	8
7	7	5
8	6	7

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

H₀: μ_d ≠ 0

H_a: μ_d = 0

H₀: μ_d > 0

H_a: μ_d ≤ 0

    

H₀: μ_d ≤ 0

H_a: μ_d = 0

H₀: μ_d = 0

H_a: μ_d ≠ 0

H₀: μ_d ≤ 0

H_a: μ_d > 0

(b)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 =

(c) State your conclusion.

Do not Reject H₀. There is sufficient evidence to conclude that seeing the commercial improves the mean potential to purchase.

Reject H₀. There is sufficient evidence to conclude that seeing the commercial improves the mean potential to purchase.    

Reject H₀. There is insufficient evidence to conclude that seeing the commercial improves the mean potential to purchase.

Do not reject H₀. There is insufficient evidence to conclude that seeing the commercial improves the mean potential to purchase.

Answer 1

Here, we have given that,

Xi: Individuals rate purchase potential of a particular product before

Yi: Individuals rate purchase potential of a particular product After

Xi (Individual Rate After)	Yi (Purchase Rate Before)	Difference (di=Yi-Xi)
1	6	5
2	6	4
3	7	4
4	4	0
5	3	-2
6	9	3
7	7	0
8	6	-2

n=number of observations=8

\bar d= Sample mean of differences = \frac{\sum di}{n}

= \frac{5+4+ ... +0+(-2)}{8}

=1.50

Sd= sample standard deviation of differences

=

=

= 2.83

(a)

Claim: To check whether the commercial improved the mean purchase potential rating.

The null and alternative hypothesis is as follows,

Versus

where =Population mean purchase potential rating difference (mean rating after - mean rating before)

This is the right one-tailed test

(b)

Now, we can find the test statistic

t-statistics =

=

= 1.50

we get the test statistics is 1.50

Now, we want to find the p-value

Degrees of freedom=n-1=8-1=7

P-value = P( T > t-statistics) as this is the right one-tailed test.

= 1 - P (T < t-statistics)

= 1 - P (T < 1.50)

=1- 0.0886 ( using EXCEL =TDIST( |t-statistics |=1.50,D.F=7,tail=1))

= 0.9114

we get the p-value is 0.9114

Decision:

= level of significance=0.05

Here, P-value (0.9114) greater than (>) 0.05 ()

Conclusion:

Do not reject Ho. There is insufficient evidence to conclude that seeing the commercial improves the mean potential to purchase.