15 A market researh f used a sanple of individuals to rate the puchase poterntia

Question

15

A market researh f used a sanple of individuals to rate the puchase poterntial of a particular paoduct befce and afte te idividuals saw a new, expeive televisioncmeicial about the paoduct. The purchase poterntial ratings were based oa a0 to 10 scale with higher values indicating more potential to purchase the produet. You want to know if the commercial impoved the mean puschase potential rating, so you set the null hypothesis to be that the nean rating "after would be less thano equal to the mean rating "before a k about v the test s set up this av If vou re ect the null, what can you acccp ? ou w 8 to test the cla n t a significance level of 0.001. To do so, vou find a sample of 39 people, vou find the mean a er "before rating to be d =-0.3 with a standard deviation of the differences of s = 0.8 what is the test statistic for this sample? test statistic = Round to 4 decimal places. b. What is the p-value for this sample? Round to 4 decimal places. p-value c. The p-value is... less than (or equal to) greater than d. This test statistic leads to a decision to... reject the null accept the null fail to eject the ull e As such, the tinal conuois that There is sufficient evidence to warrant rejection of the claim that the mean 'after-before' ratin2 1s less than 0 OTiele 1s not sufficient evidence to warrant 1 ection of the clan!! that the mean at et etore ratnx s less than 0 The sample data support the claim that the mean 'after'-'before ratig is less than 0 There is not sutficent samplvdence to suppart the caim that the m'aftrbefore ratine is less than 0

Explanation / Answer

Answer

Given,

n=39

Standard deviation(Sd)=0.8

Mean difference(d-bar)= -0.3

Null Hypothesis H0: mean rating after commercial would be less than or equal to mean rating before

Alternate Hypothesis H1: mean rating after commercial would be greater than mean rating before

A.a) test statistic(Z)

Z=d-bar/{Sd/sqrt(n)}

Z=-0.3/{0.8/sqrt(39)}

Z=-2.3419           

A.b) using Z-table for Z=-2.3419

we get p-value as 0.0096

A.c) the calculated p-value is greater than alpha(0.001)

A.d) as p-value is greater than alpha, we reject the null

A.e) the final conclusion is that, there is sufficient evidence to warrant rejection of the claim that the mean “after-before” rating is less than 0

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