a Compute d to 3 decimals and compute sd to 1 decimal b The

a. Compute d (to 3 decimals) and compute sd (to 1 decimal)

b. The p-value is:

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 = .05 and the following data to test the hypothesis and comment on the value of the commercial. Mu d __ to 0 a. Compute d (to 3 decimals) and compute sd (to 1 decimal) b. The p-value is: __ to 0 Ha: H0:

Solution

We use paired t test to do it.

Null hypothesis: the mean rating \"after\" would be less than or equal to the mean rating \"before.

Alternative hypothesis: the mean rating \"after\" would be greater to the mean rating \"before.

The test statistic is

t= mean difference/ std. error

=0.625/0.46

=1.36

Given a=0.05, the critical value is t(0.05, df=n-1=7) =1.89 (from student t table)

Since t=1.36 is less than 1.89, we do not reject Ho.


So we can not conclude that the mean rating \"after\" would be greater to the mean rating \"before.

6.000 mean After
5.375 mean Before
0.625 mean difference (After - Before)
1.302 std. dev.
0.460 std. error
8 n
7 df
a. Compute d (to 3 decimals) and compute sd (to 1 decimal) b. The p-value is: A market research firm used a sample of individuals to rate the purchase potential

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