Question

Another market researcher is trying to determine if exposure to a particular ad can change consumer’s attitudes about a new product. He collects the data below.    Use hypothesis testing and the appropriate method to see if there is a difference between attitudes before and after exposure to the ad. (α = .02).

Subject	Pre-Exposure Attitude (A1)	Post-Exposure Attitudes (A2)
1	48	55
2	20	27
3	35	38
4	48	55
5	63	61
6	88	85
7	45	45
8	34	31
9	68	72
10	70	78

Answer 1

Subject	A1	A2	Difference
1	48	55	-7
2	20	27	-7
3	35	38	-3
4	48	55	-7
5	63	61	2
6	88	85	3
7	45	45	0
8	34	31	3
9	68	72	-4
10	70	78	-8

Paired Sample t-test

For the score differences we have, mean is Dˉ=-2.8, the sample standard deviation is sD=4.4672, and the sample size is n=10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μD =0 Ha: μD ≠0 This corresponds to a Two-tailed test, for which a t-test for two paired samples be used. (2a) Critical Value Based on the information provided, the significance level is α=0.02, and the degree of freedom is n-1=10-1=9. Therefore the critical value for this Two-tailed test is tc​=2.8214. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |t|>2.8214 i.e. t>2.8214 or t<-2.8214 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0788 (5) The Decision about the null hypothesis (a) Using the traditional method Since it is observed that |t|=1.9821 < tc​=2.8214, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0788, and since p=0.0788>0.02, it is concluded that the null hypothesis is Not rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ1​ is different than μ2, at the 0.02 significance level.

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