Homework Assignment:1471515

1. A marketing specialist is assessing the effectiveness of two types of formats for TV commercials. Twenty-two regular TV watchers are randomly selected and then assigned to watch one of the two formats of a commercial for an imported car. The following data are the scores on a measure for assessing the impact of commercials. Answer the following questions using α = .10.

A	B
21	38
33	32
40	56
37	58
22	45
25	46
39	57
28	49
34	43
25	41
27	55
Sum=331	Sum=520

1. State the hypotheses.

A hypothesis is defined as an educated guess. There are two hypotheses namely the null and the alternative hypothesis. The null hypothesis shows no difference or status quo of the current situation and the treatment values while the alternative hypothesis shows the researchers’ question of interest. According to the description above, the following is the researchers’ question and the null and alternative hypothesis.

Research question: Is there a mean difference in the effectiveness of TV commercials between  A and B?

Null hypothesis Ho: There is no difference in the effectiveness of TV commercials between A and B.

Alternative Hypothesis Ha: There is a significant difference in the effectiveness of TV commercials between A and B.

• Check the homogeneity of variances assumption

Table 1: Independent Sample t-test

The homogeneity of variance is examined using Levene’s test for equality of variances (Kim, 2015). From the output above, F=0.505, p=.485. The p-value >.10 hence the result is not significant. Therefore, the equality of variance between A and B is assumed.

If the Levene test result is significant, we read from the second row under equality of variance not assumed.

• Now you need to test the above hypotheses from (1). First, set the critical values for rejecting H0 below (please specify how you selected the criteria).

The test statistic follows the t-distribution with 20 degrees of freedom (n1+n2-2)

Th formula is; t)

 t_{0.1/2 ,11+11-2} = t_{0.05 ,20}=1.725 (From student t-distribution tables)

Critical value: ±1.725

This acts as a boundary where we will check if the hypothesis is significant. Using the p-value approach, the alpha=.10 will act as the boundary to reject or accept the null hypothesis.

• Compute the standard error of the difference of the means (include your computation).

Standard deviation A;  = 6.804

Standard deviation B;  = 8.557

Standard error=

S₁=6.804, n1=11 and S₂=8.557, n2=11

Standard error=

Standard error===3.296

• Compute the test statistic (include your computation).

Test statistic=

Sample Mean A (M1); 30.09

Sample Mean B;(M2): 47.27

Test statistic==-17.18/3.296= –5.212

This is the value we will compare with the critical value of the test

• Interpret the results regarding the hypotheses from (1).

Considering that the test statistic -5.212< the critical value of -1.725, we reject the null hypothesis since the value of t is in the rejection region (Doane, Seward & Chowdhury,2020). We, therefore, conclude that there exists a significant difference in the effectiveness of TV commercials between A and B.

When checking the output for analysis provided in table 1, it is necessary to read the values in the first row under equality of variance. For instance, t(20)= -5.213,p=.00.This implies that the results are significant at 1%,5%, and 10% levels of significance. The null hypothesis is rejected and the alternative hypothesis is adopted.

References

Doane, D. P., Seward, L. E., & Chowdhury, S. (2020). Applied Statistics in Business and Economics| | SIE. McGraw-Hill Education.

Kim, T. K. (2015). T test as a parametric statistic. Korean journal of anesthesiology, 68(6), 540.