Can you help me please?

Question

A nationally known supermarket decided to promote its own brand of soft drinks on TV for two weeks. Before the ad campaign, the company randomly selected 21 of its stores accross the United States to be part of the study to measure the campaigns effectiveness. During a specified half-hour period on a certain Monday morning, all the stores in the sample counted the number of cana of its own brand pf soft drinks sold. After the campaign, a similar count was made. The average diiference was an increase of 75 cans with a standard deviation of 30 cans. Using this information, show that there is a significant difference in the soft drink sales before and after the ad campaign. Use 5% level of significance. Solve problem using the critical value approach.

Answer 1

Hello Charmaine! I usually use the comparison of the p-value approach, but I modified my post to accommodate the critical value approach per your problem. I answered this problem as if it was a free response question that requires a 4 step process : )

 

State parameters: null: mu = there is no mean difference between sodas sold before and after the marketing campaign, alternative: mu = there is a mean difference between sodas sold before and after the marketing campaign. 

 

Plan: A one sample paired t test is appropriate here because one sample is used with the pre-intervention and post-intervention results. The difference between these two interventions (post - pre) is a positive difference of x bar = 75 with a sample standard deviation of 30.

 

Do: Calculate your t test statistic by (75 - 0)/(30/square root of 21). The t test statistic, t score, or t critical value is 11.46. Since you're using the critical value approach, you'd need to compare this to the critical value where the null would be rejected. Using InvT (.025, 20)[.025 is one half of the tails and 20 represents the degrees of freedom] the critical value is 2.086 where the null be rejected. If the test statistic is larger than the critical value, the null would be rejected.

 

Conclude: Since the t-test statistics (11.46) is larger than the critical value (2.086), we'd reject the null hypothesis. There is convincing evidence that there is a mean difference between the sodas sold before and after the marketing campaign. Therefore, the results of this study were statistically significant.