PART A) The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 25 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 25 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course  the scores on his friend's course) is 3.63 and the standard deviation of the differences is 11.441. Calculate a 99% confidence interval to estimate the average difference in scores between the two courses.
PART B) A restaurant wants to test a new instore marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 7 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after  before) is 160.793 with a standard deviation of 50.528. When calculating a 90% confidence interval to estimate the true difference in nationwide sales quantity before the ad campaign and after, what is the margin of error?
PART C)Independent random samples are taken at a university to compare the average GPA of seniors to the average GPA of sophomores. Given a 99% confidence interval for the difference between the true average GPAs (seniors  sophomores) of (0.09, 1.02), what can you conclude?









PART D) A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample of individuals and splits them randomly into two groups. After the drug regimen is completed, all members of the study are given a test for memorization ability with higher scores representing a better ability to memorize. You are presented a 95% confidence interval for the difference in population mean scores (with drug  without drug) of (4.34, 18.86). What can you conclude from this interval?









Solution :
We are given that: The average difference in the scores (treated as the scores on his course  the scores on his friend's course) is 3.63 and the standard deviation of the differences is 11.441.
That is: Sample mean of differences =
Sample standard deviation of differences = S_{d} = 11.441
Sample size = n = 25
Since differences are taken on pairs of the same subject, we use a confidence interval for the paired differences.
We have to find 99% confidence interval for mean of paired differences.
Formula :
where
We have to find t_{c} value c = 99% confidence level and df = n  1 = 25  1 = 24
Two tail area = 1  c = 1  0.99 = 0.01
t critical value = 2.797
that is : t_{c} = 2.797
Thus :
Thus 99% confidence interval is :
Thus a 99% confidence interval to estimate the average difference in scores between the two courses is .
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