Question

# PART A) The owner of a golf course wants to determine if his golf course is...

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 in-store 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?

 1) We do not have enough information to make a conclusion.
 2) We are 99% confident that the average GPA of seniors is greater than the average GPA of sophomores.
 3) There is no significant difference between the true average GPA for seniors and sophomores.
 4) We are 99% confident that the average GPA of seniors is less than the average GPA of sophomores.
 5) We are 99% confident that the difference between the two sample GPAs falls within the interval.

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?

 1) There is no significant difference between the average memorization abilities for those on the drug compared to those not on the drug.
 2) We do not have enough information to make a conclusion.
 3) We are 95% confident that the average memorization ability of those not on the drug is higher than those who are on the drug.
 4) We are 95% confident that the difference between the two sample means falls within the interval.

5)

We are 95% confident that the average memorization ability of those on the drug is higher than those who are not on the drug.

PART E) A university is comparing the hourly wages of students of their university who completed a Master's Degree to students who only completed a Bachelor's Degree. Of the students surveyed, 73 had a Master's degree and 68 had a Bachelor's Degree only. The average hourly rate for those with a Master's was \$37.521 (SD = \$2.871). Among the graduates with a Bachelor's degree, the average hourly rate was \$22.48 (SD = \$8.604). You use this information to calculate a 95% confidence interval for the difference in mean hourly wage of (12.935, 17.147). Of the following statements, what is the best interpretation of this interval?

 1) We are certain that the difference between the average hourly wage of all graduates with a Master's degree and all graduates with a Bachelor's degree is between 12.935 and 17.147.
 2) We are 95% confident that the difference between the average hourly wage of all their graduates with a Master's degree and all their graduates with a Bachelor's degree surveyed is between 12.935 and 17.147.
 3) We are 95% sure that the difference in average hourly wage of all nationwide graduates with a Master's degree and all nationwide graduates with a Bachelor's degree is between 12.935 and 17.147.
 4) We are 95% confident that the difference between the average hourly wage of all their graduates with a Master's degree and all their graduates with a Bachelor's degree is between 12.935 and 17.147.
 5) We do not know the population means so we do not have enough information to make an interpretation.

PART F)

A restaurant wants to test a new in-store 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. 18 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 90% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after is (11.4, 34.2). Which of the following is the appropriate conclusion? The differences were calculated as (after ad campaign- before ad campaign).

 1) We are 90% confident that the average difference in sales quantity for all stores is positive, with the higher sales coming after the ad campaign.
 2) We are 90% confident that the average difference in sales quantity for all stores is negative, with the higher sales coming after the ad campaign.
 3) There is not a significant difference between the average sales quantity before or after the ad campaign.
 4) We are 90% confident that the average difference in sales quantity for all stores is negative, with the higher sales being before the ad campaign.
 5) We are 90% confident that the average difference in sales quantity for all stores is positive, with the higher sales being before the ad campaign.

PART G)The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 22 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 22 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 2.45 strokes and the standard deviation of the differences is 6.259 strokes. The owner uses this information to calculate the 90% confidence paired t-interval of (0.154, 4.746). Which statement is the correct interpretation of this interval?

 1) We are 90% confident that the average difference in the scores of the golfers sampled is between 0.154 and 4.746.
 2) The proportion of all golfers that had a difference in scores between the two courses is 90%.
 3) We are certain the the average difference in scores between the two courses for all golfers is between 0.154 and 4.746.
 4) We are 90% confident that the difference between the average score on the owner's golf course and the average score on the friend's golf course is between 0.154 and 4.746.
 5) We are 90% confident that the average difference in scores between the two courses for all golfers is between 0.154 and 4.746.

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 = Sd = 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 tc 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 : tc = 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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