Question

PART I

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. Those 21 participants on the drug had an average test score of 27.501 (SD = 4.089), while those 21 participants not on the drug had an average score of 40.405 (SD = 6.312). You use this information to create a 90% confidence interval for the difference in average test score of (-15.667, -10.141). Which of the following is the best interpretation of this interval?

	1) We do not know the population means so we do not have enough information to make an interpretation.
	2) We are 90% sure that the average difference in test score of all people who would take the drug and people not on the drug in the state the study was completed in is between -15.667 and -10.141.
	3) We are 90% confident that the difference between the average test score of all people who would take drug and all people not taking the drug is between -15.667 and -10.141.
	4) We are certain that the difference between the average test score of all people who would take drug and all people not taking the drug is between -15.667 and -10.141.
	5) We are 90% confident that the difference between the average test score of people who would take drug in the study and people not on the drug in the study is between -15.667 and -10.141.

PART II

Disability Services introduced a new mentorship program to help students with disabilities achieve better scholastic results. Test grades were recorded for 26 students before and after the program was introduced. The average difference in test score (after - before) was 1.794 with a standard deviation of 3.537. If Disability Services is interested in creating a 99% confidence interval for the true average difference in test scores, what is the margin of error?

	1) 1.7193
	2) 0.6937
	3) 1.9275
	4) 1.7869
	5) 1.9335

PART III

The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 9 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 9 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 7.554 and the standard deviation of the differences is 17.4972. Calculate a 99% confidence interval to estimate the average difference in scores between the two courses.

	1) (-12.016, 27.124)
	2) (-11.4003, 26.5083)
	3) (1.7216, 13.3864)
	4) (4.1986, 10.9094)
	5) (12.016, 27.124)

Answer 1

(1)

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. Those 21 participants on the drug had an average test score of 27.501 (SD = 4.089), while those 21 participants not on the drug had an average score of 40.405 (SD = 6.312). You use this information to create a 90% confidence interval for the difference in average test score of (-15.667, -10.141). Which of the following is the best interpretation of this interval?

We use confidence interval to estimate the range which will consist the true difference of the populations with some level of confidence. Therefore,

3.We are 90% confident that the difference between the average test score of all people who would take drug and all people not taking the drug is between -15.667 and -10.141.

Reasons for incorrect ans.

The (1) is unnecessay, (2) it talks only about the state where the study takes place which is not true since the test would be for the entire country as the population.

(4) There is no probability assigned to the statement. (5) This talks about the study (sample) but the estimate is for the population since the sample difference is calculated.

(2)

Disability Services introduced a new mentorship program to help students with disabilities achieve better scholastic results. Test grades were recorded for 26 students before and after the program was introduced. The average difference in test score (after - before) was 1.794 with a standard deviation of 3.537. If Disability Services is interested in creating a 99% confidence interval for the true average difference in test scores, what is the margin of error?

The subjects are same. So this will use a paired t-dist.

Margin or error = Critical value * Standard error

Std error =

=

= 0.6937

Critical value =

= .............alpha = 1 - 0.99 =0.01

= 2.787 ...........Using t-dist tables df = 25, p = 0.005

MOE = 2.787 * 0.6937

(3)

The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 9 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 9 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 7.554 and the standard deviation of the differences is 17.4972. Calculate a 99% confidence interval to estimate the average difference in scores between the two courses.

Again here since the same golf players are experimented, a paired t-dist will be used.

99% confidence interval for true average difference

Alpha= 1 - 0.99 = 0.01

Substituting the values we have