Question

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 18 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was -8.603 mg/dL with a standard deviation of 5.449 mg/dL. When creating a 90% confidence interval for the true average difference in cholesterol levels by the drug, what is the margin of error?

	1) 1.7087
	2) 2.2271
	3) 2.2342
	4) 1.2843
	5) 2.1127

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. 8 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 -198.323 with a standard deviation of 53.7248. Calculate a 99% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after.

	1) (-264.7942, -131.8518)
	2) (264.7942, -131.8518)
	3) (-262.0572, -134.5888)
	4) (-217.3176, -179.3284)
	5) (-201.8225, -194.8235)

The owner of a local golf course wants to estimate the difference between the scores of golfers on his golf course and the scores of golfers on his friend's golf course. He takes a random sample of 28 golfers and has them play a round of golf on his course and he records their scores. The next day, he has the same 28 golfers play a round of golf on his friend's course and records the scores again. He uses the scores to calculate a paired t-interval of (-20.4, -4.3) with a confidence level of 99%. Which of the following statements is the appropriate conclusion? The differences were calculated as (score on the owner's course - score on the friend's course).

	1) We are 99% confident that the average difference in scores for all golfers is positive, with the higher scores occurring on the friend's course.
	2) We are 99% confident that the average difference in scores for all golfers is positive, with the higher scores occurring on the owner's course.
	3) We are 99% confident that the average difference in scores for all golfers is negative, with the higher scores occurring on the owner's course.
	4) No significant difference exists between the average difference in scores and 0.
	5) We are 99% confident that the average difference in scores for all golfers is negative, with the higher scores occurring on the friend's course.

Researchers in the corporate office of an airline wonder if there is a significant difference between the cost of a flight on Priceline.com vs. the airline's own website. A random sample of 20 flights were tracked on Priceline and the airline's website and the average difference between the vendors was $-2.842 with a standard deviation of $6.8757. The 90% confidence paired-t interval for price (Priceline - Airline Site) was (-5.5005, -0.1835). Which of the following is the appropriate interpretation?

	1) We are 90% confident that the difference between the average price on Priceline and the average price on the airline's website is between -5.5005 and -0.1835.
	2) The proportion of all flights that had a difference in price between the two vendors is 90%.
	3) We are 90% confident that the average difference in price between the two vendors for all flights is between -5.5005 and -0.1835.
	4) We are certain the average difference in prices between the two vendors for all flights is between -5.5005 and -0.1835.
	5) We are 90% confident that the average difference in the prices of the flights sampled is between -5.5005 and -0.1835.

Answer 1

Solutions:

1. option 3) 2.2342

Explanation : n = 18, s = 5.449, 90% Confidence interval

df = 17, t = 1.7396

Margin of error = t*s/sqrt(n) = 1.7396*5.449/sqrt(18) = 2.2342
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2. option 1) (-264.7942, -131.8518)

Explanation :

Given that x = -198.323, sd = 53.7248 , 99% COnfidence interval , n = 8

df = 7, t = 3.499

99% confidence interval to estimate the true average difference in nationwide sales :
= X +/- t*s/sqrt(n)
= -198.323 +/- 3.4994*53.7248/sqrt(8)
= (-264.7926 , -131.8609)
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3) option 3)

4) option 5)