Question

A sleep disorder specialist wants to test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the night. To do so, the specialist randomly selects nine patients and records the number of hours of sleep each gets with and without the new drug. The results of the two-night study are listed below. Using this data, find the 99% confidence interval for the true difference in hours of sleep between the patients using and not using the new drug. Assume that the hours of sleep are normally distributed for the population of patients both before and after taking the new drug.

Patient	1	2	3	4	5	6	7	8	9
Hours of sleep without	4.4	6.6	5.7	2	5.1	3.1	6.1	2.8	6.8
Hours of sleep with	7.3	8.3	8.4	3.7	6.6	5.3	6.8	5.2	7.8

Step 1 of 4: Find the point estimate for the population mean of the paired differences. Let x1 be the number of hours of sleep without the new drug and x2 be the number of hours of sleep with the new drug and use the formula d=x2−x1 to calculate the paired differences. Round your answer to two decimal places.

Step 2 of 4: Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.

Step 3 of 4: Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 4 of 4: Construct the 99% confidence interval. Round your answers to two decimal places.

Answer 1

Number	Hours of sleep with	Hours of sleep without	Difference
	7.3	4.4	2.9	1.06777778
	8.3	6.6	1.7	0.02777778
	8.4	5.7	2.7	0.69444444
	3.7	2	1.7	0.02777778
	6.6	5.1	1.5	0.13444444
	5.3	3.1	2.2	0.11111111
	6.8	6.1	0.7	1.36111111
	5.2	2.8	2.4	0.28444444
	7.8	6.8	1	0.75111111
Total	59.4	42.6	16.8	4.46

part a)

Point estimate = 1.87

Part b)

Margin of Error =

Confidence Interval :-



Lower Limit =
Lower Limit = 1.0312
Upper Limit =
Upper Limit = 2.7022
99% Confidence interval is ( 1.03 , 2.70 )