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 95% confidence interval for the true difference in hours of sleep between the patients using and not using the new drug.

Let d=(hours of sleep with the new drug)−(hours of sleep without the new drug)d=(hours of sleep with the new drug)−(hours of sleep without 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 the drug	5.4	5.9	5.6	2.3	4.4	3.5	6.4	2.9	5.6
Hours of sleep with the new drug	7.4	7.6	7	4.3	5.3	4.9	7.3	4.1	7.1

Step 1 of 4 :  

Find the mean of the paired differences, d‾d‾. Round your answer to three decimal places

Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to three decimal places.

Step 4 of 4: Construct the 95% confidence interval. Round your answers to three decimal places.

Answer 1

Number	Hours of sleep with the new drug	Hours of sleep without the drug	Difference ( di )	(di - dº
	7.4	5.4	2	0.30864198
	7.6	5.9	1.7	0.06530864
	7	5.6	1.4	0.00197531
	4.3	2.3	2	0.30864198
	5.3	4.4	0.9	0.29641975
	4.9	3.5	1.4	0.00197531
	7.3	6.4	0.9	0.29641975
	4.1	2.9	1.2	0.05975309
	7.1	5.6	1.5	0.00308642
Total	55	42	13	1.34222222

Step 1

mean of the paired differences

d = Σd;/η = 13/9 = 1.444

Step 2

Critical value

( from t table )

ta/2 = +0.05/2 = 2.306

Step 3

I – u/z(p – 'p)3^ = PS


Sa= 1.3422/9-1 = 0.410

Step 4

Confidence Interval :-

d ta/2,n-1 Sa/vn


ta/2 = +0.05/2 = 2.306


1.444 10.05/2 * 0.410/V9

Lower Limit =
1.444 – 50.05/2 0.410/V9

Lower Limit = 1.129
Upper Limit =
1.444 + 0.05/2 0.410/19

Upper Limit = 1.760
95% Confidence interval is ( 1.129 , 1.760 )