A sleep disorder specialist wants to test the effectiveness of a new drug that i

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 98 % 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. Assume that the hours of sleep are normally distributed for the population of patients both before and after taking the new drug. Patient 1 2 345 678 9 Hours of sleep without the drug 5 7.13.3 5.74.1 4.14.4 5.8 3.4 Hours of sleep with the new drug 7.785.6 8.3 5.8 6. 7.48 4.3 Copy Data Step 1 of 4: Find the mean of the paired differences, d. Round your answer to two decimal places

Explanation / Answer

Since this is a paired sample,

Degrees of freedom = n - 1 = 9 - 1 = 8

Hence,

For 8 degrees of freedom and 98% confidence interval,

Critical value = 2.896

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