In: Statistics and Probability
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 the drug | 4.4 | 6.6 | 5.7 | 2.2 | 5.1 | 3.1 | 6.1 | 2.8 | 6.8 |
Hours of sleep with the new drug | 7.3 | 8.3 | 8.4 | 3.7 | 6.6 | 5.3 | 6.8 | 5.2 | 7.8 |
Step 4 of 4 :
Construct the 99% confidence interval. Round your answers to two decimal places.