Answer to Question #307496 in Statistics and Probability for secret

Question

A sample of Alzheimer's patients are tested to assess the amount of time in stage IV sleep. It has been hypothesized that individuals suffering from Alzheimer's Disease may spend less time per night in the deeper stages of sleep. Number of minutes spent in Stage IV sleep is recorded for sixty-one patients. The sample produced a mean of 34 minutes with a standard deviation of 27 minutes of stage IV sleep over a 24 hour period of time. What is the length of the 95 percent confidence interval for this data.

Answer 1

Margin of error: $ME=t_{0.025,n-1}\frac{s}{\sqrt{n}}=2.0003\frac{27}{\sqrt{61}}=6.9150.$

The length of the 95 percent confidence interval: $L=2ME=2*6.9150=13.8300.$