A sample of Alzheimer\'s patients are tested to assess the amount of time in sta
ID: 3183654 • Letter: A
Question
A sample of Alzheimer's patients are tested to assess the amount of time in stage IV sleep. Time spent (in minutes) in Stage IV sleep is recorded for 64 patients. The sample produced a mean of 48 minutes and a standard deviation of 14 minutes over a 24-hour period.
a. Compute and interpret 95 percent confidence interval for the average time of sleep for stage IV Alzheimer's patients.
b. Compute and interpret 99 percent confidence interval.
c. What do you observe on the width of confidence interval when the confidence level increases from 95 to 99 percent?
d. What do you expect on the width of confidence interval as sample size increases?
Explanation / Answer
a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=48
Standard deviation( sd )=14
Sample Size(n)=64
Confidence Interval = [ 48 ± t a/2 ( 14/ Sqrt ( 64) ) ]
= [ 48 - 1.998 * (1.75) , 48 + 1.998 * (1.75) ]
= [ 44.504,51.497 ]
b.
AT 0.01 LOS
Confidence Interval = [ 48 ± t a/2 ( 14/ Sqrt ( 64) ) ]
= [ 48 - 2.656 * (1.75) , 48 + 2.656 * (1.75) ]
= [ 43.352,52.648 ]
c.
width will be wider with 99
d.
when sample size increase width will be narrowed