It has been extensively demonstrated that adequate night sle

It has been extensively demonstrated that adequate night sleep is crucial for cognitive performance. To investigate whether first year University of Fun students are getting adequate sleep the night before a final exam, 81 randomly selected first year University of Fun students were asked to record their number of hours of sleep the night before their final exams. The mean number of hours of sleep among the 81 students was 6.2 with standard deviation of 1.5 hours.

            a. Identify the population, the sample, and the parameter of interest in this study.

b. Construct a 95% confidence interval for the mean number of hours first year University of Fun students sleep the night before an exam. Explain the meaning of this confidence interval (you are expected to explain what this confidence interval says about the parameter of interest)

c. Construct now a 99% confidence interval for the mean number of hours first year University of Fun students sleep the night before an exam. Explain the meaning of this confidence interval and contrast with the 95% confidence interval in part b.

c. Suppose that the 81 students represent a subset of a larger pool of 100 students randomly selected for the study. The 81 students are the ones who actually provided the investigators with their number of hours of sleep while the remaining 19 students did not respond when contacted by the investigators. If the 19 ‘non-responders’ were more likely to sleep more hours before an exam than a typical first year University of Fun student, how would this affect the estimate and confidence interval for the mean number of hours first year University of Fun students sleep the night before an exam?

Solution

a)
Population = Total number of Fun students from the first year University
Sample = 81 Randomly selected students
Paramter = number of hours of sleep the night before their final exams
b)
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)=6.2
Standard deviation( sd )=1.5
Sample Size(n)=81
Confidence Interval = [ 6.2 ± t a/2 ( 1.5/ Sqrt ( 81) ) ]
= [ 6.2 - 1.9901 * (0.17) , 6.2 + 1.9901 * (0.17) ]
= [ 5.87,6.53 ]

We are 95% confident that hours of sleep of each individual lies in
interval [ 5.87,6.53 ]
c)
WITH 99% of C.I
Confidence Interval = [ 6.2 ± t a/2 ( 1.5/ Sqrt ( 81) ) ]
= [ 6.2 - 2.6387 * (0.17) , 6.2 + 2.6387 * (0.17) ]
= [ 5.76,6.64 ]
We are 99% confident that hours of sleep of each individual lies in
interval [ 5.76,6.64 ]
Wider the confidence larger the interval