Because of staffing decisions, managers of the Gibson-Marimont Hotel are interes
ID: 3228024 • Letter: B
Question
Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 26 days of operation shows a sample mean of 287 rooms occupied per day and a sample standard deviation of 29 rooms.
A) What is the point estimate of the population variance?
B) Provide a 90% confidence interval estimate of the population variance (to 1 decimal).
( , )
C) Provide a 90% confidence interval estimate of the population standard deviation (to 1 decimal).
( , )
Explanation / Answer
a.
point of estimate = s.d = 29
Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size
Since aplha =0.1
^2 right = (1 - Confidence Level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 25 df are 37.6525 , 14.611
S.D( S )=29
Sample Size(n)=26
b.
Confidence Interval for variance = [ 25 * 841/37.6525 < ^2 < 25 * 841/14.611 ]
= [ 21025/37.6525 < ^2 < 21025/14.6114 ]
[ 558.3959 , 1438.9449 ]
c.
Confidence Interval for standard deviation = [ sqrt(558.3959) < < sqrt(1438.9449) ]
= [ 23.63 < < 37.933]