Question

Kelly plans to go on a vacation to Las Vegas during Christmas, but the availability of rooms in the city is a big issue between December and February due to a higher influx of tourists during that peak season.

She checks a particular motel for the number of rooms rented. She chooses 20 randomly selected dates in December and early January to examine the occupancy records for those dates. From her collected data, she found a standard deviation of 3.86 rooms rented. If the number of rooms rented is normally distributed, find the 95% confidence interval for the population standard deviation of the number of rooms rented. Write a conclusion for your finding. spss

Write: your decision:

Answer 1

CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.05
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.025 = 0.975
the two critical values ᴪ^2 left, ᴪ^2 right at 19 df are 32.852 , 8.907
s.d( s )=3.86
sample size(n)=20
confidence interval for σ^2= [ 19 * 14.9/32.852 < σ^2 < 19 * 14.9/8.907 ]
= [ 283.092/32.852 < σ^2 < 283.092/8.907 ]
[ 8.617 < σ^2 < 31.783 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (8.617) < σ < sqrt(31.783), ]
= [ 2.936 < σ < 5.638 ]
95% confidence interval for the population standard deviation of the number of rooms rented.