Respuesta :
Answer:
a) 900
b) 567 [tex]\leq[/tex] [tex]\sigma^2[/tex] [tex]\leq[/tex] 1690
c) 23.8 [tex]\leq[/tex] [tex]\sigma[/tex] [tex]\leq[/tex] 41.1
Step-by-step explanation:
Complete 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 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms.
a. What is the point estimate of the population variance (to nearest whole number)?
b. Provide a 90% confidence interval estimate of the population variance (to nearest whole number).
c. Provide a 90% confidence interval estimate of the population standard deviation (to 1 decimal).
Solution
a) Point estimate of the population variance = Square of S = S^2
30^2 =900
b) Given
n = 20
s^2 = 900
(n-1)s^2/x 0.05 [tex]\leq[/tex] [tex]\sigma^2[/tex] [tex]\leq[/tex] (n-1)s^2/x 0.95
Variation 567 [tex]\leq[/tex] [tex]\sigma^2[/tex] [tex]\leq[/tex] 1690
c) Standard Deviation
23.8 [tex]\leq[/tex] [tex]\sigma[/tex] [tex]\leq[/tex] 41.1
