Question

The durations (in days) of 14 randomly selected space shuttle flights have a sample
standard deviation of 3.54days. Assume that the durations in days is normally distributed,
construct a 95% confidence interval for the population variance and standard deviation.

Answer 1

Solution :

1) population variance interval

s = 3.54

Point estimate = s2 = 12.5316

n = 14

Degrees of freedom = df = n - 1 = 14 -1 = 13

2L = 2/2,df = 6.5861

2R = 21 - /2,df = 5.0087

The 95% confidence interval for 2 is,

(n - 1)s2 / 2/2 < 2 < (n - 1)s2 / 21 - /2

(13)12.5316/24.7356< 2 < (13)12.5316 / 5.0087

6.5861< 2 < 32.5256  

( 6.5861,32.5256 )

2) standard deviation interval

s = 3.54

s2 = 12.5316

n = 14

Degrees of freedom = df = n - 1 = 14 - 1 = 13

At 95% confidence level the 2 value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

2L = 2/2,df = 24.7356

2R = 21 - /2,df = 5.0087

The 95% confidence interval for is,

(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2

  (13)12.5316/24.7356< < (13)12.5316 /5.0087

2.5663 < < 5.7031

(2.5663 ,5.7031 )