Question

2. In order to test an advertised standard deviation on the brightness of headlights, a study was done on 51 vehicles, and a standard deviation of 30 candlepower was found. Find the 90% confidence interval of the standard deviation.

Answer 1

Solution :

Given that,

s = 30

s2 = 5.4773

n = 51

Degrees of freedom = df = n - 1 = 51 - 1 = 50

At 90% confidence level the 2 value is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

1 - / 2 = 1 - 0.05 = 0.95

2L = 2/2,df = 67.505

2R = 21 - /2,df = 34.764

The 90% confidence interval for is,

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

(50)(5.4773) / 67.505 < < (50)(5.4773) / 34.764

2.01 < < 2.81

(2.01 , 2.81)