Question

A lawn mower manufacturer is trying to determine the standard deviation of the mean life of one of its lawn mower models. To do this, it randomly selects 12 lawn mowers that sold several years ago and finds that the sample standard deviation is 3.25 years (assume it was normally distributed). Construct a 99% confidence interval for the population standard deviation, σ

please show work

Answer 1

Solution :

Given that,

s = 3.25

s2 = 1.8028

n = 12

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 99% confidence level the 2 value is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

1 - / 2 = 1 - 0.005 = 0.995

2L = 2/2,df = 26.757

2R = 21 - /2,df = 2.603

The 99% confidence interval for is,

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

(11)(1.8028) / 26.757 < < (11)(1.8028) / 2.603

0.86 < < 2.76

A 99% confidence interval for the population standard deviation, σ (0.86 , 2.76)