In: Statistics and Probability
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
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)