Question

The acceleration times (in seconds) from 0 to 60 miles per hour for 26 randomly selected sedans had a sample mean of 7.99 seconds and a sample standard deviation of 1.190. Construct a 98% confidence interval for the population standard deviation. Assume the sample is from a normal population. Enter your answer in interval notation. If your left endpoints 5 and your right endpoint is 7, your answer would be (5, 7). Round your endpoints to three decimal places.

Answer 1

The 98% confidence interval for the population standard deviation

In – 1)52 (n − 1) Tv xy2 V -a2

where

s= 1.190

n= 26

For 98% confidence ,

g= 0.02

0.99

a/2 = 0.01,1 -a/2 =

degrees of freedom = n-1 = 26-1=25

For 98% confidence with df =25

2 = x30 = 44.3141, x 2 = x3 g = 11.5240

Thus ,98% confidence interval for the population standard deviation

(26 - 11.1902 TV 44.3141 (26-111.1902 V 11.5240

=(0.894 , 1.753 )