A random sample of 25 employees of a local utility firm showed that their monthly incomes...

A random sample of 25 employees of a local utility firm showed that their monthly incomes had a sample standard deviation of $112. Provide a 90% confidence interval estimate for the standard deviation of the incomes for all the firm's employees.

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Expert Solution

Solution :

Given that,

s = 112

s2 = 10.5830

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 90% confidence level the 2 value is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

1 - / 2 = 1 - 0.05 = 0.95

2L = 2/2,df = 36.415

2R = 21 - /2,df = 13.848

The 90% confidence interval for is,

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

(24)(10.5830) / 36.415 < < (24)(10.5830) / 13.848

2.64 < < 4.28

(2.64 , 4.28)

orchestra answered 1 year ago

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