A study of arrival times for a number of international flights. The times were randomly recorded...

A study of arrival times for a number of international flights. The times were randomly recorded from a normally distributed population with the following results:

n = 19 and s = 12.1. Find the 95% confidence interval for σ.

Solutions

Expert Solution

Solution :

Given that,

s = 12.1

s2 = 3.4785

n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 95% confidence level the 2 value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

2L = 2/2,df = 31.526

2R = 21 - /2,df = 8.231

The 95% confidence interval for is,

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

(18) (3.4785) / 31.526 < < (18)(3.4785) / 8.231

1.41 < < 2.76

(1.41 , 2.76 )

orchestra answered 11 months ago

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