Question

In: Statistics and Probability

Let x = age in years of a rural Quebec woman at the time of her...

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 2.8. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.

(a) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit =

upper limit =

Solutions

Expert Solution

Solution:

Given: In the year 1941, the population variance of x was approximately σ2 = 5.1.

Sample size = n = 41

Sample Variance =  s2 = 2.8

Level of significance = 5% = 0.05

Claim: the current variance is less than 5.1.

Part a) Construct a 90% confidence interval for the population variance.

Formula:

where

and are chi-square critical values.

c = 90%

then

and

df = n - 1 = 41 - 1 = 40

Look in Chi-square table for df = 40 and right tail areas 0.95 and 0.05

and find corresponding chi-square critical values.

thus

= 55.758 and = 26.509

Thus

Thus

lower limit = 2.01

upper limit = 4.22

Part b) Use a 5% level of significance to test the claim that the current variance is less than 5.1.

90% confidence interval is between 2.01 and 4.22 which is less than 5.1.

thus at 5% level of significance, we have sufficient evidence to support the claim that the current variance is less than 5.1


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