In: Statistics and Probability
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 =
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