Question

 

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 31 women in rural Quebec gave a sample variance s² = 3.0. 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.

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

lower limit
upper limit

2.) Interpret the results in the context of the application.

We are 90% confident that σ² lies below this interval.

We are 90% confident that σ² lies within this interval.    

We are 90% confident that σ² lies above this interval.

We are 90% confident that σ² lies outside this interval

Answer 1

To Test :-

H0 :- σ² = 5.1

H1 :- σ² < 5.1

Test Statistic :-

Test Criteria :-
Reject null hypothesis if  

= 17.6471 < 18.493 , hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( > 17.6471 )
P value = 0.964
Reject null hypothesis if   P value < level of significance
Since P value = 0.964 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence to support the claim that the current variance is less than 5.1 at 5% level of significance.

Lower Limit =
Upper Limit =
90% Confidence interval is ( 2.06 , 4.87 )
( 2.06 < σ² < 4.87 )

2.) Interpret the results in the context of the application.

We are 90% confident that σ² lies within this interval ( 2.06 < σ² < 4.87 )