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 = 2.5. 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) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² < 5.1; H₁: σ² = 5.1H_o: σ² = 5.1; H₁: σ² ≠ 5.1    H_o: σ² = 5.1; H₁: σ² < 5.1H_o: σ² = 5.1; H₁: σ² > 5.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a exponential population distribution.We assume a uniform population distribution.    We assume a normal population distribution.We assume a binomial population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    

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

lower limit
upper limit

Interpret the results in the context of the application.

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

Answer 1

Part a)

α = 0.05

H_o: σ² = 5.1; H₁: σ² < 5.1

Part b)

Test Statistic :-

χ² = ( ( 31-1 ) * 2.5 ) / 5.1
χ² = 14.72

Degree of freedom = n - 1 = 31 - 1 = 30

We assume a normal population distribution.

Part c)

P value = P ( χ² > 14.7059 ) = 0.0087

0.005 < P-value < 0.010

Decision based on P value
P value = P ( χ^2 > 14.7059 )
P value = 0.0087
Reject null hypothesis if P value < α = 0.05
Since P value = 0.0087 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

Part d)

Since the P-value ≤ α, we reject the null hypothesis.

Part e)

At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    

Part f)

((n-1)S² / χ² (0.05/2)) < σ² < ((n-1)S² / χ² (1 - 0.05/2) )
(( 31-1 ) 2.5 / χ² (0.05/2) ) < σ² < ((31-1)2.5 / χ² (1 - 0.05/2) )χ^2 (0.1/2) = 43.773
χ² (0.1/2) = 43.773
χ² (1 - 0.1/2) ) = 18.4927
Lower Limit = (( 31-1 ) 2.5² / χ² (0.1/2) ) = 1.7134
Upper Limit = (( 31-1 ) 2.5² / χ² (0.1/2) ) = 4.0557
90% Confidence interval is ( 1.7134 , 4.0557 )
( 1.71 < σ² < 4.06 )

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