Let xxσ2s2(a) What is the level of significance?State the null and alternate hypotheses.Hoσ2H12Hoσ2H12Hoσ2H12H

= age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of  was approximately  = 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  = 2.4. 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.

:  = 5.1; : σ > 5.1

:  = 5.1; : σ ≠ 5.1

:  = 5.1; : σ < 5.1

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 binomial population distribution.

We assume a normal population distribution.

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

P-value > 0.100

0.050 < P-value < 0.100

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-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.