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 51 women in rural Quebec gave a sample variance s2 = 3.2. 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. Ho: σ2 = 5.1; H1: σ2 < 5.1 Ho: σ2 = 5.1; H1: σ2 ≠ 5.1 Ho: σ2 < 5.1; H1: σ2 = 5.1 Ho: 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 normal population distribution. We assume a binomial population distribution. We assume a exponential population distribution. We assume a uniform 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.

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

Answer 1

Solution:

Here, we have to use the Chi square test for the population variance. The null and alternative hypotheses for this test are given as below:

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

Null hypothesis: H₀: The current variance of age in years of a rural Quebec Woman at the time of her first marriage is 5.1.

Alternative hypothesis: H_a: The current variance of age in years of a rural Quebec Woman at the time of her first marriage is less than 5.1.

H₀: σ² = 5.1; H₁: σ² < 5.1

This is a lower tailed or left tailed (one tailed) test.

We are given

Level of significance = 0.05

Sample size = n = 51

Degrees of freedom = df = n – 1 = 51 – 1 = 50

Sample variance = S^2 = 3.2

(b) Find the value of the chi-square statistic for the sample.

We assume a normal population distribution.

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/ σ²

Chi square = (51 – 1)*3.2/5.1

Chi square = 50* 0.627451

Chi square = 31.37255

Test statistic = 31.37

Lower Critical value = 34.7643

(by using Chi square table)

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

P-value = 0.0181

(by using Chi square table)

0.010 < P-value < 0.025

α = 0.05

P-value < α

So, we reject the null hypothesis

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

There is sufficient evidence to conclude that the current variance of age in years of a rural Quebec Woman at the time of her first marriage is less than 5.1.

(f) Find the requested confidence interval for the population variance.

Now, we have to find the 90% confidence interval for the population variance.

Confidence interval for population variance

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

We are given

Confidence level = 90%

Sample size = n = 51

Sample variance = S² = 3.2

χ²_{α/2, n – 1} = 67.5048

χ²_{1 -α/2, n– 1} = 34.7643

(By using chi square table)

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

(51 – 1)*3.2/ 67.5048 < σ² < (51 – 1)*3.2/34.7643

2.3702 < σ² < 4.6024

2.37 < σ² < 4.60

Lower limit = 2.37

Upper limit = 4.60