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 = 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.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 5.1; H₁: σ² < 5.1

H_o: σ² = 5.1; H₁: σ² ≠ 5.1    

H_o: σ² < 5.1; H₁: σ² = 5.1

H_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 normal population distribution.

We assume a binomial population distribution.    

We assume a uniform population distribution.

We assume a exponential 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 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 outside 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 below this interval.

Answer 1

(a) Level of significance

Level of significance is 5% that is 0.05

Null and alternative hypothesis:

Claim: The population variance is less than 5.1 that is which goes under alternative hypothesis.

The null and alternative hypothesis are,

(b) Chi-square test statistics

The formula of chi-square test statistics is,

Where

n = sample size = 51

chi-square test statistics = 23.53

Degrees of freedom = n - 1 = 50

The assumption about the original distribution is normal distribution.

First option is correct.

We assume a normal population distribution.

(c) Estimate of P-value

To find P-value needs test statistics and degrees of freedom.

Using chi square table with the row of degrees of freedom 50, as the test statistics increases the area in right tailed decreases.

Here the alternative hypothesis contains less then sign so the test is left tailed test.

The right tailed area for the first value in the row of degrees of freedom 50 is 0.995, that is left tailed area = 1 - 0.995 = 0.005

That is the P-value for the test statistics 23.53 is less then 0.005

Last option is correct.

P-value < 0.005

(d) Decision rule:

If P-value > alpha then fail to reject the null hypothesis otherwise reject the null hypothesis.

Here P-value is less than 0.005 that is less than alpha (0.05)

So reject the null hypothesis.

Option third is correct.

Since , we reject the null hypothesis.

We assume a normal population distribution.

(e) Conclusion:

Reject the null hypothesis that means in favor of claim which is in alternative hypothesis.

Second option is correct that is,

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

(f) 90% Confidence interval for population variance

The formula of confidence interval for population variance is

Where are the chi-square critical values for the given confidence level.

Degrees of freedom = n -1 = 50

c = confidence level = 0.90

alpha = 1 - c = 0.10

alpha/2 = 0.05

The critical value using chi square table for the area 0.05 with degrees of freedom 50 is 67.505

That is

1-(alpha/2) = 1 - 0.05 = 0.95

The critical value using chi-square table for the area 0.95 with degrees of freedom 50 is 34.764

That is

The confidence interval is,

Lower limit = 1.78

Upper limit = 3.45

Interpretation:

We are 90% confidence that the population variance lies within this confidence interval.

Second option is correct.

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