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 = 2.9. 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 uniform population distribution.We assume a binomial population distribution.    We assume a exponential population distribution.We assume a normal 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 above this interval.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 outside this interval.

Answer 1

random sample size (n) =  31 , sample variance s2 = 2.9.

(a) the level of significance is 0.10

the null and alternate hypotheses.

  Ho: σ2 = 5.1; H1: σ2 < 5.1

(b) using excel>Addin>phstat>one sample test

we have

Chi-Square Test of Variance
Data
Null Hypothesis                        ^2=	5.1
Level of Significance	0.1
Sample Size	31
Sample Standard Deviation	1.70294
Intermediate Calculations
Degrees of Freedom	30
Half Area	0.05
Chi-Square Statistic	17.05885084
Lower-Tail Test
Lower Critical Value	20.59923461
p-Value	0.028137886
Reject the null hypothesis

the value of the chi-square statistic for the sample is 17.06


the degrees of freedom is 30


What assumptions are you making about the original distribution?

We assume a normal population distribution.

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

0.025 < P-value < 0.050

(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 reject the null hypothesis.

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

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.    

using excel>addin>confidence interval

we have

Confidence Interval Estimate for the Population Variance
Data
Sample Size	31
Sample Standard Deviation	1.70294
Confidence Level	90%
Intermediate Calculations
Degrees of Freedom	30
Sum of Squares	87.00014
Single Tail Area	0.05
Lower Chi-Square Value	18.49266
Upper Chi-Square Value	43.77297
Results
Interval Lower Limit for Variance	1.987531
Interval Upper Limit for Variance	4.704577
Interval Lower Limit for Standard Deviation	1.409798
Interval Upper Limit for Standard Deviation	2.169004

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

lower limit	1.99
upper limit	4.71

Interpret the results in the context of the application.

We are 90% confident that σ2 lies within this interval