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 41 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 > 5.1H_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 binomial population distribution.We assume a exponential population distribution.    We assume a normal population distribution.We assume a uniform 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

x = age in years of a rural Quebec woman at the time of her first marriage

(a)

What is the level of significance ( l.o.s) ?

Given - Use a 5% level of significance to test the claim that the current variance is less than 5.1.

Hence level of significance is 0.05

State the null and alternate hypotheses.

Since we need to test the claim that the current variance is less than 5.1.

Hence H_o: σ² = 5.1;     vs     H₁: σ² < 5.1

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

The test statistic of the chi-square test for variance is calculated as follows:

    Test Statistics .              

where

n = the sample size = 41  

s2 = the sample variance = 2.4

σ² = the population variance = 5.1

Thus   

       

             = 18.82353

Thus value of the chi-square statistic for the sample is = 18.82.

Rejection Criteria -

the rejection region or we reject null hypothesis if calculates chi-square value < Table value

What are the degrees of freedom ?

The distribution of the test statistic is the Chi-Square distribution, with n-1 degrees of freedom

Degree of freedom is n – 1 = 40           { n=41 }

What assumptions are you making about the original distribution?

Since we are using Chi-square test for variance , hence We assume a normal population distribution .

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

Now value of test statistics is 18.82 with 40 degree of freedom .

We need to find P-value i.e P( < 18.82 )

Chi-square statistical table is given below

Since We cannot find exact values from Table , we will use R-software to find P-value .

Command in R " qchisq(p, df,) " where p if given level of significance .

Given opions

i) P-value > 0.100

ii) 0.050 < P-value < 0.100    

iii) 0.025 < P-value < 0.050

iv) 0.010 < P-value < 0.025

v) 0.005 < P-value < 0.010

vi) P-value < 0.005

For option i)   -

> qchisq(0.1,40)             
[1] 29.05052

For option ii)   -

> qchisq(0.05,40) ; qchisq(0.10,40)
[1] 26.5093
[1] 29.05052

For option iii)   -

> qchisq(0.025,40) ; qchisq(0.05,40)
[1] 24.43304
[1] 26.5093

We directly move for option vi) P-value < 0.005 as we need to find P-value i.e P( < 18.82 )

> qchisq(0.005,40)
[1] 20.70654

Thus we can say correct option is P-value < 0.005

Since our alternative hypothesis is H₁: σ² < 5.1 , we need to find value corresponds to a left-tailed test

Correct option is P-value < 0.005

{ Accurate answer is

> pchisq(18.81,40)           # P( < 18.82 )
[1] 0.001752046

}

(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.

Rejection Criteria -

the rejection region or we reject null hypothesis if calculates chi-square value < Table value

calculates chi-square value = 18.82

To find table value of =

We use R-software to calculate table value at 5% of l.o.s ( Note table value can be obtain from statistical table )

Note - we need to find value corresponds to a left-tailed test , hence our level of significance will be 1 - 0.05 = 0.95

R - code and output

> qchisq(1-0.95,40)    # table value
[1] 26.5093

Thus Table value is = 26.51                 { at =5% of l.o.s }

Decision about the null hypothesis

Since it is observed that = 18.82

Table value is = 26.51

Such that = 18.82 < 26.51 , it is then concluded that the null hypothesis is rejected .

Hence null hypothesis is rejected at 5% of level of significance .

Therefore, there is enough evidence to claim that the population variance σ² is less than 5.1, at the 0.05 significance level.

Also P-value ( P-value< 0.005) which is less than = 0.05 , hence we reject null hypothesis .

Correct value of P-value is 0.001752046 ( obtained above )

Correct option is

  Since the P-value ≤ α = 0.05, we reject the null hypothesis

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

We reject null hypothesis .

Correct option

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.

{ justification given above }

(f) Find the requested confidence interval for the population variance σ². (Round your answers to two decimal places.)

A (1-α )100% confidence interval for σ² is given by .

< σ² <

n=41

s²=2.4

We are given = 10% of level of significance

     = 55.75848

= 26.5093

{

> qchisq(1-0.05,40)
[1] 55.75848

> qchisq(0.05,40)
[1] 26.5093

}

and (n-1)s² = 40*2.4 = 96

Thus

A (1-α )100% confidence interval for σ² is given by .

96 / 55.75848 < σ² < 96 / 26.5093

1.721711 < σ²   < 3.62137

Lower Limit = 1.72

Upper Limit = 3.62

We are 90% confident that σ² lies within this interval (1.72 , 3.62 )