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 = 2.8. 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 binomial population distribution. We assume a uniform 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 > 0.100 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

Here we have given that

(A)

$\alpha$ = level of significance=0.05

n= number of women in rural Quebec gave=41

$S^2$ = sample variance =2.8

Claim : To check weather The current population variance of age at first marriage is less than 5.1 at 5% level of significance.

The hypothesis are

$Ho : \sigma^2 = 5.1$

v/s

$H1: \sigma^2 < 5.1$

Here we assume that the population is normally distributed.

(B)

Now we want to find the chi-square statistic for the sample

test statistic is

$\chi ^2 = \frac{(n-1)s^2}{\sigma^2}$

$\chi ^2 = \frac{(41-1) 2.8}{5.1}$

=21.96

we get

The test statistic is 21.96

and here we assume a population is normal distribution.

Now we find the P-value

$\alpha$ = level of significance=0.05

Degrees of freedom = n-1 = 41-1 =40

P-value =0.9909 using excel=CHIDIST( $\chi ^2 STAT$ = 21.96,D.F=40)

Decision:

Here

P-value > 0.05 $\alpha$

that is here we do not reject the Null hypothesis Ho

Conclusion:

There is Insufficient evidence to conclude that The current population variance of age at first marriage is less than 5.1 at 5% level of significance

(B)

Now we want to find the 90% confidence interval for population variance

$\frac{(n-1)S^2}{\chi ^2 \frac{\alpha}{2}} < \sigma ^2 <\frac{(n-1)S^2}{\chi ^2 1- \frac{\alpha}{2}}$

1st we calculate the critical values

c= confidence level = 0.90

$\alpha$ = level of significance = 1-c = 1-0.90 = 0.10

Degress of freedom = n-1 = 41-1=40

$\chi ^2(\frac{\alpha}{2},D.F) =\chi ^2(\frac{0.10}{2},40)= \chi ^2(0.05,40)$ =55.76 using excel=CHIINV(prob=0.05,d.f=40)

$\chi ^2(1-\frac{\alpha}{D.F}=\chi ^2(1-\frac{0.10}{2},40)=\chi ^2(0.95,40)$ =26.51

Now,

$\frac{(n-1)S^2}{\chi ^2 \frac{\alpha}{2}} < \sigma ^2 <\frac{(n-1)S^2}{\chi ^2 1- \frac{\alpha}{2}}$

$\frac{(41-1) 2.8}{55.76} < \sigma ^2 <\frac{(41-1)2.8}{26.51}$

$2.009 < \sigma ^2 < 4.225$

lower limit= 2.009

Upper limit =4.225

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