Question

A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ² = 42.3. In a different section of Chaco Canyon, a random sample of 28 transects gave a sample variance s² = 50.9 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 42.3; H₁: σ² ≠ 42.3

H_o: σ² > 42.3; H₁: σ² = 42.3    

H_o: σ² = 42.3; H₁: σ² < 42.3

H_o: σ² = 42.3; H₁: σ² > 42.3

(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 exponential population distribution.

We assume a binomial 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 > 0.100

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 conclude that the variance is greater in the new section.

At the 5% level of significance, there is sufficient evidence to conclude conclude that the variance is greater in the new section.    

(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 95% confident that σ² lies below this interval.

We are 95% confident that σ² lies above this interval.    

We are 95% confident that σ² lies within this interval.

We are 95% confident that σ² lies outside this interval.

Answer 1

(a) What is the level of significance?

Level of significance = α = 0.05

State the null and alternate hypotheses.

H_o: σ² = 42.3; H₁: σ² > 42.3

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

Chi square = (n – 1)*S^2/σ^2

Chi square = (28 - 1)*50.9/42.3

Chi square = 32.48936

Chi square = 32.49

What are the degrees of freedom?

df = n – 1 = 28 – 1 = 27

df = 27

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.

We have

df = 27

Chi square = 32.49

P-value = 0.2144

(by using Chi square table)

P-value > 0.100

(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.

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

At the 5% level of significance, there is insufficient evidence to conclude conclude that the variance is greater in the new section.

(f) Find the requested 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 = 95%

Sample size = n = 28

Sample variance = S² = 50.9

χ²_{α/2, n – 1} = 43.1945

χ²_{1 -α/2, n– 1} = 14.5734

(By using chi square table)

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

(28 – 1)*50.9/43.1945 < σ² < (28 – 1)*50.9/14.5734

31.8165 < σ² < 94.3021

Lower limit = 31.82

Upper limit = 94.30

Interpret the results in the context of the application.

We are 95% confident that σ² lies within this interval.