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 20 transects gave a sample variance s² = 48.1 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 uniform population distribution.

We assume a normal population distribution.

    We assume a binomial 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 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 outside 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 below this interval.

Answer 1

Answer a)

Level of significance = α = 0.05

This corresponds to a right-tailed test, for which a Chi-Square test for one population variance will be used. So, the following null and alternative hypotheses need to be tested:

Ho: σ² = 42.3

Ha: σ² > 42.3

Correct Option: H_o: σ² = 42.3; H₁: σ² > 42.3   

Answer b)

The Chi-Squared statistic is computed as follows:

χ² = s²*(n−1)/​σ² = 48.1*(20−1)/42.3​ = 21.61

Degrees of Freedom is computed as follows:

df = n-1 = 20 - 1 = 19

A test of a single variance assumes that the underlying distribution is normal.

Correct Option: We assume a normal population distribution.

Answer c)

P-value is obtained using online calculator using χ² = 21.61 and df = 19 (Screenshot attached)

P-value is 0.3041 so it is greater than 0.100

Correct option: P-value > 0.100

Answer d) and e)

In this case, p-value is greater than α = 0.05, so we reject null hypothesis. Therefore, there is not enough evidence to claim that the population variance σ² is greater than 42.3, at the 0.05 significance level.

Correct Option: Since the P-value > α, we reject the null hypothesis.

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

We can answer 5 parts only.