Question

Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately σ² = 47.1. However, a random sample of 16 colleges and universities in Kansas showed that x has a sample variance s² = 79.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. 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 = 47.1; H₁: σ² < 47.1H_o: σ² < 47.1; H₁: σ² = 47.1    H_o: σ² = 47.1; H₁: σ² ≠ 47.1H_o: σ² = 47.1; H₁: σ² > 47.1

(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 uniform 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 the variance of annual salaries is greater in Kansas.At the 5% level of significance, there is sufficient evidence to conclude the variance of annual salaries is greater in Kansas.    

(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 above this interval.We are 95% confident that σ² lies outside this interval.    We are 95% confident that σ² lies below this interval.We are 95% confident that σ² lies within this interval.

Answer 1

(a) Significance level: The significance level is the probability of rejecting the null hypothesis, H₀ when in reality it is true.

In other words, the probability of committing a Type I error is called the significance level.

In this particular question, the significance level is given as

Null and alternative hypothesis:

(b) Test statistic is given as-

For a random sample of colleges and universities in Kansas , the sample variance is

The Chi-square statistics is calculated as

Degrees of freedom: The degrees of freedom for the chi-square distribution is given as

Assumption:

The assumption for the original distribution is that it has to be Normally distributed.

(c) We have calculated

The p-value is also be written as-

(d) Decision:

Since,

(e) Conclusion:

At the data does provide enough evidence to support the alternative hypothesis, So we reject the null hypothesis, H₀ .

In other words, At 5% level of significance, there is sufficient evidence to conclude that the variance of annual salaries is greater in Kansas.

(f) Calculation for the 95% confidence interval for the population variance

Given:

Lower limit: 43.16

Upper limit : 189.47

Interpretation: We are 95% confident that the population variance is lies within this interval, i.e.,