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 18 colleges and universities in Kansas showed that x has a sample variance s² = 80.5. 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 < 47.1

H_o: σ² = 47.1; H₁: σ² ≠ 47.1    

H_o: σ² = 47.1; H₁: σ² > 47.1

H_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 normal population distribution.We assume a uniform 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 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 within this interval.   

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

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

Answer 1

H_o: σ² = 47.1; H₁: σ² > 47.1

n = 18

Test Statistic :-

Test Criteria :-
Reject null hypothesis if

= 29.0552 > 27.587 , hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( > 29.0552 )
P value = 0.034
Reject null hypothesis if P value <
Since P value = 0.034 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

the degrees of freedom = 19 - 1 = 17

We assume a normal population distribution.

0.025 < P-value < 0.050

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

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

Lower Limit =
Upper Limit =
95% Confidence interval is ( 45.33 , 180.92 )
( 45.33 < < 180.92 )

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