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 15 colleges and universities in Kansas showed that x has a sample variance s² = 85.4. 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 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 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 within this interval. 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.

Answer 1

(a) What is the level of significance?

evel of significance[\=alpha=0.05

Ho: σ2 = 47.1; H1: σ2 > 47.1

Solutionb:

Chi sq=(n-1)*s^2/sigma^2

chi sq=(15-1)*85.4/47.1

=14*85.4/47.1

=1195.6/47.1

=25.38429

=25.38

ANSWERS:TEST STATISTIC=25.38

What are the degrees of freedom?

Degrees of freedom=n-1=15-1=14

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.

usinge xcel and

=CHISQ.DIST.RT(25.38;14)

=0.030996

p less than alpha

Reject null hypothesis

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

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

. 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.)

confidence interval for variance is

45.78<sigma^2<212.41

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