In: Statistics and Probability
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 17 colleges and universities in Kansas showed that x has a sample variance s2 = 82.6. 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.
Ho: σ2 < 47.1; H1: σ2 = 47.1Ho: σ2 = 47.1; H1: σ2 < 47.1 Ho: σ2 = 47.1; H1: σ2 > 47.1Ho: 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 normal population distribution. We assume a exponential population distribution.We assume a uniform 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 outside this interval.We are 95% confident that σ2 lies within this interval. We are 95% confident that σ2 lies below this interval.We are 95% confident that σ2 lies above this interval.
(a) What is the level of significance α = 0.05
Claim : variance for colleges and universities in Kansas is greater than 47.1
H0 : σ2 = 47.1
H1 : σ2 > 47.1
Given : S2 = 82.6 , n = 17
Test statistic:
Χ2 = =
b) Χ2 = 28.06
Degrees of freedom (df) = n – 1 = 17 – 1 = 16
We assume a normal population distribution.
(c) Find or estimate the P-value of the sample test statistic.
As Χ2 = 28.06 lies between 26.30 and 28.85 in the row for df = 16,
So 0.025 < P-value < 0.050
(e) Interpret your conclusion in the context of the application.
As p value is less than 0.05, we reject H0
At the 5% level of significance, there is sufficient evidence to conclude the variance of annual salaries is greater in Kansas.
f) 95% confidence interval for the population variance :
Lower bound = ; Χ2L is critical value Χ2( 1 - α )/2 , n-1
Upper bound = ; Χ2U is critical value Χ2( 1 +α)/2 , n-1
We are given level of confidence c = 0.95
( 1 - c )/2 = 0.025 and ( 1 + c )/2 = 0.975
Degrees of freedom (df) = n – 1 = 12 – 1 = 11
Therefore, critical value Χ2(0.025,16) = 28.85 and critical value Χ2(0.975,16) = 6.91 ----from chi square table.
Lower bound = = 45.81
Upper bound = = 191.26
We are 95% confident that σ2 lies within this interval.