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 = 86.8. 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 = 47.1
Ho: σ2 = 47.1; H1: σ2 ≠ 47.1
Ho: σ2 = 47.1; H1: σ2 > 47.1
Ho: 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 uniform population distribution.
We assume a exponential population distribution.
We assume a binomial population distribution.
We assume a normal 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 outside this interval.
We are 95% confident that σ2 lies below this interval.
We are 95% confident that σ2 lies above this interval.
We are 95% confident that σ2 lies within this interval.
a) level of significance =0.05
Ho: σ2 = 47.1; H1: σ2 > 47.1
b)
value of the chi-square statistic for the sample =(n-1)*(s2/σ2 )=(18-1)*(86.8/47.1)=31.33
degrees of freedom =n-1 =18-1 =17
We assume a normal population distribution.
c) 0.010 < P-value < 0.025
d) Since the P-value ≤ α, we reject the null hypothesis.
e) At the 5% level of significance, there is sufficient evidence to conclude the variance of annual salaries is greater in Kansas.
f) lower limit =48.88
upper limit =195.08
We are 95% confident that σ2 lies within this interval
Let x represent the average annual salary of college and university professors (in thousands of dollars)...
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