Let's solve the problem step by step.

Part (a) What is the level of significance?

The level of significance is $\alpha = 0.05$ (5%).

Null and Alternative Hypotheses:

• $H_0: \sigma^2 = 47.1$ (The variance is equal to 47.1.)
• $H_1: \sigma^2 > 47.1$ (The variance is greater than 47.1.)

This is a one-tailed chi-square test for variance.

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Part (b) Chi-Square Test Statistic

The formula for the chi-square test statistic is:

$\chi^2 = \frac{(n-1) \cdot s^2}{\sigma^2}$

Where:

• $n = 17$ (sample size),
• $s^2 = 80.5$ (sample variance),
• $\sigma^2 = 47.1$ (population variance).

Substitute the values:

$\chi^2 = \frac{(17 - 1) \cdot 80.5}{47.1} = \frac{16 \cdot 80.5}{47.1} \approx 27.36$

Degrees of Freedom (df): $\text{df} = n - 1 = 16$

Assumption: We assume the population follows a normal distribution.

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Part (c) P-Value of the Sample Test Statistic

Using the chi-square table or software, we determine the p-value for $\chi^2 = 27.36$ with $df = 16$ in a one-tailed test.

• From tables or software: $0.010 < p < 0.025$.

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Part (d) Decision: Reject or Fail to Reject $H_0$?

• The decision rule is: Reject $H_0$ if $p \leq \alpha$.
• Since $p < 0.05$, we reject the null hypothesis.

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Part (e) Interpretation

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

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Final Answers:

(a) Level of Significance: $\alpha = 0.05$.
$H_0: \sigma^2 = 47.1$, $H_1: \sigma^2 > 47.1$.

(b) Chi-Square Test Statistic: $\chi^2 = 27.36$, $df = 16$.
Assumption: Normal population distribution.

(c) P-Value: $0.010 < p < 0.025$.

(d) Decision: Reject $H_0$.

(e) Interpretation: There is sufficient evidence to conclude the variance is greater in Kansas.

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Would you like a detailed breakdown of any step? Here are 5 questions to expand:

1. How is the chi-square formula derived for variance testing?
2. Can you explain how to use the chi-square distribution table?
3. Why is it necessary to assume a normal population distribution?
4. What happens if $\alpha$ is changed to a more strict value, e.g., 1%?
5. What are other statistical tests for variance aside from chi-square?

Tip: Always double-check your $df$ and p-values with tables/software for precision!