To determine whether the average salary of state employees is less than that of federal employees, we can perform a hypothesis test.
Let's set up the hypotheses:
- Null Hypothesis (H0): μ_state >= μ_federal (State employees earn on average at least as much as federal employees)
- Alternative Hypothesis (H1): μ_state < μ_federal (State employees earn on average less than federal employees)
Where:
- μ_state = mean salary of state employees
- μ_federal = mean salary of federal employees ($59,593)
Given the information:
- Sample mean of state employees (x̄) = $58,800
- Federal employees mean (μ_federal) = $59,593
- Sample size (n) = 30
- Standard deviation (σ) of state employees = $1,500
- Significance level (α) = 0.01
Next, we need to calculate the test statistic using the formula for the z-test for a single sample mean:
\[ z = \frac{x̄ - μ_federal}{σ / \sqrt{n}} \]
Substituting the known values:
\[ z = \frac{58,800 - 59,593}{1,500 / \sqrt{30}} \]
Calculating the standard deviation of the sample mean (σ / √n):
\[ σ / \sqrt{n} = \frac{1,500}{\sqrt{30}} \approx 274.07 \]
Now, calculate the z-value:
\[ z = \frac{58,800 - 59,593}{274.07} \approx \frac{-793}{274.07} \approx -2.89 \]
Next, we need to find the critical z-value for a one-tailed test at the 0.01 significance level. This can be found using a z-table or calculator:
The critical z-value for α = 0.01 (one-tailed) is approximately -2.33.
Now, we can compare our calculated z-value with the critical z-value:
- Calculated z-value: -2.89
- Critical z-value: -2.33
Since -2.89 is less than -2.33, we reject the null hypothesis.
Based on our results:
- Rejecting the null hypothesis suggests there is sufficient evidence to support the alternative hypothesis that state employees earn on average less than federal employees.
Thus, the correct conclusion is:
c) The sample data support the claim that state employees earn on average less than federal employees.
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