The mean salary of federal government employees on the General Schedule is $59,593. The average salary of 30 state employees who do similar work is $58,800 with

To determine if we can conclude that state employees earn on average less than federal employees, we will conduct a one-sample t-test.

### Given Data:
- Mean salary of federal government employees (\( \mu_0 \)): $59,593
- Mean salary of state employees (\( \bar{x} \)): $58,800
- Sample standard deviation of state employees (\( s \)): $1,500
- Sample size of state employees (\( n \)): 30
- Significance level (\( \alpha \)): 0.01

### Hypotheses:
- Null Hypothesis (\( H_0 \)): \( \mu \geq \mu_0 \) (state employees earn on average at least as much as federal employees)
- Alternative Hypothesis (\( H_a \)): \( \mu < \mu_0 \) (state employees earn on average less than federal employees)

### Test Statistic Calculation:
We will use the t-test formula for one sample:

\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]

Where:
- \(\bar{x} = 58,800\)
- \(\mu_0 = 59,593\)
- \(s = 1,500\)
- \(n = 30\)

Substituting the values:

\[
t = \frac{58,800 - 59,593}{1,500 / \sqrt{30}} = \frac{-793}{1,500 / 5.477} = \frac{-793}{273.86} \approx -2.90
\]

### Critical Value:
Now we need to find the critical value for a one-tailed t-test at \( \alpha = 0.01 \) with \( n - 1 = 30 - 1 = 29 \) degrees of freedom.

Using a t-distribution table or calculator, the critical value \( t_{critical} \) at \( \alpha = 0.01 \) for 29 degrees of freedom is approximately -2.457.

### Decision Rule:
- If \( t < t_{critical} \), we reject the null hypothesis.
- If \( t \geq t_{critical} \), we fail to reject the null hypothesis.

### Conclusion:
Calculate \( -2.90 < -2.457 \).

Since the calculated t-value (-2.90) is less than the critical t-value (-2.457), we reject the null hypothesis.

### Final Answer:
**Reject the null hypothesis.**

This suggests that there is enough evidence to conclude that state employees earn on average less than federal employees at the 0.01 level of significance.