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

Question

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 
= $1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees?

What is the critical value? Round your answer to the nearest hundredths.

GPT-4o mini · Answer

To determine whether state employees earn on average less than federal employees, we need to perform a hypothesis test. Given the information provided, we can set up the hypotheses as follows:

Null Hypothesis ($H_0$): $\mu_{state} \geq \mu_{federal}$ (State employees earn at least as much on average as federal employees).
Alternative Hypothesis ($H_a$): $\mu_{state} < \mu_{federal}$ (State employees earn less on average than federal employees).

Where:

$\mu_{state}$ is the mean salary of state employees.
$\mu_{federal} = 59,593$.

Given Data:

Mean salary of state employees ($\bar{x}$) = $58,800
Population mean salary of federal employees ($\mu_{federal}$) = $59,593
Standard deviation ($s$) of state employees = $1500
Sample size ($n$) of state employees = 30

Test Statistic:

We will use a one-sample t-test since we're comparing the sample mean to a known population mean. The test statistic is computed as:

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

Where:

$ \bar{x} $ = sample mean = $58,800
$ \mu $ = population mean = $59,593
$ s $ = sample standard deviation = $1500
$ n $ = sample size = 30

Substituting the values into the formula:

\[ t = \frac{58,800 - 59,593}{1500 / \sqrt{30}} \]

Calculating the standard error ($SE$):

\[ SE = \frac{1500}{\sqrt{30}} \approx \frac{1500}{5.477} \approx 273.86 \]

Now substituting back into the $t$ equation:

\[ t = \frac{-793}{273.86} \approx -2.895 \]

Critical Value:

To find the critical value for the one-tailed t-test at the 0.01 level of significance, we need the degrees of freedom, which is $n - 1 = 30 - 1 = 29$.

Using a t-distribution table or calculator, we look for the critical t-value for $df = 29$ at the 0.01 level of significance.

For a one-tailed test, the critical t-value approximately is:

\[ t_{critical} \approx -2.462 \]

Conclusion:

Thus, the critical value at the 0.01 significance level for a one-tailed test with 29 degrees of freedom is approximately -2.46 (rounded to the nearest hundredths).

You would then compare your test statistic $-2.895$ with the critical value if $-2.895 < -2.462$, you would reject the null hypothesis.