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?
The test is
= $1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees?
The test is
Answers
Answered by
GPT-4o mini
To determine if state employees earn on average less than federal employees, we can use a one-sample t-test for the mean in this scenario since we're comparing the mean salary of state employees to a known population mean (the federal government's mean salary).
### Given Information:
- Population mean salary of federal employees (μ) = $59,593
- Sample mean salary of state employees (\( \bar{x} \)) = $58,800
- Sample standard deviation (s) = $1,500
- Sample size (n) = 30
- Significance level (α) = 0.01
### Hypotheses:
- Null Hypothesis (\( H_0 \)): \( \mu = 59,593 \) (State employees earn the same as federal employees)
- Alternative Hypothesis (\( H_a \)): \( \mu < 59,593 \) (State employees earn less than federal employees)
### Test Statistic:
Since we have the sample mean, sample standard deviation, and sample size, we can calculate the t-statistic using the formula:
\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]
Where:
- \( \bar{x} \) = sample mean
- \( \mu \) = population mean
- \( s \) = sample standard deviation
- \( n \) = sample size
### Plugging in the Values:
\[
t = \frac{58,800 - 59,593}{1,500 / \sqrt{30}}
\]
First, calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{1500}{\sqrt{30}} \approx \frac{1500}{5.477} \approx 273.86
\]
Now calculate the t-statistic:
\[
t = \frac{-793}{273.86} \approx -2.90
\]
### Degrees of Freedom:
The degrees of freedom (df) for this test is \( n - 1 \):
\[
df = 30 - 1 = 29
\]
### Critical Value:
For a one-tailed t-test at the 0.01 significance level with 29 degrees of freedom, we can look up the t-table or use a calculator.
The critical t-value (for one-tailed) at 0.01 significance level and df = 29 is approximately:
\[
t_{critical} \approx -2.462
\]
### Decision Rule:
- If \( t < t_{critical} \), we reject the null hypothesis.
- If \( t \geq t_{critical} \), we fail to reject the null hypothesis.
### Conclusion:
Since our calculated t-value (-2.90) is less than the critical value (-2.462), we reject the null hypothesis.
### Final Conclusion:
At the 0.01 level of significance, there is sufficient evidence to conclude that state employees earn, on average, less than federal government employees.
### Given Information:
- Population mean salary of federal employees (μ) = $59,593
- Sample mean salary of state employees (\( \bar{x} \)) = $58,800
- Sample standard deviation (s) = $1,500
- Sample size (n) = 30
- Significance level (α) = 0.01
### Hypotheses:
- Null Hypothesis (\( H_0 \)): \( \mu = 59,593 \) (State employees earn the same as federal employees)
- Alternative Hypothesis (\( H_a \)): \( \mu < 59,593 \) (State employees earn less than federal employees)
### Test Statistic:
Since we have the sample mean, sample standard deviation, and sample size, we can calculate the t-statistic using the formula:
\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]
Where:
- \( \bar{x} \) = sample mean
- \( \mu \) = population mean
- \( s \) = sample standard deviation
- \( n \) = sample size
### Plugging in the Values:
\[
t = \frac{58,800 - 59,593}{1,500 / \sqrt{30}}
\]
First, calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{1500}{\sqrt{30}} \approx \frac{1500}{5.477} \approx 273.86
\]
Now calculate the t-statistic:
\[
t = \frac{-793}{273.86} \approx -2.90
\]
### Degrees of Freedom:
The degrees of freedom (df) for this test is \( n - 1 \):
\[
df = 30 - 1 = 29
\]
### Critical Value:
For a one-tailed t-test at the 0.01 significance level with 29 degrees of freedom, we can look up the t-table or use a calculator.
The critical t-value (for one-tailed) at 0.01 significance level and df = 29 is approximately:
\[
t_{critical} \approx -2.462
\]
### Decision Rule:
- If \( t < t_{critical} \), we reject the null hypothesis.
- If \( t \geq t_{critical} \), we fail to reject the null hypothesis.
### Conclusion:
Since our calculated t-value (-2.90) is less than the critical value (-2.462), we reject the null hypothesis.
### Final Conclusion:
At the 0.01 level of significance, there is sufficient evidence to conclude that state employees earn, on average, less than federal government employees.