Forbes magazine recently reported that the mean entry level salary for a Nurse Practitioner in the U.S. is $120,000. A recruiter has reason to suspect that the salary is significantly more in their region of the country.  They randomly select 40 new Nurse Practitioner hires in their region and determine their mean salary to be $123,550 with a standard deviation of $7,000. Using a 10% significance level, is this significant evidence that the salary is more in their region?

To determine if the mean salary for Nurse Practitioners in this region is significantly higher than the national average of $120,000, we can perform a one-sample t-test. Here's how we can approach it:

### Hypotheses:
- Null Hypothesis ($H_0$): $\mu = 120,000$ (The mean salary in this region is equal to the national average)
- Alternative Hypothesis ($H_1$): $\mu > 120,000$ (The mean salary in this region is greater than the national average)

### Given Data:
- Sample mean ($\bar{x}$) = $123,550
- Sample size ($n$) = 40
- Sample standard deviation ($s$) = $7,000
- Population mean ($\mu_0$) = $120,000
- Significance level ($\alpha$) = 0.10

### Test Statistic:
We will use the t-test statistic for this hypothesis test. The formula for the t-test statistic is:

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

Substituting the given values:

\[ t = \frac{123,550 - 120,000}{7,000 / \sqrt{40}} \]

First, calculate the standard error of the mean:

\[ SE = \frac{7,000}{\sqrt{40}} \approx \frac{7,000}{6.3246} \approx 1,106.80 \]

Now calculate the t-statistic:

\[ t \approx \frac{3,550}{1,106.80} \approx 3.21 \]

### Degrees of Freedom:
The degrees of freedom (df) for this test is $ n - 1 = 40 - 1 = 39 $.

### Critical Value and Decision:
At a 10% significance level ($\alpha = 0.10$), we need to find the critical t-value for a one-tailed test with 39 degrees of freedom. Using a t-table or calculator:

\[ t_{critical} \approx 1.304 \]

### Conclusion:
- If the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis.

Since $ t \approx 3.21 $ is greater than $ t_{critical} \approx 1.304 $, we reject the null hypothesis.

### Interpretation:
There is significant evidence at the 10% significance level to conclude that the mean entry-level salary for Nurse Practitioners in this region is significantly higher than the national average of $120,000.

Would you like more details or have any questions? Here are some questions you might consider next:

1. How does changing the significance level affect the conclusion?
2. What is the p-value for this test, and how does it compare to the significance level?
3. How would a larger sample size impact the test results?
4. Can you explain how to interpret the standard deviation in this context?
5. What assumptions must be met for the t-test to be valid?

**Tip:** Always ensure your sample is random and representative of the population to draw valid conclusions from hypothesis tests.

Explore whether Nurse Practitioner salaries are significantly higher in a specific region compared to the national average using a one-sample t-test. Includes step-by-step calculation of t-statistic, critical value determination, and conclusion interpretation.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation