Problem 6
Question
Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?
Step-by-Step Solution
Verified Answer
Yes, it is reasonable; the test statistic is significant at \( \alpha = 0.02 \), and the p-value supports it.
1Step 1: State the Hypotheses
We need to determine whether union nurses earn more than nonunion nurses. We set up our hypotheses as follows: \( H_0: \mu_1 = \mu_2 \) (Union and nonunion nurses earn the same) \( H_a: \mu_1 > \mu_2 \) (Union nurses earn more than nonunion nurses). This is a one-tailed test.
2Step 2: Identify the Level of Significance
The significance level provided is \( \alpha = 0.02 \). This means we will reject the null hypothesis if our test statistic falls into the critical region defined by \( \alpha = 0.02 \).
3Step 3: Calculate the Test Statistic
We will use a two-sample t-test to compare the means:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Where:\( \bar{x}_1 = 20.75 \), \( \bar{x}_2 = 19.80 \)\( s_1 = 2.25 \), \( s_2 = 1.90 \)\( n_1 = 40 \), \( n_2 = 45 \)Plug in the values:\[ t = \frac{20.75 - 19.80}{\sqrt{\frac{2.25^2}{40} + \frac{1.90^2}{45}}} \]
4Step 4: Calculate Degrees of Freedom
Use the formula for the degrees of freedom for a two-sample t-test:\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]Plug in the values to find \( df \).
5Step 5: Determine Critical Value and Compare the Test Statistic
With the calculated degrees of freedom, find the critical t-value at \( \alpha = 0.02 \) using a t-distribution table. Compare it with the calculated test statistic from Step 3. If the test statistic is greater than the critical value, reject the null hypothesis.
6Step 6: Calculate the p-value
Use a t-distribution table or statistical software to find the p-value associated with the calculated t statistic and the degrees of freedom. This p-value will indicate the probability of observing the computed statistic if the null hypothesis is true.
7Step 7: Conclusion
Based on the comparison in Step 5 and the p-value from Step 6, determine if \( H_0 \) should be rejected or not. If the p-value is less than \( \alpha = 0.02 \), reject \( H_0 \) and conclude that union nurses earn more.
Key Concepts
Two-Sample t-TestSignificance LevelDegrees of Freedomp-Value
Two-Sample t-Test
When comparing the wages of union and non-union nurses, a two-sample t-test is an effective tool. This statistical test helps us determine if there is a significant difference between the means of two independent groups. Here, we compare the mean wage of union nurses (\( \bar{x}_1 = 20.75 \)) with that of non-union nurses (\( \bar{x}_2 = 19.80 \)).
The formula for the two-sample t-test statistic is given by:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]- \( s_1 \) and \( s_2 \) are the standard deviations of the union and non-union groups, respectively.- \( n_1 \) and \( n_2 \) are their respective sample sizes.
This test assumes that the data from both groups have a normal distribution and that the variances are equal. This helps in asserting if the observed difference in means is statistically significant or just due to random chance.
The formula for the two-sample t-test statistic is given by:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]- \( s_1 \) and \( s_2 \) are the standard deviations of the union and non-union groups, respectively.- \( n_1 \) and \( n_2 \) are their respective sample sizes.
This test assumes that the data from both groups have a normal distribution and that the variances are equal. This helps in asserting if the observed difference in means is statistically significant or just due to random chance.
Significance Level
In hypothesis testing, the significance level (\( \alpha \)) is crucial. It represents the probability of rejecting the null hypothesis when it is actually true, also known as the Type I error.
Mary Jo decided on a significance level of 0.02 for her test. This means she is willing to accept a 2% risk of concluding that union wages are greater when they are not.
Mary Jo decided on a significance level of 0.02 for her test. This means she is willing to accept a 2% risk of concluding that union wages are greater when they are not.
- A smaller significance level means a more conservative test, reducing the risk of a false positive result.
- This choice impacts decisions guided by probability tables, where we compare our test statistic to a critical value.
Degrees of Freedom
Degrees of freedom (df) are a concept important in calculating the test statistic for the two-sample t-test. They show how many values in calculating a statistic are free to vary, taking into account the constraints in the problem.
The calculation of degrees of freedom for a two-sample t-test is given by:\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]This formula helps provide an accurate critical value during the hypothesis testing process.
The degrees of freedom adjust based on the differing sample sizes and standard deviations of the two groups, influencing the shape and spread of the t-distribution used to determine critical values and p-values.
The calculation of degrees of freedom for a two-sample t-test is given by:\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]This formula helps provide an accurate critical value during the hypothesis testing process.
The degrees of freedom adjust based on the differing sample sizes and standard deviations of the two groups, influencing the shape and spread of the t-distribution used to determine critical values and p-values.
p-Value
The p-value is a measure that helps in decision-making during hypothesis testing. It tells us the probability of observing test results at least as extreme as those observed, assuming the null hypothesis is true.
Once the test statistic is calculated, the p-value is obtained by comparing it against a standard t-distribution.
Once the test statistic is calculated, the p-value is obtained by comparing it against a standard t-distribution.
- If the p-value is less than the chosen significance level (\( \alpha = 0.02 \) in this case), then we reject the null hypothesis.
- A smaller p-value indicates stronger evidence against the null hypothesis.
- In this exercise, if the p-value is below 0.02, it suggests strong evidence that union nurses indeed earn more than non-union nurses.
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