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Chapter 10: Problem 14

Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? Salary.com provided salary data showing staff nurses in Tampa earn less than staff nurses in Dallas (The Tampa Tribune, January 15,2007 ). Suppose that in a follow-up study of 40 staff nurses in Tampa and 50 staff nurses in Dallas you obtain the following results. $$\begin{array}{cc} \text { Tampa } & \text { Dallas } \\ n_{1}=40 & n_{2}=50 \\ \bar{x}_{1}=\$ 56,100 & \bar{x}_{2}=\$ 59,400 \\ s_{1}=\$ 6000 & s_{2}=\$ 7000 \end{array}$$ a. Formulate hypotheses so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in Tampa are significantly lower than for those in Dallas. Use \(\alpha=.05\) b. What is the value of the test statistic? c. What is the \(p\) -value? d. What is your conclusion?

Short Answer

Expert verified
Tampa salaries are significantly lower than Dallas if we reject the null hypothesis at \( \alpha = 0.05 \). The test statistic \( t = -2.09 \) and p-value \( \leq 0.05 \) support rejecting the null hypothesis.

Step by step solution

01

Formulate Hypotheses

To test if the nursing salaries in Tampa are significantly lower than those in Dallas, formulate the null and alternative hypotheses as follows: - Null Hypothesis ( H_0 ): ar{x}_{1} eq ar{x}_{2} (The mean salary in Tampa is equal to or greater than in Dallas) - Alternative Hypothesis ( H_a ): ar{x}_{1} < ar{x}_{2} (The mean salary in Tampa is less than in Dallas)
02

Determine the Test Statistic Formula

Since we are comparing the means of two independent samples with known standard deviations, we'll use a two-sample t-test. The test statistic is calculated using the formula: \[ t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }} \]Substitute the given values to find the test statistic.
03

Calculate the Test Statistic

Substitute the given values into the test statistic formula:\[ t = \frac{56100 - 59400}{\sqrt{ \frac{6000^2}{40} + \frac{7000^2}{50} }} \]Calculate the result to find the value of the test statistic.
04

Compute the Test Statistic Value

Upon calculating, we find:\[ t = \frac{-3300}{\sqrt{ \frac{36000000}{40} + \frac{49000000}{50} }} \] = \frac{-3300}{1579.68} = -2.09. Thus, the test statistic value is \( t = -2.09 \).
05

Determine the p-value

Using a t-distribution table or a calculator, find the p-value corresponding to the test statistic \( t = -2.09 \) for a left-tailed test with degrees of freedom calculated using:\[ df = min(n_1 - 1, n_2 - 1) = min(39, 49) = 39. \]Look up the p-value for \( t = -2.09 \) with \( 39 \) degrees of freedom.
06

Find the p-value

The p-value for \( t = -2.09 \) with 39 degrees of freedom can be found using statistical software or a t-table, which shows approximately \( p \leq 0.05 \). This suggests the p-value is small enough to consider rejecting the null hypothesis.
07

Conclusion

Since the p-value is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. This concludes that the salaries for staff nurses in Tampa are significantly lower than those for their counterparts in Dallas.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-sample t-test
The two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. This test is ideal for situations where we have two sets of data that are unrelated. It helps us compare the means from both groups to see if they differ by a significant margin, considering any natural variance within the samples.
  • We apply a two-sample t-test when the data comes from a normal distribution.
  • This test is particularly useful when we have small sample sizes.
  • It considers the standard deviation and sample size to determine if differences in means are due to true differences or just random variation.
In the context of comparing nursing salaries, we use a two-sample t-test to see if the average salaries in Tampa differ from those in Dallas. The computations involve considering each group's sample size, mean, and standard deviation to calculate a test statistic that tells us the likelihood of observing such a difference by chance.
Nursing Salaries
Nursing salaries can vary significantly based on geographic location, employer, experience, and education level. Among these factors, location often plays a critical role, as cost of living and demand for healthcare services can affect wages. In the exercise, we're analyzing salaries from two different cities鈥擳ampa, Florida, and Dallas, Texas.
  • Nurses in Tampa earn an average salary of \(\\(56,100\) with a standard deviation of \(\\)6000\).
  • Nurses in Dallas earn an average salary of \(\\(59,400\) with a standard deviation of \(\\)7000\).
By comparing these figures, we aim to understand if geographical differences significantly impact the average wages of nursing staff between these two locations. Conducting a statistical test helps provide data-driven insights into these differences beyond simple salary comparisons.
p-value calculation
The p-value is a fundamental concept in hypothesis testing, providing the probability of obtaining the observed results, or more extreme results, assuming that the null hypothesis is true. It helps us determine the significance of our test results. A smaller p-value indicates stronger evidence against the null hypothesis.
  • A p-value less than the significance level \( \alpha = 0.05 \) suggests rejecting the null hypothesis.
  • For the exercise, we calculate the test statistic value as \( t = -2.09 \).
  • Using 39 degrees of freedom, we look up \( t = -2.09 \) in the t-distribution table to find the corresponding p-value.
In this case, the p-value is found to be approximately \( p \leq 0.05 \). This value indicates there is statistically significant evidence to reject the null hypothesis. Therefore, the difference in average nursing salaries isn't likely due to random chance alone.
Null and Alternative Hypotheses
In statistical analysis, formulating null and alternative hypotheses is the first step in hypothesis testing. These hypotheses provide the framework for testing our assumptions about data using statistical tests.
  • The null hypothesis \(H_0\) asserts that any observed difference is due to random sampling, stating no effect or no difference.
  • The alternative hypothesis \(H_a\) contradicts the null hypothesis, suggesting a real effect or difference.
For our exercise, the null hypothesis is that the mean salaries in Tampa are equal to or greater than those in Dallas \((\bar{x}_1 \geq \bar{x}_2)\). The alternative hypothesis claims the mean salaries in Tampa are less than those in Dallas \((\bar{x}_1 < \bar{x}_2)\). By testing these hypotheses, we determine whether there's significant statistical evidence to support that nursing salaries differ between Tampa and Dallas. Through the analysis, if our p-value is below \(0.05\), we reject the null hypothesis, supporting the claim that Tampa's average salary is indeed lower.

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Most popular questions from this chapter

Injuries to Major League Baseball players have been increasing in recent years. For the period 1992 to 2001 , league expansion caused Major League Baseball rosters to increase \(15 \%\) However, the number of players being put on the disabled list due to injury increased \(32 \%\) over the same period (USA Today, July 8,2002 ). A research question addressed whether Major League Baseball players being put on the disabled list are on the list longer in 2001 than players put on the disabled list a decade earlier.a. Using the population mean number of days a player is on the disabled list, formulate null and alternative hypotheses that can be used to test the research question. b. Assume that the following data apply: \(\begin{array}{lll} & \mathbf{2 0 0 1} \text { Season } & \mathbf{1 9 9 2} \text { Season } \\ \text { Sample size } & n_{1}=45 & n_{2}=38 \\ \text { Sample mean } & \bar{x}_{1}=60 \text { days } & \bar{x}_{2}=51 \text { days } \\\ \text { Sample standard deviation } & s_{1}=18 \text { days } & s_{2}=15 \text { days }\end{array}\) What is the point estimate of the difference between population mean number of days on the disabled list for 2001 compared to \(1992 ?\) What is the percentage increase in the number of days on the disabled list? c. Use \(\alpha=.01 .\) What is your conclusion about the number of days on the disabled list? What is the \(p\) -value? d. Do these data suggest that Major League Baseball should be concerned about the situation?

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: \(\mathrm{SST}=10,800 ; \mathrm{SSTR}=4560\) a. Set up the ANOVA table for this problem. b. Use \(\alpha=.05\) to test for any significant difference in the means for the three assembly methods.

StreetInsider.com reported 2002 earnings per share data for a sample of major companies (February 12,2003 ). Prior to 2002 , financial analysts predicted the 2002 earnings per share for these same companies (Barron's, September 10,2001 ). Use the following data to comment on differences between actual and estimated earnings per share. $$\begin{array}{lcc} \text { Company } & \text { Actual } & \text { Predicted } \\ \text { AT\&T } & 1.29 & 0.38 \\ \text { American Express } & 2.01 & 2.31 \\ \text { Citigroup } & 2.59 & 3.43 \\ \text { Coca-Cola } & 1.60 & 1.78 \\ \text { DuPont } & 1.84 & 2.18 \\ \text { ExxonMobil } & 2.72 & 2.19 \\ \text { General Electric } & 1.51 & 1.71 \\ \text { Johnson \& Johnson } & 2.28 & 2.18 \\ \text { McDonald's } & 0.77 & 1.55 \\ \text { Wal-Mart } & 1.81 & 1.74 \end{array}$$ a. Use \(\alpha=.05\) and test for any difference between the population mean actual and population mean estimated earnings per share. What is the \(p\) -value? What is your conclusion? b. What is the point estimate of the difference between the two means? Did the analysts tend to underestimate or overestimate the earnings? c. At \(95 \%\) confidence, what is the margin of error for the estimate in part (b)? What would you recommend based on this information?

In a study conducted to investigate browsing activity by shoppers, each shopper was initially classified as a nonbrowser, light browser, or heavy browser. For each shopper, the study obtained a measure to determine how comfortable the shopper was in a store. Higher scores indicated greater comfort. Suppose the following data were collected. Use \(\alpha=.05\) to test for differences among comfort levels for the three types of browsers. $$\begin{array}{ccc} & \text { Light } & \text { Heavy } \\ \text { Nonbrowser } & \text { Browser } & \text { Browser } \\ 4 & 5 & 5 \\ 5 & 6 & 7 \\ 6 & 5 & 5 \\ 3 & 4 & 7 \\ 3 & 7 & 4 \\ 4 & 4 & 6 \\ 5 & 6 & 5 \\ 4 & 5 & 7 \end{array}$$

The average expenditure on Valentine's Day was expected to be \(\$ 100.89\) (USA Today. February 13,2006 ). Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 male consumers was \(\$ 135.67,\) and the average expenditure in a sample survey of 30 female consumers was \(\$ 68.64 .\) Based on past surveys, the standard deviation for male consumers is assumed to be \(\$ 35,\) and the standard deviation for female consumers is assumed to be \(\$ 20\) a. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females? b. \(\quad\) At \(99 \%\) confidence, what is the margin of error? c. Develop a \(99 \%\) confidence interval for the difference between the two population means.
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