Help!!! ### Testing the Claim: Do Nurses in Tampa Earn More Than Nurses in Dallas?#### BackgroundThe Florida Tribune published an article that claimed that nurses in Tampa earn more than nurses in Dallas. As a manager at a Tampa hospital, you want to test the claim of the Tribune. You wish to test the claim at a significance level of $\alpha = 0.05$. To do this, you conduct a follow-up survey of 41 nurses in Tampa and 50 nurses in Dallas. The results of the survey are as follows:- **Sample from Tampa** (Sample 1): - Average Salary: $63,200 - Standard Deviation: $5,300 - Sample Size: 41- **Sample from Dallas** (Sample 2): - Average Salary: $60,700 - Standard Deviation: $4,800 - Sample Size: 50#### Hypotheses- Null Hypothesis ($H_0$): $\mu_1 - \mu_2 = 0$- Alternative Hypothesis ($H_a$): $\mu_1 - \mu_2 > 0$_(In this setup, sample 1 is the group of nurses from Tampa and sample 2 is the group of nurses from Dallas)_#### Questions**a) What is the test statistic?**To find the test statistic, we use the formula for the two-sample Z-test for means when the population variances are unknown but assumed to be equal.\[ Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]Where:\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]Let's plug in the values:\[ s_p^2 = \frac{(41 - 1) \cdot 5300^2 + (50 - 1) \cdot 4800^2}{41 + 50 - 2} \]\[ s_p^2 = \frac{40 \cdot

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Description: Help!!! ### Testing the Claim: Do Nurses in Tampa Earn More Than Nurses in Dallas?#### BackgroundThe Florida Tribune published an article that claimed that nurses in Tampa earn more than nurses in Dallas. As a manager at a Tampa hospital, you want to

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The Florida Tribune published an article that claimed that nurses in Tampa earn more than nurses in Dallas. You are a manager at a Tampa hospital and want to test the claim of the Tribune. You wish to test the claim at a significance level of a=0.05. So you conduct a follow up survey of 41 nurses in Tampa and 50 nurses in Dallas. You find that the sample from Tampa showed an average salary of $63200 with a standard deviation of 5300, and the sample from Dallas showed an average salary of $60700 with a standard deviation of 4800. Our claims are: Ho: µ1 – Hz = 0 На: И — Н2 > 0 (in this setup, sample 1 is the group of nurses from Tampa and sample 2 is the group of nurses from Dallas) a) What is the test statistic? b) What is the p-value? c) State your conclusion (in the context of the problem).

The Florida Tribune published an article that claimed that nurses in Tampa earn more than nurses in Dallas. You are a manager at a Tampa hospital and want to test the claim of the Tribune. You wish to test the claim at a significance level of a=0.05. So you conduct a follow up survey of 41 nurses in Tampa and 50 nurses in Dallas. You find that the sample from Tampa showed an average salary of $63200 with a standard deviation of 5300, and the sample from Dallas showed an average salary of $60700 with a standard deviation of 4800. Our claims are: Ho: µ1 – Hz = 0 На: И — Н2 > 0 (in this setup, sample 1 is the group of nurses from Tampa and sample 2 is the group of nurses from Dallas) a) What is the test statistic? b) What is the p-value? c) State your conclusion (in the context of the problem).

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Help!!!

### Testing the Claim: Do Nurses in Tampa Earn More Than Nurses in Dallas?

#### Background

The Florida Tribune published an article that claimed that nurses in Tampa earn more than nurses in Dallas. As a manager at a Tampa hospital, you want to test the claim of the Tribune. You wish to test the claim at a significance level of $\alpha = 0.05$. To do this, you conduct a follow-up survey of 41 nurses in Tampa and 50 nurses in Dallas. The results of the survey are as follows:

- **Sample from Tampa** (Sample 1):
- Average Salary: $63,200
- Standard Deviation: $5,300
- Sample Size: 41

- **Sample from Dallas** (Sample 2):
- Average Salary: $60,700
- Standard Deviation: $4,800
- Sample Size: 50

#### Hypotheses
- Null Hypothesis ($H_0$): $\mu_1 - \mu_2 = 0$
- Alternative Hypothesis ($H_a$): $\mu_1 - \mu_2 > 0$

_(In this setup, sample 1 is the group of nurses from Tampa and sample 2 is the group of nurses from Dallas)_

#### Questions

**a) What is the test statistic?**

To find the test statistic, we use the formula for the two-sample Z-test for means when the population variances are unknown but assumed to be equal.

\[ Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]

Where:

\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]

Let's plug in the values:

\[ s_p^2 = \frac{(41 - 1) \cdot 5300^2 + (50 - 1) \cdot 4800^2}{41 + 50 - 2} \]
\[ s_p^2 = \frac{40 \cdot

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