### Hypothesis Testing for Nurse Salaries in ConnecticutThis exercise investigates whether the mean salary of nurses in Connecticut is statistically significantly lower than the national average. The given data is as follows:- **National Mean Salary (μ)**: $62,500- **Standard Deviation (σ)**: $6,000- **Sample Size (n)**: 36 nurses in Connecticut- **Sample Mean ($\bar{x}$)**: $60,400We will conduct a hypothesis test at the 5% significance level to determine if the mean salary for nurses in Connecticut is lower than the national mean salary.#### 1. State the Hypotheses- Null Hypothesis ($H_0$): μ = $62,500 (The mean salary for nurses in Connecticut is equal to the national average)- Alternative Hypothesis ($H_1$): μ < $62,500 (The mean salary for nurses in Connecticut is lower than the national average)#### 2. Determine the Test StatisticSince the standard deviation of the population is known, we use the Z-test for the hypothesis.The Z-test statistic is calculated as:\[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Plugging in the values:\[ Z = \frac{60,400 - 62,500}{\frac{6,000}{\sqrt{36}}} \]\[ Z = \frac{-2,100}{1,000} = -2.1 \]#### 3. Determine the Critical ValueAt the 5% significance level for a one-tailed test (lower tail), the critical value of Z is -1.645.#### 4. Make a DecisionCompare the calculated Z-value to the critical value:- If $ Z \leq -1.645 $, reject the null hypothesis $H_0$- If $ Z > -1.645 $, fail to reject the null hypothesis $H_0$Since -2.1 ≤ -1.645, we reject the null hypothesis.#### 5. ConclusionThere is sufficient evidence at the 5% significance level to conclude that the mean salary of nurses in Connecticut is lower than the national average.This finding is important for policymakers and stakeholders in the healthcare sector, as it may indicate a need for salary adjustments to ensure competitiveness and fair

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Description: ### Hypothesis Testing for Nurse Salaries in ConnecticutThis exercise investigates whether the mean salary of nurses in Connecticut is statistically significantly lower than the national average. The given data is as follows:- **National Mean Salary

The random variable salary follows normal distribution. In this problem, we know the population…

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9. The mean salary for nurses nation wide is u=$62,500 with o-6,000. A sample of 36 nurses in Connecticut have a mean salary of x $60,400. At the 5 % significance level, does the data provide sufficient evidence to show that nurse's salaries in Connecticut are lower than the national average? Give an explanation for your answer.

9. The mean salary for nurses nation wide is u=$62,500 with o-6,000. A sample of 36 nurses in Connecticut have a mean salary of x $60,400. At the 5 % significance level, does the data provide sufficient evidence to show that nurse's salaries in Connecticut are lower than the national average? Give an explanation for your answer.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Concept explainers

Inverse Normal Distribution

The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.

Mean, Median, Mode

It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.

Z-Scores

A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.

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Question

### Hypothesis Testing for Nurse Salaries in Connecticut

This exercise investigates whether the mean salary of nurses in Connecticut is statistically significantly lower than the national average. The given data is as follows:

- **National Mean Salary (μ)**: $62,500
- **Standard Deviation (σ)**: $6,000
- **Sample Size (n)**: 36 nurses in Connecticut
- **Sample Mean ($\bar{x}$)**: $60,400

We will conduct a hypothesis test at the 5% significance level to determine if the mean salary for nurses in Connecticut is lower than the national mean salary.

#### 1. State the Hypotheses
- Null Hypothesis ($H_0$): μ = $62,500 (The mean salary for nurses in Connecticut is equal to the national average)
- Alternative Hypothesis ($H_1$): μ < $62,500 (The mean salary for nurses in Connecticut is lower than the national average)

#### 2. Determine the Test Statistic
Since the standard deviation of the population is known, we use the Z-test for the hypothesis.

The Z-test statistic is calculated as:

\[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Plugging in the values:

\[ Z = \frac{60,400 - 62,500}{\frac{6,000}{\sqrt{36}}} \]

\[ Z = \frac{-2,100}{1,000} = -2.1 \]

#### 3. Determine the Critical Value
At the 5% significance level for a one-tailed test (lower tail), the critical value of Z is -1.645.

#### 4. Make a Decision
Compare the calculated Z-value to the critical value:

- If $ Z \leq -1.645 $, reject the null hypothesis $H_0$
- If $ Z > -1.645 $, fail to reject the null hypothesis $H_0$

Since -2.1 ≤ -1.645, we reject the null hypothesis.

#### 5. Conclusion
There is sufficient evidence at the 5% significance level to conclude that the mean salary of nurses in Connecticut is lower than the national average.

This finding is important for policymakers and stakeholders in the healthcare sector, as it may indicate a need for salary adjustments to ensure competitiveness and fair

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