Podcast
Questions and Answers
What is the significance level of the test?
What is the significance level of the test?
What is the null hypothesis for this test?
What is the null hypothesis for this test?
What is the sample size of the test?
What is the sample size of the test?
What is the standard deviation of the sample?
What is the standard deviation of the sample?
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What is the sample mean of the starting salary for the nurses?
What is the sample mean of the starting salary for the nurses?
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What is the alternative hypothesis for this test?
What is the alternative hypothesis for this test?
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What is the type of test to be conducted in this problem?
What is the type of test to be conducted in this problem?
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What is the critical region for this test?
What is the critical region for this test?
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How many degrees of freedom does the t-distribution have in this test?
How many degrees of freedom does the t-distribution have in this test?
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What is the test statistic for this test?
What is the test statistic for this test?
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Study Notes
Job Placement Director's Claim
- The average starting salary for nurses is claimed to be $24,000.
- This claim is made by a job placement director.
Sample Data
- A sample of 10 nurses was used for analysis.
- The sample mean salary is $23,450.
- The sample standard deviation of salaries is $400.
Hypothesis Testing
- Null Hypothesis (H0): The average salary for nurses is $24,000.
- Alternative Hypothesis (H1): The average salary for nurses is less than $24,000.
Significance Level
- The significance level (α) is set at 0.05, indicating a 5% threshold for rejecting the null hypothesis.
Test Statistic Calculation
- Use the formula for the t-test statistic:
- ( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} )
- where ( \bar{x} ) is the sample mean, ( \mu ) is the population mean under the null hypothesis, ( s ) is the sample standard deviation, and ( n ) is the sample size.
- Substituting the values:
- ( t = \frac{23,450 - 24,000}{400/\sqrt{10}} )
Degrees of Freedom
- Degrees of freedom (df) equal ( n - 1 ), which is 9 for this sample.
Critical Value
- For a one-tailed test at α = 0.05 with 9 degrees of freedom, the critical t-value can be found from t-distribution tables.
Decision Rule
- If the calculated t-value is less than the critical t-value, reject the null hypothesis.
Conclusion Evaluation
- Based on the analysis, evaluate whether there is enough statistical evidence to reject the job placement director's claim about the average starting salary for nurses.
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Description
Determine if there's enough evidence to reject the claim that the average starting salary for nurses is $24,000. Apply hypothesis testing concepts to a sample of 10 nurses with a given mean and standard deviation. Test your knowledge at a 0.05 significance level.
