Chapter 9: Problem 25
Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 45 \\ H_{\mathrm{a}}: \mu<45 \end{array} \\] A sample of 36 is used. Identify the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=44\) and \(s=5.2\) b. \(\bar{x}=43\) and \(s=4.6\) c. \(\bar{x}=46\) and \(s=5.0\)
Short Answer
Step by step solution
Define the hypothesis
Understand the test statistic
Calculate the test statistic for part a
Find the p-value for part a
Conclusion for part a
Calculate the test statistic for part b
Find the p-value for part b
Conclusion for part b
Calculate the test statistic for part c
Find the p-value for part c
Conclusion for part c
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the p-value in Hypothesis Testing
Think of the p-value like a scale:
- If the p-value is less than a predefined significance level (\( \alpha \)), we reject \( H_0 \) because the evidence is strong.
- If it's greater, we fail to reject \( H_0 \) since there isn't enough evidence.
- For \( \bar{x} = 44 \), the p-value is approximately 0.1251.
- For \( \bar{x} = 43 \), the p-value is approximately 0.0046.
- For \( \bar{x} = 46 \), the p-value is approximately 0.8849.
Exploring the Z-Score
The z-score formula is:
\[z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]
where:
- \( \bar{x} \) is the sample mean.
- \( \mu_0 \) is the hypothesized population mean.
- \( s \) is the sample standard deviation.
- \( n \) is the sample size.
Significance Level Decoded
Here's how to think about significance levels:
- A significance level of 0.01 implies that there is only a 1% risk of rejecting the null hypothesis when it is actually true. This is known as a Type I error.
- The lower the significance level, the stronger the evidence must be to reject the null.