Suppose you are using a = 0.01 to test the claim that μ≤36 using a P-value. You are given the sample statistics n = 39, x = 37.9, and s=4.2. Find the P-value. 0.0195 X 0.0024 0.1014 0.9976
Suppose you are using a = 0.01 to test the claim that μ≤36 using a P-value. You are given the sample statistics n = 39, x = 37.9, and s=4.2. Find the P-value. 0.0195 X 0.0024 0.1014 0.9976
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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**Example: Calculating P-Value for Hypothesis Testing**
**Context:**
Given a significance level (α) of 0.01, we are testing the claim that the population mean (μ) is less than or equal to 36. You are provided with the following sample statistics:
- Sample size (n) = 39
- Sample mean (\( \bar{x} \)) = 37.9
- Sample standard deviation (s) = 4.2
**Question:**
Find the P-value associated with this test.
**Options:**
- \(0.0195\)
- \(0.0024\)
- \(0.1014\)
- **\(0.9976\) (selected)**
**Explanation:**
To find the P-value, follow these steps:
1. Set up the null hypothesis \(H_0: \mu \le 36\) and the alternative hypothesis \(H_1: \mu > 36\).
2. Compute the test statistic using the formula for a t-test:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
Where:
- \( \bar{x} = 37.9 \)
- \( \mu_0 = 36 \)
- \( s = 4.2 \)
- \( n = 39 \)
3. Simplify and calculate the t-statistic:
\[
t = \frac{37.9 - 36}{4.2 / \sqrt{39}}
\]
\[
t \approx 2.63
\]
4. Determine the P-value corresponding to the calculated t-statistic and degrees of freedom (df = n - 1 = 38) from the t-distribution table or using statistical software.
In this case, the P-value calculated is approximately \(0.9976\).
**Conclusion:**
Since the P-value \(0.9976\) is much greater than the significance level \( \alpha = 0.01\), we fail to reject the null hypothesis.
---
By incorporating this example, students can gain practical understanding of the steps required to calculate P-values in hypothesis testing scenarios.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08ad6c20-d691-4a3f-8949-0fe4518b5718%2F60651883-9eb8-4e8a-89df-a38486390405%2Foo7sgzt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:On an educational website, the transcription and explanation of the image would be as follows:
---
**Example: Calculating P-Value for Hypothesis Testing**
**Context:**
Given a significance level (α) of 0.01, we are testing the claim that the population mean (μ) is less than or equal to 36. You are provided with the following sample statistics:
- Sample size (n) = 39
- Sample mean (\( \bar{x} \)) = 37.9
- Sample standard deviation (s) = 4.2
**Question:**
Find the P-value associated with this test.
**Options:**
- \(0.0195\)
- \(0.0024\)
- \(0.1014\)
- **\(0.9976\) (selected)**
**Explanation:**
To find the P-value, follow these steps:
1. Set up the null hypothesis \(H_0: \mu \le 36\) and the alternative hypothesis \(H_1: \mu > 36\).
2. Compute the test statistic using the formula for a t-test:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
Where:
- \( \bar{x} = 37.9 \)
- \( \mu_0 = 36 \)
- \( s = 4.2 \)
- \( n = 39 \)
3. Simplify and calculate the t-statistic:
\[
t = \frac{37.9 - 36}{4.2 / \sqrt{39}}
\]
\[
t \approx 2.63
\]
4. Determine the P-value corresponding to the calculated t-statistic and degrees of freedom (df = n - 1 = 38) from the t-distribution table or using statistical software.
In this case, the P-value calculated is approximately \(0.9976\).
**Conclusion:**
Since the P-value \(0.9976\) is much greater than the significance level \( \alpha = 0.01\), we fail to reject the null hypothesis.
---
By incorporating this example, students can gain practical understanding of the steps required to calculate P-values in hypothesis testing scenarios.
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