uppose you are using a = 0.01 to test the claim that u≤36 using a P-value. You are given the sample statistics n = 39, x = 37.9, nd s=4.2. Find the P-value. 0.0195 0.0024 0.1014 0.9976
uppose you are using a = 0.01 to test the claim that u≤36 using a P-value. You are given the sample statistics n = 39, x = 37.9, nd s=4.2. Find the P-value. 0.0195 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
Problem 1P
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Question
Find the p-value.
![### Hypothesis Testing and P-Value Calculation
#### Problem Statement
Suppose you are using \(\alpha = 0.01\) to test the claim that \(\mu \leq 36\) using a P-value. You are given the following sample statistics:
- \( n = 39 \) (sample size)
- \( \bar{x} = 37.9 \) (sample mean)
- \( s = 4.2 \) (sample standard deviation)
Find the P-value.
#### Potential P-value Options
1. 0.0195
2. 0.0024
3. 0.1014
4. 0.9976
#### Solution Explanation
To determine the P-value, follow these steps:
1. **State the Null and Alternative Hypotheses:**
- Null Hypothesis, \(H_0: \mu \leq 36\)
- Alternative Hypothesis, \(H_1: \mu > 36\)
2. **Calculate the Test Statistic:**
Use the formula for the t-test statistic:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
Where:
- \(\bar{x} = 37.9\)
- \(\mu_0 = 36\)
- \(s = 4.2\)
- \(n = 39\)
Plug in these values to get:
\[
t = \frac{37.9 - 36}{4.2 / \sqrt{39}} \approx 2.722
\]
3. **Determine the P-value:**
Since this is a one-tailed test, we need to find the area to the right of the calculated t-statistic in the t-distribution with \(n - 1 = 38\) degrees of freedom.
Using a t-distribution table or statistical software, the P-value associated with \(t \approx 2.722\) for 38 degrees of freedom is approximately 0.0195.
4. **Compare the P-value with \(\alpha\):**
- If P-value < \(\alpha\), reject the null hypothesis.
- In this case, 0.0195 > 0.01, so we do not reject the null hypothesis](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08ad6c20-d691-4a3f-8949-0fe4518b5718%2F9381fd15-fdfc-4f56-986d-ce9475b326c0%2F0jzota_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing and P-Value Calculation
#### Problem Statement
Suppose you are using \(\alpha = 0.01\) to test the claim that \(\mu \leq 36\) using a P-value. You are given the following sample statistics:
- \( n = 39 \) (sample size)
- \( \bar{x} = 37.9 \) (sample mean)
- \( s = 4.2 \) (sample standard deviation)
Find the P-value.
#### Potential P-value Options
1. 0.0195
2. 0.0024
3. 0.1014
4. 0.9976
#### Solution Explanation
To determine the P-value, follow these steps:
1. **State the Null and Alternative Hypotheses:**
- Null Hypothesis, \(H_0: \mu \leq 36\)
- Alternative Hypothesis, \(H_1: \mu > 36\)
2. **Calculate the Test Statistic:**
Use the formula for the t-test statistic:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
Where:
- \(\bar{x} = 37.9\)
- \(\mu_0 = 36\)
- \(s = 4.2\)
- \(n = 39\)
Plug in these values to get:
\[
t = \frac{37.9 - 36}{4.2 / \sqrt{39}} \approx 2.722
\]
3. **Determine the P-value:**
Since this is a one-tailed test, we need to find the area to the right of the calculated t-statistic in the t-distribution with \(n - 1 = 38\) degrees of freedom.
Using a t-distribution table or statistical software, the P-value associated with \(t \approx 2.722\) for 38 degrees of freedom is approximately 0.0195.
4. **Compare the P-value with \(\alpha\):**
- If P-value < \(\alpha\), reject the null hypothesis.
- In this case, 0.0195 > 0.01, so we do not reject the null hypothesis
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