A sample mean, sample size, and population standard deviation are given. Uso ihe one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical-value approach. x= 51, n= 46 , a= 3.6, Ho: u= 50; H > 50, a = 0.01
A sample mean, sample size, and population standard deviation are given. Uso ihe one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical-value approach. x= 51, n= 46 , a= 3.6, Ho: u= 50; H > 50, a = 0.01
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Hypothesis Testing Using One-Mean Z-Test
#### Problem Statement
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical-value approach.
#### Given Data
- Sample mean (\( \overline{x} \)): 51
- Sample size (\( n \)): 46
- Population standard deviation (\( \sigma \)): 3.6
- Null hypothesis (\( H_0 \)): \( \mu = 50 \)
- Alternative hypothesis (\( H_a \)): \( \mu > 50 \)
- Significance level (\( \alpha \)): 0.01
#### Choices
1. \( z = 0.28 \); critical value = 2.33; do not reject \( H_0 \)
2. \( z = 1.884 \); critical value = 2.33; reject \( H_0 \)
3. \( z = 1.884 \); critical value = 1.33; reject \( H_0 \)
4. \( z = 1.884 \); critical value = 2.33; do not reject \( H_0 \)
#### Explanation:
1. **Choice A**:
- \( z = 0.28 \)
- Given the critical value of 2.33.
- Since \( 0.28 < 2.33 \), we do not reject \( H_0 \).
2. **Choice B**:
- \( z = 1.884 \)
- Given the critical value of 2.33.
- Since \( 1.884 < 2.33 \), we do not reject \( H_0 \).
3. **Choice C**:
- \( z = 1.884 \)
- Given the critical value of 1.33.
- Since \( 1.884 > 1.33 \), we reject \( H_0 \).
4. **Choice D**:
- \( z = 1.884 \)
- Given the critical value of 2.33.
- Since \( 1.884 < 2.33 \), we do not reject \( H_0 \).
#### Conclusion:
Based on the given choices, the correct conclusion is provided when \( z \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5c84145-a33d-498c-9246-5dc8646df0e1%2Ff31d2a7a-bee0-4182-96de-ea832cf2379c%2F4861mr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing Using One-Mean Z-Test
#### Problem Statement
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical-value approach.
#### Given Data
- Sample mean (\( \overline{x} \)): 51
- Sample size (\( n \)): 46
- Population standard deviation (\( \sigma \)): 3.6
- Null hypothesis (\( H_0 \)): \( \mu = 50 \)
- Alternative hypothesis (\( H_a \)): \( \mu > 50 \)
- Significance level (\( \alpha \)): 0.01
#### Choices
1. \( z = 0.28 \); critical value = 2.33; do not reject \( H_0 \)
2. \( z = 1.884 \); critical value = 2.33; reject \( H_0 \)
3. \( z = 1.884 \); critical value = 1.33; reject \( H_0 \)
4. \( z = 1.884 \); critical value = 2.33; do not reject \( H_0 \)
#### Explanation:
1. **Choice A**:
- \( z = 0.28 \)
- Given the critical value of 2.33.
- Since \( 0.28 < 2.33 \), we do not reject \( H_0 \).
2. **Choice B**:
- \( z = 1.884 \)
- Given the critical value of 2.33.
- Since \( 1.884 < 2.33 \), we do not reject \( H_0 \).
3. **Choice C**:
- \( z = 1.884 \)
- Given the critical value of 1.33.
- Since \( 1.884 > 1.33 \), we reject \( H_0 \).
4. **Choice D**:
- \( z = 1.884 \)
- Given the critical value of 2.33.
- Since \( 1.884 < 2.33 \), we do not reject \( H_0 \).
#### Conclusion:
Based on the given choices, the correct conclusion is provided when \( z \)
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