A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with µ = 524. The teacher obtains a random sample of 2000 students, puts them through the review class, and finds that the mean math score of the 2000 students is 532 with a standard deviation of 117. Complete parts (a) through (d) below. ... Find the P-value. The P-value is (Round to three decimal places as needed.)
A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with µ = 524. The teacher obtains a random sample of 2000 students, puts them through the review class, and finds that the mean math score of the 2000 students is 532 with a standard deviation of 117. Complete parts (a) through (d) below. ... Find the P-value. The P-value is (Round to three decimal places as needed.)
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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![### Statistical Analysis of a Review Course's Effect on College Entrance Exam Scores
A math teacher claims to have developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, it is known that exam scores are normally distributed with a mean (µ) of 524.
To test the effectiveness of the review course, the teacher conducted an experiment:
- **Sample Size (n):** 2000 students
- **Mean Score after Review Course:** 532
- **Standard Deviation (σ):** 117
The steps below outline the method to test the teacher's claim by finding the P-value.
### Problem Statement
**Objective:** Determine if the mean score of the students who took the review class is significantly higher than the population mean score.
#### Given Data:
- Population Mean (µ): 524
- Sample Mean (x̄): 532
- Standard Deviation (σ): 117
- Sample Size (n): 2000
#### Steps to Find the P-value:
1. **State the Hypotheses:**
- Null Hypothesis (H₀): μ = 524 (The review course does not increase scores)
- Alternative Hypothesis (H₁): μ > 524 (The review course increases scores)
2. **Calculate the Test Statistic:**
\[
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
Substitute the given values:
\[
z = \frac{532 - 524}{\frac{117}{\sqrt{2000}}}
\]
Simplify inside the denominator:
\[
z = \frac{532 - 524}{\frac{117}{44.72}} \approx \frac{8}{2.615} \approx 3.06
\]
3. **Find the P-value:**
Using the Z-table or standard normal distribution calculator, find the P-value corresponding to the calculated test statistic \( z = 3.06 \).
### Calculator or Z-table Usage
The P-value for \( z = 3.06 \) is approximately \( 0.001 \) (rounded to three decimal places).
#### Conclusion:
If the P-value is less than the significance level (typically α = 0.05), we reject the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfee5922-5e9b-4d5c-bb5d-6eee324be123%2Fc9043156-d2cc-40e2-b741-3252cbf18623%2Fh94ygb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Statistical Analysis of a Review Course's Effect on College Entrance Exam Scores
A math teacher claims to have developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, it is known that exam scores are normally distributed with a mean (µ) of 524.
To test the effectiveness of the review course, the teacher conducted an experiment:
- **Sample Size (n):** 2000 students
- **Mean Score after Review Course:** 532
- **Standard Deviation (σ):** 117
The steps below outline the method to test the teacher's claim by finding the P-value.
### Problem Statement
**Objective:** Determine if the mean score of the students who took the review class is significantly higher than the population mean score.
#### Given Data:
- Population Mean (µ): 524
- Sample Mean (x̄): 532
- Standard Deviation (σ): 117
- Sample Size (n): 2000
#### Steps to Find the P-value:
1. **State the Hypotheses:**
- Null Hypothesis (H₀): μ = 524 (The review course does not increase scores)
- Alternative Hypothesis (H₁): μ > 524 (The review course increases scores)
2. **Calculate the Test Statistic:**
\[
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
Substitute the given values:
\[
z = \frac{532 - 524}{\frac{117}{\sqrt{2000}}}
\]
Simplify inside the denominator:
\[
z = \frac{532 - 524}{\frac{117}{44.72}} \approx \frac{8}{2.615} \approx 3.06
\]
3. **Find the P-value:**
Using the Z-table or standard normal distribution calculator, find the P-value corresponding to the calculated test statistic \( z = 3.06 \).
### Calculator or Z-table Usage
The P-value for \( z = 3.06 \) is approximately \( 0.001 \) (rounded to three decimal places).
#### Conclusion:
If the P-value is less than the significance level (typically α = 0.05), we reject the
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