Construct a 95% confidence interval estimate for the percentage of students who prefer online learning.
Construct a 95% confidence interval estimate for the percentage of students who prefer online learning.
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
![**Constructing a Confidence Interval for Population Proportion**
**Objective:**
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion \( p \). Round your answer to two decimals.
**16) Problem Statement:**
According to the results of a new national study of 1000 randomly selected students, 760 of the students surveyed prefer online learning rather than traditional classrooms. Construct a 95% confidence interval estimate for the percentage of students who prefer online learning.
**Solution Steps:**
1. **Identify the Sample Proportion \( \hat{p} \):**
\[
\hat{p} = \frac{x}{n}
\]
where \( x \) is the number of students who prefer online learning (760) and \( n \) is the total number of students surveyed (1000).
\[
\hat{p} = \frac{760}{1000} = 0.76
\]
2. **Determine the standard error \( SE \):**
\[
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]
\[
SE = \sqrt{\frac{0.76(1-0.76)}{1000}} = \sqrt{\frac{0.76 \times 0.24}{1000}} = \sqrt{\frac{0.1824}{1000}} = \sqrt{0.0001824} \approx 0.0135
\]
3. **Find the critical value for a 95% confidence level:**
For a 95% confidence level, the critical value \( Z \) is approximately 1.96.
4. **Construct the confidence interval:**
\[
\text{Margin of Error} = Z \times SE
\]
\[
\text{Margin of Error} = 1.96 \times 0.0135 \approx 0.0265
\]
5. **Calculate the lower and upper bounds of the confidence interval:**
\[
\text{Lower bound} = \hat{p} - \text{Margin of Error} = 0.76 - 0.0265 = 0.7335
\]
\[
\text{Upper bound} = \hat{p} + \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8edc575-09a4-4784-8812-8f207ff38f0b%2F43fb2144-2b8c-4f1a-8296-8a50e82e8152%2F4csoqpj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Constructing a Confidence Interval for Population Proportion**
**Objective:**
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion \( p \). Round your answer to two decimals.
**16) Problem Statement:**
According to the results of a new national study of 1000 randomly selected students, 760 of the students surveyed prefer online learning rather than traditional classrooms. Construct a 95% confidence interval estimate for the percentage of students who prefer online learning.
**Solution Steps:**
1. **Identify the Sample Proportion \( \hat{p} \):**
\[
\hat{p} = \frac{x}{n}
\]
where \( x \) is the number of students who prefer online learning (760) and \( n \) is the total number of students surveyed (1000).
\[
\hat{p} = \frac{760}{1000} = 0.76
\]
2. **Determine the standard error \( SE \):**
\[
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]
\[
SE = \sqrt{\frac{0.76(1-0.76)}{1000}} = \sqrt{\frac{0.76 \times 0.24}{1000}} = \sqrt{\frac{0.1824}{1000}} = \sqrt{0.0001824} \approx 0.0135
\]
3. **Find the critical value for a 95% confidence level:**
For a 95% confidence level, the critical value \( Z \) is approximately 1.96.
4. **Construct the confidence interval:**
\[
\text{Margin of Error} = Z \times SE
\]
\[
\text{Margin of Error} = 1.96 \times 0.0135 \approx 0.0265
\]
5. **Calculate the lower and upper bounds of the confidence interval:**
\[
\text{Lower bound} = \hat{p} - \text{Margin of Error} = 0.76 - 0.0265 = 0.7335
\]
\[
\text{Upper bound} = \hat{p} + \
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