st survey of 1,068,000 students taking a standardized test revealed that 7.9% of the students were planning on studying engineering in t survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence ence between proportions p₁ - P₂ by using the following inequality. Assume the samples are random and independent. P191 P292 P₁-P₂) P191 P292 n₁ n₂ + ·
st survey of 1,068,000 students taking a standardized test revealed that 7.9% of the students were planning on studying engineering in t survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence ence between proportions p₁ - P₂ by using the following inequality. Assume the samples are random and independent. P191 P292 P₁-P₂) P191 P292 n₁ n₂ + ·
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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![**Understanding Confidence Intervals for Proportions: A Case Study**
A past survey of 1,068,000 students taking a standardized test revealed that 7.9% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence interval for the difference between proportions \( p_1 - p_2 \) by using the following inequality. Assume the samples are random and independent.
\[ \left( \hat{p}_1 - \hat{p}_2 \right) - z_c \sqrt{ \frac{\hat{p}_1 q_1}{n_1} + \frac{\hat{p}_2 q_2}{n_2} } < p_1 - p_2 < \left( \hat{p}_1 - \hat{p}_2 \right) + z_c \sqrt{ \frac{\hat{p}_1 q_1}{n_1} + \frac{\hat{p}_2 q_2}{n_2} } \]
where:
- \(\hat{p}_1\) is the sample proportion from the first survey (7.9% or 0.079),
- \(\hat{p}_2\) is the sample proportion from the second survey (9.2% or 0.092),
- \(q_1 = 1 - \hat{p}_1\),
- \(q_2 = 1 - \hat{p}_2\),
- \(n_1\) is the sample size from the first survey (1,068,000),
- \(n_2\) is the sample size from the second survey (1,476,000),
- \(z_c\) is the critical value corresponding to the desired confidence level (for 90% confidence level, \(z_c \approx 1.645\)).
**Calculating the Confidence Interval**
Using the formula provided, calculate the confidence interval as follows:
\[ \left(0.079 - 0.092\right) - 1.645 \sqrt{ \frac{0.079 \cdot (1 - 0.079)}{1,068,000} + \frac{0.092 \cdot (1 - 0.092](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fac4255ba-2787-4b8b-b63d-664d949df058%2F635bbcdc-9f03-415a-98ef-1467af0bb636%2Fsjevri_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Confidence Intervals for Proportions: A Case Study**
A past survey of 1,068,000 students taking a standardized test revealed that 7.9% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence interval for the difference between proportions \( p_1 - p_2 \) by using the following inequality. Assume the samples are random and independent.
\[ \left( \hat{p}_1 - \hat{p}_2 \right) - z_c \sqrt{ \frac{\hat{p}_1 q_1}{n_1} + \frac{\hat{p}_2 q_2}{n_2} } < p_1 - p_2 < \left( \hat{p}_1 - \hat{p}_2 \right) + z_c \sqrt{ \frac{\hat{p}_1 q_1}{n_1} + \frac{\hat{p}_2 q_2}{n_2} } \]
where:
- \(\hat{p}_1\) is the sample proportion from the first survey (7.9% or 0.079),
- \(\hat{p}_2\) is the sample proportion from the second survey (9.2% or 0.092),
- \(q_1 = 1 - \hat{p}_1\),
- \(q_2 = 1 - \hat{p}_2\),
- \(n_1\) is the sample size from the first survey (1,068,000),
- \(n_2\) is the sample size from the second survey (1,476,000),
- \(z_c\) is the critical value corresponding to the desired confidence level (for 90% confidence level, \(z_c \approx 1.645\)).
**Calculating the Confidence Interval**
Using the formula provided, calculate the confidence interval as follows:
\[ \left(0.079 - 0.092\right) - 1.645 \sqrt{ \frac{0.079 \cdot (1 - 0.079)}{1,068,000} + \frac{0.092 \cdot (1 - 0.092
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