A past survey of 1,068,000 students taking a standardized test revealed that 9.6% 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, -P2 by using the following inequality. Assume the samples are random and independent. AA P191 P292 P292
A past survey of 1,068,000 students taking a standardized test revealed that 9.6% 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, -P2 by using the following inequality. Assume the samples are random and independent. AA P191 P292 P292
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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![A past survey of 1,068,000 students taking a standardized test revealed that 9.6% 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.
\[
(\hat{p}_1 - \hat{p}_2) - z_c \sqrt{\frac{\hat{p}_1 \hat{q}_1}{n_1} + \frac{\hat{p}_2 \hat{q}_2}{n_2}} < p_1 - p_2 < (\hat{p}_1 - \hat{p}_2) + z_c \sqrt{\frac{\hat{p}_1 \hat{q}_1}{n_1} + \frac{\hat{p}_2 \hat{q}_2}{n_2}}
\]
The confidence interval is \( [ \, \, \, ] < p_1 - p_2 < [ \, \, \, ] \).
(Round to three decimal places as needed.)
**Explanation of Formula:**
The expression given is used to calculate the confidence interval for the difference between two population proportions.
- \( \hat{p}_1 \) and \( \hat{p}_2 \) represent the sample proportions from each survey.
- \( \hat{q}_1 \) and \( \hat{q}_2 \) are the complements of these proportions (i.e., \( \hat{q}_1 = 1 - \hat{p}_1 \) and \( \hat{q}_2 = 1 - \hat{p}_2 \)).
- \( n_1 \) and \( n_2 \) are the sample sizes for each survey.
- \( z_c \) is the critical value for the specified confidence level (90% in this case).
This model assumes random and independent samples and aims to estimate the range in which the true difference between the population proportions lies with a 90% level of confidence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7a24eb4-57b8-4308-8ec2-e6ee4c7ba268%2Fc8bed0c3-86e5-4d59-9b7e-2c42454d1141%2Falhoxih_processed.png&w=3840&q=75)
Transcribed Image Text:A past survey of 1,068,000 students taking a standardized test revealed that 9.6% 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.
\[
(\hat{p}_1 - \hat{p}_2) - z_c \sqrt{\frac{\hat{p}_1 \hat{q}_1}{n_1} + \frac{\hat{p}_2 \hat{q}_2}{n_2}} < p_1 - p_2 < (\hat{p}_1 - \hat{p}_2) + z_c \sqrt{\frac{\hat{p}_1 \hat{q}_1}{n_1} + \frac{\hat{p}_2 \hat{q}_2}{n_2}}
\]
The confidence interval is \( [ \, \, \, ] < p_1 - p_2 < [ \, \, \, ] \).
(Round to three decimal places as needed.)
**Explanation of Formula:**
The expression given is used to calculate the confidence interval for the difference between two population proportions.
- \( \hat{p}_1 \) and \( \hat{p}_2 \) represent the sample proportions from each survey.
- \( \hat{q}_1 \) and \( \hat{q}_2 \) are the complements of these proportions (i.e., \( \hat{q}_1 = 1 - \hat{p}_1 \) and \( \hat{q}_2 = 1 - \hat{p}_2 \)).
- \( n_1 \) and \( n_2 \) are the sample sizes for each survey.
- \( z_c \) is the critical value for the specified confidence level (90% in this case).
This model assumes random and independent samples and aims to estimate the range in which the true difference between the population proportions lies with a 90% level of confidence.
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