Consider a large population with p = 0.24. Assuming ≤ 0.05, find the mean and standar proportion for a sample size of 840. Round your answer for standard deviation to three decimal places. = 6A = Mi Mi
Consider a large population with p = 0.24. Assuming ≤ 0.05, find the mean and standar proportion for a sample size of 840. Round your answer for standard deviation to three decimal places. = 6A = Mi Mi
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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![### Calculating the Mean and Standard Deviation of the Sample Proportion
Consider a large population with the proportion \( p = 0.24 \). Assuming \( \frac{n}{N} \leq 0.05 \), where \( n \) is the sample size and \( N \) is the population size, we aim to find the mean and standard deviation of the sample proportion for a sample size of 840.
To find these values, follow the steps below:
#### Mean of the Sample Proportion
The mean of the sample proportion (\( \mu_{\hat{p}} \)) is equal to the population proportion (\( p \)). Therefore,
\[
\mu_{\hat{p}} = p = 0.24
\]
#### Standard Deviation of the Sample Proportion
The standard deviation of the sample proportion (\( \sigma_{\hat{p}} \)) is given by the formula:
\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
\]
Substituting the given values:
\[
p = 0.24, \quad n = 840
\]
Calculate the standard deviation as follows:
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.24 (1-0.24)}{840}}
\]
Simplify the expression inside the square root:
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.24 \times 0.76}{840}}
\]
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.1824}{840}}
\]
\[
\sigma_{\hat{p}} \approx \sqrt{0.000217142857}
\]
\[
\sigma_{\hat{p}} \approx 0.01473
\]
Finally, round the standard deviation to three decimal places:
\[
\sigma_{\hat{p}} \approx 0.015
\]
### Summary
The mean and standard deviation of the sample proportion for a sample size of 840 are:
\[
\mu_{\hat{p}} = 0.24
\]
\[
\sigma_{\hat{p}} = 0.015
\]
These values are useful in various statistical analyses, particularly when working with proportions in large sample sizes.
#### Diagram Description
In the provided interface:
- There are two fields labeled](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcce67d60-ba9c-4b65-a73f-671ae19935ed%2Faa0fc91e-50cf-468f-93ca-524b08b48a76%2F50gjoil_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Mean and Standard Deviation of the Sample Proportion
Consider a large population with the proportion \( p = 0.24 \). Assuming \( \frac{n}{N} \leq 0.05 \), where \( n \) is the sample size and \( N \) is the population size, we aim to find the mean and standard deviation of the sample proportion for a sample size of 840.
To find these values, follow the steps below:
#### Mean of the Sample Proportion
The mean of the sample proportion (\( \mu_{\hat{p}} \)) is equal to the population proportion (\( p \)). Therefore,
\[
\mu_{\hat{p}} = p = 0.24
\]
#### Standard Deviation of the Sample Proportion
The standard deviation of the sample proportion (\( \sigma_{\hat{p}} \)) is given by the formula:
\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
\]
Substituting the given values:
\[
p = 0.24, \quad n = 840
\]
Calculate the standard deviation as follows:
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.24 (1-0.24)}{840}}
\]
Simplify the expression inside the square root:
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.24 \times 0.76}{840}}
\]
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.1824}{840}}
\]
\[
\sigma_{\hat{p}} \approx \sqrt{0.000217142857}
\]
\[
\sigma_{\hat{p}} \approx 0.01473
\]
Finally, round the standard deviation to three decimal places:
\[
\sigma_{\hat{p}} \approx 0.015
\]
### Summary
The mean and standard deviation of the sample proportion for a sample size of 840 are:
\[
\mu_{\hat{p}} = 0.24
\]
\[
\sigma_{\hat{p}} = 0.015
\]
These values are useful in various statistical analyses, particularly when working with proportions in large sample sizes.
#### Diagram Description
In the provided interface:
- There are two fields labeled
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