estimate how much parents spend on their kids birt opulation standard deviation is approximately o restimate is within 4 dollar(s) of average spending u have to sample?
estimate how much parents spend on their kids birt opulation standard deviation is approximately o restimate is within 4 dollar(s) of average spending u have to sample?
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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![### Sample Size Determination for Estimating Parental Spending on Kids' Birthday Parties
**Objective:**
To estimate how much parents spend on their kids' birthday parties using a sample size that ensures a specified confidence level and margin of error.
**Scenario:**
You want to obtain a sample to estimate how much parents spend on their kids' birthday parties. Based on a previous study, you believe the population standard deviation is approximately \( \sigma = 34.6 \) dollars. You would like to be 90% confident that your estimate is within 4 dollar(s) of average spending on the birthday parties.
**Question:**
How many parents do you have to sample?
**Formula:**
To find the necessary sample size \( n \), we use the following formula derived from the margin of error for a population mean:
\[ n = \left ( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right )^2 \]
Where:
- \( Z_{\alpha/2} \) is the Z-value that corresponds to the desired confidence level.
- \( \sigma \) is the population standard deviation.
- \( E \) is the margin of error.
For a 90% confidence level, the corresponding \( Z_{\alpha/2} \) value is approximately 1.645. Given:
- Population standard deviation, \( \sigma = 34.6 \) dollars.
- Desired margin of error, \( E = 4 \) dollars.
You can plug these values into the formula to calculate the required sample size \( n \).
**Calculation:**
\[ n = \left ( \frac{1.645 \cdot 34.6}{4} \right )^2 \]
**Result:**
\[ n = \left ( \frac{56.977}{4} \right )^2 = \left ( 14.24425 \right )^2 \approx 203 \]
Thus, you need to sample approximately **203 parents** to estimate how much parents spend on their kids' birthday parties with 90% confidence and a margin of error of 4 dollars.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc260e632-e4e2-45c7-853f-2d07e95220d0%2F14d9877b-b140-45cf-868a-f632eb19e6ee%2F1cndheq_processed.png&w=3840&q=75)
Transcribed Image Text:### Sample Size Determination for Estimating Parental Spending on Kids' Birthday Parties
**Objective:**
To estimate how much parents spend on their kids' birthday parties using a sample size that ensures a specified confidence level and margin of error.
**Scenario:**
You want to obtain a sample to estimate how much parents spend on their kids' birthday parties. Based on a previous study, you believe the population standard deviation is approximately \( \sigma = 34.6 \) dollars. You would like to be 90% confident that your estimate is within 4 dollar(s) of average spending on the birthday parties.
**Question:**
How many parents do you have to sample?
**Formula:**
To find the necessary sample size \( n \), we use the following formula derived from the margin of error for a population mean:
\[ n = \left ( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right )^2 \]
Where:
- \( Z_{\alpha/2} \) is the Z-value that corresponds to the desired confidence level.
- \( \sigma \) is the population standard deviation.
- \( E \) is the margin of error.
For a 90% confidence level, the corresponding \( Z_{\alpha/2} \) value is approximately 1.645. Given:
- Population standard deviation, \( \sigma = 34.6 \) dollars.
- Desired margin of error, \( E = 4 \) dollars.
You can plug these values into the formula to calculate the required sample size \( n \).
**Calculation:**
\[ n = \left ( \frac{1.645 \cdot 34.6}{4} \right )^2 \]
**Result:**
\[ n = \left ( \frac{56.977}{4} \right )^2 = \left ( 14.24425 \right )^2 \approx 203 \]
Thus, you need to sample approximately **203 parents** to estimate how much parents spend on their kids' birthday parties with 90% confidence and a margin of error of 4 dollars.
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