Suppose that you wanted to estimate the population mean amount spent eating out per week by Gainesville families. In 2010, a survey found the sample standard deviation was 200 dollars. What size sample would you need to collect in order to make a 95% confidence interval with a margin of error of 15 dollars? O 683 O 27 O783 O 583
Suppose that you wanted to estimate the population mean amount spent eating out per week by Gainesville families. In 2010, a survey found the sample standard deviation was 200 dollars. What size sample would you need to collect in order to make a 95% confidence interval with a margin of error of 15 dollars? O 683 O 27 O783 O 583
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 Population Mean**
*Question:*
Suppose that you wanted to estimate the population mean amount spent eating out per week by Gainesville families. In 2010, a survey found the sample standard deviation was 200 dollars. What size sample would you need to collect in order to make a 95% confidence interval with a margin of error of 15 dollars?
*Options:*
- 683
- 27
- 783
- 583
**Explanation:**
- The problem involves a sample size calculation for estimating a population mean.
- Given data:
- Sample standard deviation (s) = $200
- Desired margin of error (E) = $15
- Confidence level = 95%
To determine the sample size \[ n \] for a confidence interval, use the formula:
\[ n = \left( \frac{Z \cdot s}{E} \right)^2 \]
Where:
- Z is the Z-value from the standard normal distribution corresponding to the desired confidence level (for a 95% confidence level, Z ≈ 1.96).
By plugging in the given values:
\[ n = \left( \frac{1.96 \cdot 200}{15} \right)^2 \]
\[ n = \left( \frac{392}{15} \right)^2 \]
\[ n \approx (26.13)^2 \]
\[ n \approx 682.75 \]
Rounding up to the nearest integer:
\[ n \approx 683 \]
Hence, the correct answer is:
- 683](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8f0470c7-e9db-41c2-b89b-22c87c01f8e0%2Fa8a7becb-2550-42bc-aae2-43e40cc8abe5%2F9fvds6sc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Sample Size Determination for Estimating Population Mean**
*Question:*
Suppose that you wanted to estimate the population mean amount spent eating out per week by Gainesville families. In 2010, a survey found the sample standard deviation was 200 dollars. What size sample would you need to collect in order to make a 95% confidence interval with a margin of error of 15 dollars?
*Options:*
- 683
- 27
- 783
- 583
**Explanation:**
- The problem involves a sample size calculation for estimating a population mean.
- Given data:
- Sample standard deviation (s) = $200
- Desired margin of error (E) = $15
- Confidence level = 95%
To determine the sample size \[ n \] for a confidence interval, use the formula:
\[ n = \left( \frac{Z \cdot s}{E} \right)^2 \]
Where:
- Z is the Z-value from the standard normal distribution corresponding to the desired confidence level (for a 95% confidence level, Z ≈ 1.96).
By plugging in the given values:
\[ n = \left( \frac{1.96 \cdot 200}{15} \right)^2 \]
\[ n = \left( \frac{392}{15} \right)^2 \]
\[ n \approx (26.13)^2 \]
\[ n \approx 682.75 \]
Rounding up to the nearest integer:
\[ n \approx 683 \]
Hence, the correct answer is:
- 683
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