From a population with a variance of 1296, a sample of 225 items is selected. At 95% confidence, the margin of error is O 0.31 169.34 4.70 1.96 Clear my choice
From a population with a variance of 1296, a sample of 225 items is selected. At 95% confidence, the margin of error is O 0.31 169.34 4.70 1.96 Clear my choice
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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![**Confidence Interval and Margin of Error**
When working with statistical data, one of the essential concepts to understand is the margin of error when estimating population parameters based on sample data. This helps in determining how precise our estimate is and gives us a range in which we expect the true population parameter to lie.
In this example, we will explore how to calculate the margin of error at a 95% confidence level given specific parameters.
**Problem Statement:**
From a population with a variance of 1296, a sample of 225 items is selected. At 95% confidence, the margin of error is to be determined.
There are four answer options provided:
- 0.31
- 169.34
- 4.70 (this option is selected)
- 1.96
**Understanding the Options:**
To calculate the margin of error (E), we use the following formula:
\[ E = z \times \frac{\sigma}{\sqrt{n}} \]
where
- \( z \) is the z-score corresponding to the desired confidence level (For a 95% confidence level, the z-score is approximately 1.96)
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size
Given:
- Population variance (\( \sigma^2 \)) = 1296
- Sample size (\( n \)) = 225
First, we need to find the population standard deviation (\( \sigma \)):
\[ \sigma = \sqrt{1296} = 36 \]
Next, substitute the values into the formula:
\[ E = 1.96 \times \frac{36}{\sqrt{225}} \]
Calculate the denominator:
\[ \sqrt{225} = 15 \]
Therefore:
\[ E = 1.96 \times \frac{36}{15} = 1.96 \times 2.4 = 4.704 \]
Thus, the margin of error, rounded to two decimal places, is approximately 4.70.
**Conclusion:**
The margin of error at a 95% confidence level for a sample of 225 items from a population with a variance of 1296 is 4.70. Thus, the correct answer in the given options is 4.70.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe12f3ad-f419-427a-9045-294545f74979%2F22e5053a-ec7b-438c-8f55-1a788f94ebf1%2Fkoupj5m_processed.png&w=3840&q=75)
Transcribed Image Text:**Confidence Interval and Margin of Error**
When working with statistical data, one of the essential concepts to understand is the margin of error when estimating population parameters based on sample data. This helps in determining how precise our estimate is and gives us a range in which we expect the true population parameter to lie.
In this example, we will explore how to calculate the margin of error at a 95% confidence level given specific parameters.
**Problem Statement:**
From a population with a variance of 1296, a sample of 225 items is selected. At 95% confidence, the margin of error is to be determined.
There are four answer options provided:
- 0.31
- 169.34
- 4.70 (this option is selected)
- 1.96
**Understanding the Options:**
To calculate the margin of error (E), we use the following formula:
\[ E = z \times \frac{\sigma}{\sqrt{n}} \]
where
- \( z \) is the z-score corresponding to the desired confidence level (For a 95% confidence level, the z-score is approximately 1.96)
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size
Given:
- Population variance (\( \sigma^2 \)) = 1296
- Sample size (\( n \)) = 225
First, we need to find the population standard deviation (\( \sigma \)):
\[ \sigma = \sqrt{1296} = 36 \]
Next, substitute the values into the formula:
\[ E = 1.96 \times \frac{36}{\sqrt{225}} \]
Calculate the denominator:
\[ \sqrt{225} = 15 \]
Therefore:
\[ E = 1.96 \times \frac{36}{15} = 1.96 \times 2.4 = 4.704 \]
Thus, the margin of error, rounded to two decimal places, is approximately 4.70.
**Conclusion:**
The margin of error at a 95% confidence level for a sample of 225 items from a population with a variance of 1296 is 4.70. Thus, the correct answer in the given options is 4.70.
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