A simple random sample of 60 items resulted in a sample mean of 25. The population standard deviation is a = 71 (Round your answers to two decimal places.) (a) What is the standard error of the mean, σ-? (b) At 95% confidence, what is the margin of error?
A simple random sample of 60 items resulted in a sample mean of 25. The population standard deviation is a = 71 (Round your answers to two decimal places.) (a) What is the standard error of the mean, σ-? (b) At 95% confidence, what is the margin of error?
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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![### Statistical Analysis and Confidence Intervals
**Problem Statement:**
A simple random sample of 60 items resulted in a sample mean of 25. The population standard deviation is \( \sigma = 7 \). (Round your answers to two decimal places.)
(a) What is the standard error of the mean, \( \sigma_{\bar{x}} \)?
(b) At 95% confidence, what is the margin of error?
**Instructions for Students:**
1. **Calculating the Standard Error of the Mean (SEM):**
- The formula for the standard error of the mean is:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
- Where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
2. **Calculating the Margin of Error (ME) at 95% Confidence Level:**
- Use the Z-score associated with the 95% confidence level, which is 1.96.
- The formula for the margin of error is:
\[
ME = Z \times \sigma_{\bar{x}}
\]
- Where \( Z \) is the Z-score and \( \sigma_{\bar{x}} \) is the standard error of the mean.
**Example Calculations:**
1. **Standard Error of the Mean:**
- Given \( \sigma = 7 \) and \( n = 60 \),
\[
\sigma_{\bar{x}} = \frac{7}{\sqrt{60}} \approx \frac{7}{7.75} \approx 0.90
\]
2. **Margin of Error:**
- At 95% confidence level, \( Z = 1.96 \),
\[
ME = 1.96 \times 0.90 \approx 1.76
\]
These calculations help determine the accuracy and reliability of the sample mean as an estimate of the population mean.
**Need Help?**
Click the "Read It" button for further explanation and examples.
---
### Visual Aids and Diagrams
**No visual aids or diagrams were included in the provided text.**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5018757-df34-47e5-8789-7c8e64b31b6d%2Fe5d0d1b0-6ac7-4be4-8bac-cb3972814d23%2F463ytb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Statistical Analysis and Confidence Intervals
**Problem Statement:**
A simple random sample of 60 items resulted in a sample mean of 25. The population standard deviation is \( \sigma = 7 \). (Round your answers to two decimal places.)
(a) What is the standard error of the mean, \( \sigma_{\bar{x}} \)?
(b) At 95% confidence, what is the margin of error?
**Instructions for Students:**
1. **Calculating the Standard Error of the Mean (SEM):**
- The formula for the standard error of the mean is:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
- Where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
2. **Calculating the Margin of Error (ME) at 95% Confidence Level:**
- Use the Z-score associated with the 95% confidence level, which is 1.96.
- The formula for the margin of error is:
\[
ME = Z \times \sigma_{\bar{x}}
\]
- Where \( Z \) is the Z-score and \( \sigma_{\bar{x}} \) is the standard error of the mean.
**Example Calculations:**
1. **Standard Error of the Mean:**
- Given \( \sigma = 7 \) and \( n = 60 \),
\[
\sigma_{\bar{x}} = \frac{7}{\sqrt{60}} \approx \frac{7}{7.75} \approx 0.90
\]
2. **Margin of Error:**
- At 95% confidence level, \( Z = 1.96 \),
\[
ME = 1.96 \times 0.90 \approx 1.76
\]
These calculations help determine the accuracy and reliability of the sample mean as an estimate of the population mean.
**Need Help?**
Click the "Read It" button for further explanation and examples.
---
### Visual Aids and Diagrams
**No visual aids or diagrams were included in the provided text.**
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