**Problem Statement:** Bob draws a simple random sample from an original population. The sample size is \( n = 51 \). The sample mean turns out to be \( \bar{x} = 78 \), and the sample standard deviation turns out to be \( s = 14 \). Find the 95% confidence interval for the population mean \( \mu \). **Options:** A. (74.062, 81.938) B. (74.062, 83.948) C. (74.062, 85.958) D. (74.062, 87.968) **Explanation:** The task is to calculate a 95% confidence interval for the population mean based on the given sample data. The confidence interval can be determined using the formula for the confidence interval of the mean when the population standard deviation is not known, and the sample size is greater than 30: \[ \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \] Where: - \( \bar{x} \) is the sample mean - \( s \) is the sample standard deviation - \( n \) is the sample size - \( t_{\alpha/2} \) is the t-score that corresponds to the desired level of confidence (from t-distribution tables) For a 95% confidence level and generally large sample size, \( t \) is approximately 2.009 for \( n-1 \) degrees of freedom, which can be used to find the margin of error for the interval. Users will need to calculate the interval and choose the correct one from the provided options.
**Problem Statement:** Bob draws a simple random sample from an original population. The sample size is \( n = 51 \). The sample mean turns out to be \( \bar{x} = 78 \), and the sample standard deviation turns out to be \( s = 14 \). Find the 95% confidence interval for the population mean \( \mu \). **Options:** A. (74.062, 81.938) B. (74.062, 83.948) C. (74.062, 85.958) D. (74.062, 87.968) **Explanation:** The task is to calculate a 95% confidence interval for the population mean based on the given sample data. The confidence interval can be determined using the formula for the confidence interval of the mean when the population standard deviation is not known, and the sample size is greater than 30: \[ \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \] Where: - \( \bar{x} \) is the sample mean - \( s \) is the sample standard deviation - \( n \) is the sample size - \( t_{\alpha/2} \) is the t-score that corresponds to the desired level of confidence (from t-distribution tables) For a 95% confidence level and generally large sample size, \( t \) is approximately 2.009 for \( n-1 \) degrees of freedom, which can be used to find the margin of error for the interval. Users will need to calculate the interval and choose the correct one from the provided options.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
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![**Problem Statement:**
Bob draws a simple random sample from an original population. The sample size is \( n = 51 \). The sample mean turns out to be \( \bar{x} = 78 \), and the sample standard deviation turns out to be \( s = 14 \). Find the 95% confidence interval for the population mean \( \mu \).
**Options:**
A. (74.062, 81.938)
B. (74.062, 83.948)
C. (74.062, 85.958)
D. (74.062, 87.968)
**Explanation:**
The task is to calculate a 95% confidence interval for the population mean based on the given sample data. The confidence interval can be determined using the formula for the confidence interval of the mean when the population standard deviation is not known, and the sample size is greater than 30:
\[
\bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right)
\]
Where:
- \( \bar{x} \) is the sample mean
- \( s \) is the sample standard deviation
- \( n \) is the sample size
- \( t_{\alpha/2} \) is the t-score that corresponds to the desired level of confidence (from t-distribution tables)
For a 95% confidence level and generally large sample size, \( t \) is approximately 2.009 for \( n-1 \) degrees of freedom, which can be used to find the margin of error for the interval.
Users will need to calculate the interval and choose the correct one from the provided options.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F989374ce-3c35-4f4d-b4d6-005e01ca096b%2F18a2eb8a-2272-4979-8d45-e8984c8cd15f%2Fgxbr7zp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Bob draws a simple random sample from an original population. The sample size is \( n = 51 \). The sample mean turns out to be \( \bar{x} = 78 \), and the sample standard deviation turns out to be \( s = 14 \). Find the 95% confidence interval for the population mean \( \mu \).
**Options:**
A. (74.062, 81.938)
B. (74.062, 83.948)
C. (74.062, 85.958)
D. (74.062, 87.968)
**Explanation:**
The task is to calculate a 95% confidence interval for the population mean based on the given sample data. The confidence interval can be determined using the formula for the confidence interval of the mean when the population standard deviation is not known, and the sample size is greater than 30:
\[
\bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right)
\]
Where:
- \( \bar{x} \) is the sample mean
- \( s \) is the sample standard deviation
- \( n \) is the sample size
- \( t_{\alpha/2} \) is the t-score that corresponds to the desired level of confidence (from t-distribution tables)
For a 95% confidence level and generally large sample size, \( t \) is approximately 2.009 for \( n-1 \) degrees of freedom, which can be used to find the margin of error for the interval.
Users will need to calculate the interval and choose the correct one from the provided options.
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