**Title: Applying the Central Limit Theorem to Sampling Distributions** **Objective:** Learn how to use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. **Problem Statement:** The weights of people in a certain population are normally distributed with: - Mean (\(\mu\)) = 155 lb - Standard Deviation (\(\sigma\)) = 22 lb When random samples of size 8 are drawn from the population, identify the mean and standard deviation of the sampling distribution of sample means with sample size \(n\). **Instructions:** - Provide exact answers when possible. If not exact, round to the nearest hundredth. **Calculations:** 1. **Mean of the Sampling Distribution (\(\mu_{\bar{x}}\)):** - The mean of the sampling distribution of the sample means is the same as the mean of the population. - \(\mu_{\bar{x}} = \mu = 155 \text{ lb}\) 2. **Standard Error of the Mean (SEM or \(\sigma_{\bar{x}}\)):** - The standard error is calculated as the population standard deviation divided by the square root of the sample size. - \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{8}}\) **Results:** - **Mean (\(\mu_{\bar{x}}\)):** 155 lb - **Standard Error (\(\sigma_{\bar{x}}\)):** Calculate \(\frac{22}{\sqrt{8}}\) and round to the nearest hundredth. **Conclusion:** Using the Central Limit Theorem, we can estimate the mean and standard error of the sampling distribution for any given sample size, provided the initial population is normally distributed or the sample size is sufficiently large. This understanding is crucial for making inferences about population parameters based on sample statistics.
**Title: Applying the Central Limit Theorem to Sampling Distributions** **Objective:** Learn how to use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. **Problem Statement:** The weights of people in a certain population are normally distributed with: - Mean (\(\mu\)) = 155 lb - Standard Deviation (\(\sigma\)) = 22 lb When random samples of size 8 are drawn from the population, identify the mean and standard deviation of the sampling distribution of sample means with sample size \(n\). **Instructions:** - Provide exact answers when possible. If not exact, round to the nearest hundredth. **Calculations:** 1. **Mean of the Sampling Distribution (\(\mu_{\bar{x}}\)):** - The mean of the sampling distribution of the sample means is the same as the mean of the population. - \(\mu_{\bar{x}} = \mu = 155 \text{ lb}\) 2. **Standard Error of the Mean (SEM or \(\sigma_{\bar{x}}\)):** - The standard error is calculated as the population standard deviation divided by the square root of the sample size. - \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{8}}\) **Results:** - **Mean (\(\mu_{\bar{x}}\)):** 155 lb - **Standard Error (\(\sigma_{\bar{x}}\)):** Calculate \(\frac{22}{\sqrt{8}}\) and round to the nearest hundredth. **Conclusion:** Using the Central Limit Theorem, we can estimate the mean and standard error of the sampling distribution for any given sample size, provided the initial population is normally distributed or the sample size is sufficiently large. This understanding is crucial for making inferences about population parameters based on sample statistics.
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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Question
Transcribed Image Text:**Title: Applying the Central Limit Theorem to Sampling Distributions**
**Objective:**
Learn how to use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.
**Problem Statement:**
The weights of people in a certain population are normally distributed with:
- Mean (\(\mu\)) = 155 lb
- Standard Deviation (\(\sigma\)) = 22 lb
When random samples of size 8 are drawn from the population, identify the mean and standard deviation of the sampling distribution of sample means with sample size \(n\).
**Instructions:**
- Provide exact answers when possible. If not exact, round to the nearest hundredth.
**Calculations:**
1. **Mean of the Sampling Distribution (\(\mu_{\bar{x}}\)):**
- The mean of the sampling distribution of the sample means is the same as the mean of the population.
- \(\mu_{\bar{x}} = \mu = 155 \text{ lb}\)
2. **Standard Error of the Mean (SEM or \(\sigma_{\bar{x}}\)):**
- The standard error is calculated as the population standard deviation divided by the square root of the sample size.
- \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{8}}\)
**Results:**
- **Mean (\(\mu_{\bar{x}}\)):** 155 lb
- **Standard Error (\(\sigma_{\bar{x}}\)):** Calculate \(\frac{22}{\sqrt{8}}\) and round to the nearest hundredth.
**Conclusion:**
Using the Central Limit Theorem, we can estimate the mean and standard error of the sampling distribution for any given sample size, provided the initial population is normally distributed or the sample size is sufficiently large. This understanding is crucial for making inferences about population parameters based on sample statistics.
Expert Solution
Step 1
Obtain the mean of the sampling distribution of the sample mean.
The mean of the sampling distribution of the sample mean is obtained below as follows:
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