**Topic 6.1.47: Sample Size Determination****Objective:** Determine the minimum sample size required to achieve a specified confidence interval for a sample mean.**Problem Statement:**Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean. Assume the population has a standard deviation ($\sigma$) of 11.3 and is normally distributed.**Task:**Calculate the required sample size for a 95% confidence level.**Instructions:**- Round up to the nearest whole number as needed. _Note: The problem is purely textual and has no accompanying graphs or diagrams._

Answered: Determine the minimum sample size…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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**Topic 6.1.47: Sample Size Determination**

**Objective:** Determine the minimum sample size required to achieve a specified confidence interval for a sample mean.

**Problem Statement:**

Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean. Assume the population has a standard deviation ($\sigma$) of 11.3 and is normally distributed.

**Task:**

Calculate the required sample size for a 95% confidence level.

**Instructions:**

- Round up to the nearest whole number as needed.

_Note: The problem is purely textual and has no accompanying graphs or diagrams._

Expert Solution

Step 1

The sample size of a population is calculated as follows,

$\begin{array}{rcl} E & = & (\frac{z_{α / 2} \times σ}{\sqrt{n}}) \end{array}$

Where,

$\begin{array}{rcl} E & = & Margin of error \\ z_{α / 2} & = & z score \\ σ & = & Standard deviation \\ n & = & Sample size \end{array}$

Step 2

To find the sample size at 95% confidence interval as follows,

$\begin{array}{rcl} σ & = & 11.3 \\ α & = & 1 - 95 % = 1 - 0.95 = 0.05 \\ \frac{α}{2} & = & \frac{0.05}{2} = 0.025 \\ z_{α / 2} & = & z_{0.025} = 1.96 (from z table) \\ E & = & 1 \end{array}$

Substituting the values in following equation,

$\begin{array}{rcl} n & = & {(\frac{z_{α / 2} \times σ}{E})}^{2} \\ = & {(\frac{1.96 \times 11.3}{1})}^{2} \\ = & 490.5 \\ n & = & 491 \end{array}$

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