Q 8.103. - Elementary Statistics

Question

Suppose that a simple random sample is taken from a normal population having a standard deviation of $10$ for the purpose of obtaining a $95\%$ confidence interval for the mean of the population.

a. If the sample size is $4$ , obtain the margin of error.

b. Repeat part (a) for a sample size of $16$

c. Can you guess the margin of error for a sample size of $64$ ? Explain your reasoning.

Step-by-Step Solution

Answer

Part (a) The margin of error by using the formula is $9.8$

Part (b) The margin of error by using the formula is $4.9$

Part (c) The margin of error changes from $4.9$ to $2.45$

Part (a) Step 1: Given information

For the aim of producing a $95\%$ confidence interval for the population mean, a simple random sample is obtained from a normal population with a standard deviation of $10$

Part (a) Step 2: Concept

The formula used $\text{ Margin of error}\left(E\right)=z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Part (a) Step 3: Calculation

Using a formula, calculate the margin of error.

Consider $n=4, \sigma =10$, and confidence level is $95 \%$

The needed value of $z_{\frac{a}{2}}$ with a $95\%$ confidence level is $1.96$, according to "Table II Areas under the standard normal curve."

The margin of error is,

$$\begin{gathered} \text{ Margin of error}\left(E\right)=z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right) \\ =1.96\left(\frac{10}{\sqrt{4}}\right) \\ =1.96\left(5\right) \\ =9.8 \end{gathered}$$

Thus, 

Part (b) Step 1: Calculation

Using a formula, calculate the margin of error.

Consider $n=16, \sigma =10$, and confidence level is $95\%$

The margin of error is, 

$$\begin{gathered} Margin\text{ of error }\left(E\right)=z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right) \\ =1.96\left(\frac{10}{\sqrt{16}}\right) \\ =1.96(2.5) \\ =4.9 \end{gathered}$$

Thus, the margin of error by using formula is $4.9$

Part (c) Step 1: Calculation

The margin of error diminishes as the sample size is raised while the confidence level remains constant. The sample size is increased from $16$ to $64$, resulting in a lower margin of error. As a result, the margin of error has increased from $4.9$ to $2.45$