Question

Suppose that you plan to apply the one-mean $z-$interval procedure to obtain a $90\%$confidence interval for a population mean, $\mu$You know that $\sigma =12$ and that you are going to use a sample of size $9$

a. What will be your margin of error?

b. What else do you need to know in order to obtain the confidence interval?

Short answer

Part (a)$6.5794$

Part (b)$(\overline{x}-E,\overline{x}+E)$

Step-by-step solution

Part (a) Step 1: Given information

Population s.d.$\sigma =12$

Sample size $n=9$

Part (a) Step 2: Concept

The formula used:$E=Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

Population s.d.$\sigma =12$

Sample size $n=9$

Confidence level $=90\%=100\times 0.90\%$

$$\begin{gathered} \therefore 1-\alpha =0.90 \\ \Rightarrow \alpha =1-0.90 \\ \Rightarrow \alpha =0.10 \\ \Rightarrow \alpha /2=0.05 \end{gathered}$$

The margin of error, $E$, for a $100(1-\alpha )\%$ confidence interval is given by

$E=Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}$

As a result, for a$90\%$ confidence interval, the margin of error is

$$\begin{gathered} E=Z_{0.05}\frac{\sigma }{\sqrt{n}} \\ =1.645\times \frac{12}{\sqrt{9}}\left[\because Z_{0.05}=1.645\right] \\ =6.5794 \end{gathered}$$

Part (b) Step 1: Calculation

Because the confidence interval is given by $(\overline{x}-E,\overline{x}+E)$, we need to know the sample mean $\overline{x}$ to get the confidence interval.