Question

A $95\%$ confidence interval for a population mean was reported to be $152$ to $160$. If standard deviation is $15$,
what sample size was used in this study?

Answer 1

Find the sample size used in this study:

Given that :

$95\%$ confidence interval for a population mean was reported to be 152 to 160.

So that the mean is :

$\begin{array}{rcl}Mean& =& \frac{152+160}{2}\\ & \Rightarrow & Mean=\frac{312}{2}\\ & \Rightarrow & Mean=156\end{array}$

Also we have standard deviation $\left(\sigma \right)$ is $15$.

Then by central limit theorem :

$\overline{X}=\mu +\frac{Z_{\alpha }\times \sigma }{\sqrt{n}}$, where $\overline{X}=160,\mu =156and\sigma =15$.

Also we have $95\%$ confidence level. So we choose $Z_{\alpha }=1.96$ and $n$ is the sample size.

Then :

$\begin{array}{rcl}160& =& 156+\frac{1.96\left(15\right)}{\sqrt{n}}\\ & \Rightarrow & 160-156=\frac{29.4}{\sqrt{n}}\\ \Rightarrow 4\sqrt{n}& =& 29.4\\ \Rightarrow \sqrt{n}& =& \frac{29.4}{4}\\ \Rightarrow n& =& {(7.35)}^2\end{array}$

So that : $n\approx 54$.

Hence, the required sample size is $\mathit{n}\mathbf{=}\mathbf{54}$.