Question

Suppose you want to estimate a population mean,$\mathit{\mu }$,and,$\overline{\mathbf{x}}\mathbf{=}\mathbf{422}$,$\mathit{s}\mathbf{=}\mathbf{14}$,$\mathit{N}\mathbf{=}\mathbf{375}$and$\mathit{n}\mathbf{=}\mathbf{40}$.Find an approximate 95% confidence interval for$\mathit{\mu }$.

Short answer

The approximate confidence interval for$\mu$ is$(\text{417.8164},\text{426.1836})$ .

Step-by-step solution

Given information

$\begin{array}{l}\overline{x}=422\\ s=14\\ N=375\\ n=40\end{array}$

Calculate the approximate confidence interval forμ

$\begin{array}{c}\frac{n}{N}=\frac{40}{375}\\ =\text{0.1066}\end{array}$

Since,$\frac{n}{N}=\text{0.1066>0.05}$

The finite population correction factor is used.

The estimated standard error is${\widehat{\sigma }}_{\overline{x}}$

$\begin{array}{c}{\widehat{\sigma }}_{\overline{x}}=\frac{s}{\sqrt{n}}\sqrt{\frac{(N-n)}{N}}\\ =\frac{14}{\sqrt{40}}\sqrt{\frac{(375-40)}{375}}\\ =\text{2.2136}\times \text{0.945}1\\ =\text{2.0920}\end{array}$

The approximate$95\%$ confidence interval is$\overline{x}\pm 2{\widehat{\sigma }}_{\overline{x}}$

$\begin{array}{c}\overline{x}\pm 2{\widehat{\sigma }}_{\overline{x}}=422\pm (2\times 2.092)\\ =422\pm \text{4.184}\\ \text{=}(\text{417.8164},\text{426.1836})\end{array}$

Therefore, the approximate $95\%$confidence interval for$\mu$is$(\text{417.8164},\text{426.1836})$