Question

Suppose you want to estimate a population proportion,,$\mathit{p}$and,$\widehat{\mathbf{p}}\mathbf{=}\mathbf{.}\mathbf{42}$,$\mathit{N}\mathbf{=}\mathbf{6000}$and$\mathit{n}\mathbf{=}\mathbf{1600}$.Find an approximate 95% confidence interval for$\mathit{p}$.

Short answer

The approximate confidence interval for $p$is$(\text{0.399},\text{0.441})$

Step-by-step solution

Given information

$\begin{array}{l}\widehat{p}=.42\\ N=6000\\ n=1600\end{array}$

Since,$\frac{n}{N}=\text{0.2666>0.05}$

The finite population correction factor is used.

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

$\begin{array}{c}{\widehat{\sigma }}_{\widehat{p}}=\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\sqrt{\frac{N-n}{N}}\\ =\sqrt{\frac{0.42(1-0.42)}{1600}}\sqrt{\frac{6000-1600}{6000}}\\ =\sqrt{\frac{\text{0.2436}}{1600}}\sqrt{\frac{\text{4,400}}{6000}}\\ =\text{0.0123}\times \text{0.8563}\\ =\text{0.0105}\end{array}$

The approximate confidence interval is$\widehat{p}\pm 2{\widehat{\sigma }}_{\widehat{p}}$

$\begin{array}{c}\widehat{p}\pm 2{\widehat{\sigma }}_{\widehat{p}}=0.42\pm (2\times 0.0105)\\ =0.42\pm \text{0.021}\\ =(\text{0.399},\text{0.441})\end{array}$

Therefore, the approximate $95\%$confidence interval for$p$is$(\text{0.399},\text{0.441})$