# A bill is being debated. Representative Williams wants to estimate within 5 percentage points and...

## Question:

A bill is being debated. Representative Williams wants to estimate within five percentage points and with {eq}95\% {/eq} confidence the difference in the proportion of her male and female constituents who favor the bill. What sample size should she obtain?

A) {eq}n_1 = n_2 = 769 {/eq}.

B) {eq}n_1 = n_2 = 542 {/eq}.

C) {eq}n_1 = n_2 = 385 {/eq}.

D) {eq}n_1 = n_2 = 39 {/eq}.

## Confidence interval

The 95% confidence interval interpreted as there is a 95% chance that the value of population mean or the value of population parameters lies within the confidence interval. The value of the margin of error is the value of confidence width divided by 2.

## Answer and Explanation:

Given information

• Margin of error: 0.05

Let assume {eq}{p_1} = {p_2} = 0.5{/eq}

The sample size is calculated as follow.

{eq}\begin{align*} M.O.E &= {Z_{\alpha /2}}*\sqrt {\dfrac{{{p_1}\left( {1 - {p_1}} \right)}}{{{n_1}}} + \dfrac{{{p_2}\left( {1 - {p_2}} \right)}}{{{n_2}}}} \\ 0.05 &= 1.96*\sqrt {\dfrac{{0.5\left( {1 - 0.5} \right)}}{{{n_1}}} + \dfrac{{0.5\left( {1 - 0.5} \right)}}{{{n_2}}}} \\ {0.05^2} &= {1.96^2}*\dfrac{{0.5}}{n}\,\,\,\,\left( {{\rm{By \ taking \ square \ both \ side}}} \right)\\ n &= 768.32 \simeq 769 \end{align*}{/eq}

Therefore, required sample size is 769

option (A) is correct.