Question

You want to create a $95\%$ confidence interval with a margin of error of no more than $0.05$ for a population proportion.

The historical data indicate that population has remained constant at about $0.55$.

What is the minimum size random sample you need to construct this interval $?$

• A. $223$
• B. $324$
• C. $378$
• D. $381$

Answer 1

The correct option is D

$381$

Explanation for correct option:

Option $\left(D\right)$:

Find the minimum size random sample:

Use Margin of error formula to find the sample size:

$\mathbf{E}\mathbf{=}{\mathbf{z}}^{\mathbf{\ast }}\sqrt{\frac{{\displaystyle \hat{\mathbf{p}}}\mathbf{(}\mathbf{1}\mathbf{-}{\displaystyle \hat{\mathbf{p}}}}{\mathbf{n}}}$

$\mathrm{E}$ is the margin of error for population proportion, ${\mathrm{z}}^{\ast }$ is the critical value, $\hat{\mathrm{p}}$ is the sample proportion and $\mathrm{n}$ is the sample size.

(For $95\%$ confidence level,${\mathrm{z}}^{\ast }$ is $1.96$)

It is given that:

$\mathrm{E}$ $=$$0.05$, ${\mathrm{z}}^{\ast }$$=$$1.96$, $\hat{\mathrm{p}}$$=$$0.55$.

Substitute the above values in the margin of error formula:

$$\begin{gathered} 0.05=1.96\sqrt{\frac{0.55(1-0.55)}{\mathrm{n}}} \\ \Rightarrow \frac{0.05}{1.96}=\sqrt{\frac{0.55\times 0.45}{\mathrm{n}}} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & \frac{25}{38416}=\frac{0.2475}{n}\\ \therefore \mathrm{n}& =& 380.3184\end{array}$

Hence, minimum size random sample is $\mathbf{\approx }\mathbf{381}\mathbf{.}$

Explanation for incorrect option:

Since the minimum size sample is $381$, not equivalent to option $\left(\mathrm{A}\right)$, option $\left(\mathrm{A}\right)$ is incorrect.

Since the minimum size sample is $381$, not equivalent to option $\left(\text{B}\right)$, option $\left(\text{B}\right)$ is incorrect.

Since the minimum size sample is $381$, not equivalent to option $\left(\mathrm{C}\right)$, option $\left(\mathrm{C}\right)$ is incorrect.

Hence, the only correct option is $\left(D\right)$.