Question

what is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 in the following scenarios? (round your answers up the nearest whole number.) a button hyperlink to the salt program that reads: use salt. (a) a preliminary estimate for p is 0.33 (b) there is no preliminary estimate for p g

Answer 1

The minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 $=n=97$

As per the question Margin of Error=0.1

$E=0.1\\ Z_{\frac{\alpha}{2} }=1.96 \\\\(a) \dot{\mathbf{p}}=0.27$

We have to determine the sample size :

$\mathrm{n}=\left(\frac{Z_{\frac{a}{2}}}{\mathrm{E}}\right)^2 \times \hat{\mathrm{p}} \times(1-\hat{\mathrm{p}})=\left(\frac{1.96}{0.1}\right)^2 \times 0.27 \times 0.73=76$

$n=76$

At $95 \%$ confidence level the $z$ is,

$\begin{aligned}& \alpha=1-95 \%=1-0.95=0.05 \\& \frac{\alpha}{2} =\frac{0.05}{2} =0.025 \\& Z_{\frac{\alpha}{2} }=Z_{0.025}=1.96\end{aligned}$

$\hat{\mathrm{p}}=0.5$

Sample size is,

$\mathrm{n}=\left(\frac{1.96}{0.1}\right)^2 \times 0.5 \times 0.5=97$

Therefore the sample size is $n=97$

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