In each case, find the approximate sample size required to

Step 1 of 2

Now we have to find the approximate sample size required to a 95% confidence interval for p that has of margin of error of ME = 0.08.

Our goal is :

a). We assume that p is near 0.2.

b). We assume that you have no prior knowledge about p, but you wish to be certain that your sample is large enough to achieve the specified accuracy for the same.

a). We assume that p is near 0.2.

Now we have to compute sample size n.

The formula for n is

\(\mathrm{n}=\frac{\left(z_{\alpha / 2}\right)^{2} p q}{M E^{2}}\)

We know that level of significance \(\alpha=0.05\)

Then,

\(\begin{array}{l}
\frac{\alpha}{2}=\frac{0.05}{2} \\
\frac{\alpha}{2}=0.025
\end{array}\)

Using the normal distribution tables,

\(z_{0.025}=1.96\)

We know that p = 0.2 and

q = 1 - p

q = 1 - 0.2

q = 0.8

Therefore, q = 0.8.

Then given ME = 0.08.

\(\begin{aligned}
\mathrm{n} & =\frac{(1.96)^{2}(0.2)(0.8)}{(0.08)^{2}} \\
\mathrm{n} & =\frac{(3.8416)(0.16)}{0.0064} \\
\mathrm{n} & =\frac{0.6146}{0.0064} \\
\mathrm{n} & =96.04 \\
\mathrm{n} & \approx 97
\end{aligned}\)

Hence we would need to take a sample of size is 97.