Question

​It's believed that as many as 22​% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group.

Suppose we want to cut the margin of error to

6%.What is the necessary sample​ size?

Answer 1

It is given that the proportion of adults who never graduated from high school \(\hat{p}=0.22\).

Let the level of significance be \(\alpha=0.1\).

The margin of error is \(M E=0.06\)

The margin of error is given as follows:

\(M \cdot E=z_{\frac{a}{2}} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

The critical value of \(z_{\frac{\alpha}{2}}=z_{\frac{01}{2}}=z_{005}=1.645\).

The required sample size is obtained as follows:

$$ \begin{aligned} &M . E=z_{\frac{a}{2}} \times \sqrt{\frac{p(1-\bar{p})}{n}} \\ &\Rightarrow 0.06=1.645 \times \sqrt{\frac{0.22(1-0.22)}{n}} \\ &\Rightarrow \frac{0.06}{1.645}=\sqrt{\frac{0.22(1-0.22)}{n}} \\ &\Rightarrow\left(\frac{0.06}{1.645}\right)^{2}=\frac{0.22(1-0.22)}{n} \\ &\Rightarrow n=0.22(1-0.22) \times\left(\frac{1.645}{0.06}\right)^{2} \\ &\Rightarrow n-128.98 \sim 129 \end{aligned} $$

Therefore, the required sample size is \(n=129\).

Similarly, if the level of significance is \(\alpha=0.05\).

The margin of error is \(M . E=0.06\)

The margin of error is given as follows:

\(M \cdot E=z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) The critical value of \(z_{\frac{\alpha}{2}}=z_{\frac{0.05}{2}}=z_{0.025}=1.96 .\)

The required sample size is obtained as follows:

$$ \begin{aligned} &M \cdot E=z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ &\Rightarrow 0.06=1.96 \times \sqrt{\frac{0.22(1-0.22)}{n}} \\ &\Rightarrow \frac{0.06}{1.96}=\sqrt{\frac{0.22(1-0.22)}{n}} \\ &\Rightarrow\left(\frac{0.06}{1.96}\right)^{2}=\frac{0.22(1-0.22)}{n} \\ &\Rightarrow n=0.22(1-0.22) \times\left(\frac{1.96}{0.06}\right)^{2} \\ &\Rightarrow n=183.11 \sim 184 \end{aligned} $$

Therefore, the required sample size is \(n=184\).