Chapter 9, Problem 9.66SE

To determine

The common sample size $n$ for a test

Answer to Problem 9.66SE

The common sample size $n$ for a test is $310$

Explanation of Solution

Given:

Maximum value of $p\left(1-p\right)=0.25$

Significance level: $\alpha =0.01$

Calculation:

Consider:

  $H_0:$ The proportion of adolescent African American $\left(p_1\right)$ who state that their SES is ‘about the same’ is same as the proportion of adolescent Hispanic American $\left(p_2\right)$ with the same opinion.

  $p_1-p_2=0$

  $H_0:$ The proportions of adolescent African American $\left(p_1\right)$ who state that their SES is ‘about the same’ exceed that for adolescent Hispanic American $\left(p_2\right)$ .

  $p_1-p_2>0$

The test statistics under null hypothesis is calculated as:

  $z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{pq\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Need to consider the common sample size: $n_1=n_2=n$

Then above eq. rewrite as:

  $\begin{array}{l}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{pq\left(\frac{1}{n}+\frac{1}{n}\right)}}\\ z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{pq\left(\frac{2}{n}\right)}}\end{array}$

At the significance level of the critical value of $z_a=1.645$

Thus, the critical region is $z>1.645$

  $\begin{array}{l}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{pq\left(\frac{2}{n}\right)}}\\ \frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-0}{\sqrt{pq\left(\frac{2}{n}\right)}}>1.645\end{array}$

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)>1.645\times \sqrt{pq\left(\frac{2}{n}\right)}$  .....(1)

It is given that the maximum value of $p\left(1-p\right)=0.25$

  $\begin{array}{l}p\left(1-p\right)=0.25\\ pq=0.25\end{array}$

Now, the $\left({\widehat{p}}_1-{\widehat{p}}_2\right)$ can be rewritten by using eq. (1) as:

  $\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)>1.645\times \sqrt{0.25\left(\frac{2}{n}\right)}\\ \left({\widehat{p}}_1-{\widehat{p}}_2\right)>1.645\times \sqrt{\left(\frac{0.5}{n}\right)}\\ \left({\widehat{p}}_1-{\widehat{p}}_2\right)>\frac{1.1632}{\sqrt{n}}\end{array}$

It given that, in fact the $p_1$ exceed $p_2$ by $0.1$

That is: $p_1-p_2=0.1$

The power of the test $\left(1-\beta \right)$ can be written as:

  $\begin{array}{l}\beta \le 0.20\\ \left(1-\beta \right)\ge 0.80\end{array}$

When the null hypothesis is true, the power of the test is the probability of rejecting the null hypothesis $H_0$ .

  $\begin{array}{l}P\left(\left({\widehat{p}}_1-{\widehat{p}}_2\right)>\frac{1.1632}{\sqrt{n}}\right)\ge 0.80\\ P\left(\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{pq\left(\frac{2}{n}\right)}}>\frac{\frac{1.1632}{\sqrt{n}}-0.1}{\sqrt{pq\left(\frac{2}{n}\right)}}>\frac{\frac{1.1632-0.1\sqrt{n}}{\sqrt{n}}}{\sqrt{0.25\left(\frac{2}{n}\right)}}\right)\ge 0.80\\ P\left(Z>\frac{\frac{1.1632-0.1\sqrt{n}}{\sqrt{n}}}{\sqrt{0.25\left(\frac{2}{n}\right)}}\right)\ge 0.80\end{array}$

From standard normal tables, the value corresponding to $0.80$ cumulative probability is $-0.84162$

Then, $\begin{array}{l}\frac{\frac{1.1632-0.1\sqrt{n}}{\sqrt{n}}}{\sqrt{0.25\left(\frac{2}{n}\right)}}=-0.84162\\ \sqrt{n}=\frac{1.1632+\left(0.84162\times \sqrt{0.5}\right)}{0.1}\\ n=309.1672\\ n\approx 310\end{array}$

The common sample size is estimated as $n\approx 310$

Conclusion:

Therefore, the common sample size is estimated as $n\approx 310$