Solution Found!
Assuming that n1 = n2, find the sample sizes needed to
Chapter 8, Problem 61E(choose chapter or problem)
Assuming that \(n_1 = n_2\), find the sample sizes needed to estimate (\(p_1 - p_2\)) for each of the following situations:
a. Margin of error = .01 with 99% confidence. Assume that \(p_1\ \approx\ .4\) and \(p_2\ \approx\ .7\).
b. A 90% confidence interval of width .05. Assume that there is no prior information available to obtain approximate values of \(p_1\) and \(p_2\).
c. Margin of error = .03 with 90% confidence. Assume that \(p_1\ \approx\ .2\ \text{and}\ p_2\ \approx\ .3\).
Questions & Answers
QUESTION:
Assuming that \(n_1 = n_2\), find the sample sizes needed to estimate (\(p_1 - p_2\)) for each of the following situations:
a. Margin of error = .01 with 99% confidence. Assume that \(p_1\ \approx\ .4\) and \(p_2\ \approx\ .7\).
b. A 90% confidence interval of width .05. Assume that there is no prior information available to obtain approximate values of \(p_1\) and \(p_2\).
c. Margin of error = .03 with 90% confidence. Assume that \(p_1\ \approx\ .2\ \text{and}\ p_2\ \approx\ .3\).
ANSWER:Step 1 of 3
a) Here we have the margin of error as ME = 0.01 with 99% con?dence and we
assumed that \(p_{1} \approx 0.4\) and \(p_{2} \approx 0.7\).
Also we assumed that \(n_{1}=n_{2}=n[\text { say })\).
From the given data, we have
\(\begin{array}{l}
1-\alpha=0.99 \\
\Rightarrow \alpha=0.01
\end{array}\)
Thus
\(\begin{array}{l}
\mathrm{z}_{\alpha / 2}=\mathrm{z}_{0.01 / 2} \\
=\mathrm{z}_{0.005} \\
=2.58
\end{array}\)
By definitions, the sample size needed to estimate \(\left(p_{1}-p_{2}\right)\) in this situation is
given as follows:
\(\begin{array}{l}
\begin{aligned}
n_{1} \text { or } n_{2} & =\frac{\left(z_{a, 2}\right)^{2}\left(p_{1} q_{1}+p_{2} q_{2}\right)}{(M E)^{2}} \\
& =\frac{(2.58)^{2}(0.4+0.7 \times 0.3)}{(0.01)^{2}} \\
& =\frac{6.6564 \times 0.45}{0.0001}
\end{aligned}\\
\quad \ \ \ \ \ \ \ \ \ \ =29.953 .8
\end{array}\)
\(n_{1} \text { or } n_{2} \approx 29,954\)