Question

Conducting a political poll. A pollster wants to estimate the difference between the proportions of men and women who favor a particular national candidate using a 90% confidence interval of width .04. Suppose the pollster has no prior information about the proportions. If equal numbers of men and women are to be polled, how large should the sample sizes be?

Short answer

The required sample size is 3383.

Step-by-step solution

Given Information

With a 95% confidence interval, a pollster intends to calculate the actual difference between the proportion of men and women.

From the given information

$$\begin{gathered} 2SE=0.04 \\ SE=0.02 \end{gathered}$$

Z-value

For $\alpha =0.1$

The z-value is given by

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{0.05} \\ =1.645 \end{gathered}$$

Compute the sample

For, $z=1.645,p_1=p_2=0.5 andSE=0.02$

The sample is calculated as

$$\begin{gathered} n_1=n_2 \\ =\frac{{\left(z_{0.05}\right)}^2\times \left[p_1\left(1-p_1\right)+p_2\left(1-p_2\right)\right]}{{\left(SE\right)}^2} \\ =\frac{{1.645}^2\times \left[0.25+0.25\right]}{{0.02}^2} \\ =\frac{2.706025\times .5}{0.0004} \\ =\frac{1.3530125}{0.0004} \\ =3382.53 \\ \approx 3383 \end{gathered}$$

Therefore, the required sample size is 3383.