Chapter PVII, Problem 5RE

To determine

To find:the number of such arbitrary checks would have to do to calculate the true percentage of people that misrepresent their backgrounds to within with 98 percent confidence.

Answer to Problem 5RE

406

Explanation of Solution

Given:

  $ME=0.05$ .

  $\widehat{p}=0.25$ .

  $\alpha =0.02$ .

Formula used:

  $ME=z^{\ast }\times SE\left(\widehat{p}\right)$

  $n={\left[z^{\ast }\frac{\sqrt{\widehat{p}\widehat{q}}}{E}\right]}^2$

Calculation:

Use the excel, at $\alpha /2=0.01$ two-tailed critical value is,

  $\begin{array}{c}{z}^{\ast }=\left[\text{NORMSINV}\left(0.01\right)\right]\\ =\pm 2.326\end{array}$

The margin of error is,

  $\begin{array}{c}ME=z^{\ast }\times SE\left(\widehat{p}\right)\\ =z^{\ast }\times \sqrt{\frac{\widehat{p}\widehat{q}}{n}}\end{array}$

Now, the formula for sample size will be,

  $n={\left[z^{\ast }\frac{\sqrt{\widehat{p}\widehat{q}}}{E}\right]}^2$

Finding the value of n.

Put the known values in the formula.

  $\begin{array}{c}n={\left[2.326\times \frac{\sqrt{\left(0.25\right)\left(0.75\right)}}{0.05}\right]}^2\\ =405.8928242\\ =406\end{array}$

Therefore, they will have to perform 406 random checks.