Question

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confidence level: 99%; p and q unknown

Answer 1

Answer:

The minimum sample size is $n = 2123$

Step-by-step explanation:

From the question we are told that

    The margin of error is  $E = 0.028$

   

Given that the confidence level is  99% then the level of significance is evaluated as

        $\alpha = 100 - 99$

        $\alpha = 1 \%$

        $\alpha =0.01$

Next we obtain the critical value of  $\frac{ \alpha }{2}$ from the normal distribution table

   The  value is  $Z_{\frac{ \alpha }{2} } = 2.58$

Now let  assume that the sample proportion is  $\r p = 0.5$

  hence  $\r q = 1 - \r p$

=>            $\r q = 0.50$

   Generally the sample size is mathematically represented as

                $n =[ \frac{Z_{\frac{ \alpha }{2} }}{ E} ]^2 * \r p * \r q$

                $n =[ \frac{2.58}{ 0.028} ]^2 * 0.5 * 0.5$

              $n = 2123$