Question

One country reports that 46% of it's labor force is female. Another country wants to know if their labor force is the same so they get the records of 10,000 female workers to estimate the percentage of females working in the country. Without going through all of the records, and using a margin of error of +/- 5% with 95% confidence, how many records should they review? *

Answer 1

Answer:

We are given that a country reports that 46% of its labor force is female.

Also, the margin of error = 0.05 and confidence level = 0.95.

Therefore, the number of records that they should review is:

$n=\hat{p}(1-\hat{p}) \left( \frac{z_{\frac{0.05}{2}}}{E}\right)^{2}$

Where:

$z_{\frac{0.05}{2}} = 1.96$ is the critical value at 0.05 significance level.

$E=0.05$ is the margin of error.

Therefore, we have:

$n=0.46(1-0.46) \left( \frac{1.96}{0.05}\right)^{2}$

      $=381.70 \approx 382$

Hence, 382 records should be reviewed.