# Solved:The proportion of public accountants who have changed companies within the last three years is to be estimated within 3%. The 90% level of confidence

#### ByDr. Raju Chaudhari

Aug 23, 2020

The proportion of public accountants who have changed companies within the last three years is to be estimated within 3%. The 90% level of confidence is to be used. A study conducted several years ago revealed that the percent of public accountants changing companies within three years was 21?(Use Z Distribution Table.) (Round the z-values to 2 decimal places. Round up your answers to the next whole number.)

a. To update this study, the files of how many public accountants should be studied?
b. How many public accountants should be contacted if no previous estimates of the population proportion are available?

#### Solution

The formula to estimate the sample size required to estimate the proportion is

\begin{aligned} n &=p*(1-p)\bigg(\frac{z}{E}\bigg)^2 \end{aligned}
where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

a. Given that $p = 0.21$ and the margin of error $E =0.03$. The confidence coefficient is $0.9$. Assume that the proportion is $p =0.21$.

The critical value of $Z$ is $Z_{\alpha/2} = 1.64$.

The minimum sample size required to estimate the proportion is
\begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.21(1-0.21)\bigg(\frac{1.64}{0.03}\bigg)^2\\ &=495.7829\\ &\approx 496. \end{aligned}

Thus, the sample of size $n=496$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.03$.

b. Assume that $p$ is unknown. Let $p =0.5$.

Given that margin of error $E =0.03$. The confidence coefficient is $0.9$.

The critical value of $Z$ is $Z_{\alpha/2} = 1.64$.

The minimum sample size required to estimate the proportion is
\begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{1.64}{0.03}\bigg)^2\\ &=747.1111\\ &\approx 748. \end{aligned}

Thus, the sample of size $n=748$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.03$.