# Solved: The U.S. Customs Service conducted a random check of Miami longshoremen and found that 36 of 50 had arrest records. Construct a 90 percent

#### ByDr. Raju Chaudhari

Feb 23, 2021

The U.S. Customs Service conducted a random check of Miami longshoremen and found that 36 of 50 had arrest records. Construct a 90 percent confidence interval for the true proportion. (See U.S. News & World Report 123, no. 22 [December 8, 1997].)

#### Solution

Let $X$ denote the number of longshoremen who had arrest records.

Given that sample size $n = 50$, observed $X = 36$.

Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{36}{50}=0.72$.

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

The estimate of the population proportion is $\hat{p} =\frac{X}{n} =\frac{36}{50}=0.72$.

$100(1-\alpha)$% confidence interval for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$. Z-critical value

Thus $Z_{\alpha/2} = Z_{0.05} = 1.645$.

The margin of error for proportions is
 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.645 \sqrt{\frac{0.72*(1-0.72)}{50}}\\ & =0.104. \end{aligned}

$90$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.72 - 0.104 & \leq p \leq 0.72 + 0.104\\ 0.6155 & \leq p \leq 0.8245. \end{aligned}
Thus, $90$% confidence interval estimate of the true proportion of proportion is $(0.6155,0.8245)$.