# Solved: A random sample of 539 households from a certain mid-western city was selected, and it was determined that 133 of these households owned at least one rearm

#### ByDr. Raju Chaudhari

Feb 23, 2021

A random sample of 539 households from a certain mid-western city was selected, and it was determined that 133 of these households owned at least one rearm ( The Social Determinants of Gun Ownership: Self-Protection in an Urban Environment, Criminology, 1997: 629-640). Using a 95% confidence level, calculate a lower confidence bound for the proportion of all households in this city that own at least one rearm.

#### Solution

Given that sample size $n = 539$, observed $X = 133$.

Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{133}{539}=0.247$.

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

Given that sample size $n =539$, observed number of successes $X=133$.

The estimate of the proportion of success is $\hat{p} =\frac{X}{n} =\frac{133}{539}=0.247$.

$100(1-\alpha)$% 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 $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

The margin of error for proportions is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.247*(1-0.247)}{539}}\\ & =0.036. \end{aligned}

$95$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.247 - 0.036 & \leq p \leq 0.247 + 0.036\\ 0.2104 & \leq p \leq 0.2831. \end{aligned}

Thus, a lower confidence bound for the proportion of all households in this city that own at least one rearm is $0.2104$.