A trucking firm has a large inventory of spare parts that have been in storage for a long time. It knows that some proportion of these spare parts will have deteriorated to the point where they are no longer usable. In order to estimate this proportion, the firm tests 75 parts, and finds that 19 of them are not usable. Find the 95% confidence interval for the population proportion of parts that are not usable.


Step 1 Specify the confidence level $(1-\alpha)$

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

Step 2 Given information

Given that sample size $n =75$, observed value of $X$ is $X=19$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{19}{75}=0.253$.

Step 3 Specify the formula

$100(1-\alpha)$% 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.

Step 4 Determine the critical value

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

Z-critical value
Z-critical value

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

Step 5 Compute the margin of error

The margin of error for proportions is

$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.253*(1-0.253)}{75}}\\ & =0.098. \end{aligned} $$

Step 6 Determine the confidence interval

$95$% confidence interval estimate for population proportion is

$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.253 - 0.098 & \leq p \leq 0.253 + 0.098\\ 0.1549 & \leq p \leq 0.3518. \end{aligned} $$

Thus, $95$% confidence interval estimate for population proportion $p$ is $(0.1549,0.3518)$.

Further Reading