The highway department wants to estimate the proportion of vehicles on Interstate 25 between the hours of midnight and 5:00 A.M. that are 18-wheel tractor trailers. The estimate will be used to determine highway repair and construction considerations and in highway patrol planning. Suppose researchers for the highway department counted vehicles at different locations on the interstate for several nights during this time period. Of the 3,471 vehicles counted, 945 were 18-wheelers.
a. Determine the point estimate for the proportion of vehicles traveling Interstate 25 during this time period that are 18-wheelers.
b. Construct a 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers.
Solution
Given that sample size $n = 3471$, observed $X = 945$.
Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{945}{3471}=0.272$.
a. The point estimate for the proportion of vehicles traveling interstate 25 during this time period that are 18 wheelers is
$\hat{p}=0.272$.
b. The procedure for $0.99$ % confidence interval for the proportion of vehicles on interstate 25 during this time period that are 18 wheelers is as follows:
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.
Step 2 Given information
Given that sample size $n =3471$, observed number of successes $X=945$.
The estimate of the proportion of success is $\hat{p} =\frac{X}{n} =\frac{945}{3471}=0.272$.
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 $Z_{\alpha/2}$.
Thus $Z_{\alpha/2} = Z_{0.005} = 2.576$.
Step 5 Compute the margin of error
The margin of error for proportions is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 2.576 \sqrt{\frac{0.272*(1-0.272)}{3471}}\\ & =0.019. \end{aligned} $$
Step 6 Determine the confidence interval
$99$% confidence interval estimate for population proportion is
$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.272 - 0.019 & \leq p \leq 0.272 + 0.019\\ 0.2528 & \leq p \leq 0.2917. \end{aligned} $$
Thus, $99$% confidence interval estimate for population proportion $p$ is $(0.2528,0.2917)$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
