Answer to Question #248633 in Statistics and Probability for Blessing

Question

The Gauteng traffic department records show that 25% of all drivers wear seatbelts. In a random sample of 400 cars stopped at a roadblock in Gauteng, 152 of the drivers were wearing seatbelts. What is the probability that at most 18% of the drivers in Gauteng wear seatbelts? A 90% confidence interval for the proportion who wear seatbelts is given by

Answer 1

Given $n=400, p=0.25, q=1-p=1-0.25=0.75.$

$$np=400(0.25)=100\geq10, nq=400(0.75)=300\geq10$$

Then $\hat{p}$ has approximately a normal distribution with $\mu_{\hat{p}}=p=0.25$ and $\sigma_{\hat{p}}=\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{0.25(0.75)}{400}}=\dfrac{\sqrt{3}}{80}.$

Then

$$P(\hat{p}<0.18)=P(Z<\dfrac{0.18-0.25}{\sqrt{3}/80})$$

$$\approx P(Z<-3.23316)\approx0.000612$$

The sample proportion is computed as follows, based on the sample size $N = 400$ and the number of favorable cases $X = 152:$

$$\hat p = \displaystyle \frac{X}{N} = \displaystyle \frac{152}{400} = 0.38$$

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, ​ ​$$

$$\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(0.38-1.6449\sqrt{\dfrac{0.38(1-0.38)}{400}},$$

$$0.38+1.6449\sqrt{\dfrac{0.38(1-0.38)}{400}})$$

$$=(0.34, 0.42)$$

Therefore, based on the data provided, the $90\%$  confidence interval for the population proportion is $0.34 < p < 0.42,$ which indicates that we are $90\%$ confident that the true population proportion $p$ is contained by the interval $(0.34, 0.42).$