Crash risk of using cell phones while driving. Studies have shown that drivers who use cell phones while operating a motor passenger vehicle increase their risk of an accident. To quantify this risk, the New England Journal of Medicine (January 2, 2014) reported on the risk of a crash (or near crash) for both novice and expert drivers when using a cell phone. In a sample of 371 cases of novices using a cell phone while driving, 24 resulted in a crash (or near crash). In a sample of 1,467 cases of experts using a cell phone while driving, 67 resulted in a crash (or near crash).

a. Give a point estimate of p, the true crash risk (probability) for novice drivers who use a cell phone while driving.

b. Find a 95% confidence interval for p.

c. Give a practical interpretation of the interval, part b.

d. Repeat parts a–c for expert drivers.

In a sample of 371 cases of novices using a cell phone while driving, 24 resulted in a crash (or near crash). In a sample of 1,467 cases of experts using a cell phone while driving, 67 resulted in a crash (or near crash). We need to n compute the following:

a. a point estimate of p, the true cash risk for novice drivers who use a cell phone while driving.

b. a 95% confidence interval for p.

c. a practical interpretation of the interval, part b

d. repetition of parts a-cfor expertdrivers.

Sample size = n= 371

Number of successes x =24

Sample proportion

$\begin{array}{rcl}\hat{p}& =& \frac{x}{n}\\ & =& \frac{24}{371}\\ & =& 0.0647\\ & & \end{array}$

A point estimate pthe true crash risk (probability) for novice drivers who use a cell phone while driving.is 0.0647.

Here the confidence coefficient is 95%, hence $\alpha =5\%$.Now from the standard normal distribution table $z_{\frac{\alpha }{2}}=1.96$,

The margin of error

$\begin{array}{rcl}E& =& z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\\ & =& 1.96\times \sqrt{\frac{0.0647\times \left(1-0.0647\right)}{371}}\\ & =& 0.0250\\ & & \end{array}$

Hence 95% lower limit:$\hat{p}-E=0.0647-0.0250=0.0397$

95% upper limit:$\hat{p}+E=0.0647+0.0250=0.0897$

Hence the 95% confidence interval is$\left(0.0397,0.0897\right)$

We are 95% confident that the true crash risk (probability) for novice drivers who use a cell phone while driving.is between$\left(0.0397,0.0897\right)$

Sample size = n= 1467

Number of successes x = 67

Sample proportion

$\begin{array}{rcl}\hat{p}& =& \frac{x}{n}\\ & =& \frac{67}{1467}\\ & =& 0.0457\\ & & \end{array}$

A point estimate pthe true crash risk (probability) for expert drivers who use a cell phone while driving.is 0.0457

Here the confidence coefficient is 95%, hence $\alpha =5\%$.Now from the standard normal distribution table,$z_{\frac{\alpha }{2}}=1.96$

The margin of error

$\begin{array}{rcl}E& =& z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\\ & =& 1.96\times \sqrt{\frac{0.0457\times \left(1-0.0457\right)}{1467}}\\ & =& 0.0107\\ & & \end{array}$

Hence 95% lower limit:$\hat{p}-E=0.0457-0.0107=0.0350$

95% upper limit:$\hat{p}+E=0.0457+0.0107=0.0564$

We are 95% confident that the true crash risk (probability) for expert drivers who use a cell phone while driving.is between$\left(0.0350,0.0564\right)$