Cell phone use by drivers. Studies have shown that driverswho use cell phones while operating a motor passenger vehicleincrease their risk of an accident. Nevertheless, driverscontinue to make cell phone calls whiledriving. A June2011 Harris Pollof 2,163 adults found that 60% (1,298adults) use cell phones while driving.

1. Give a point estimate of p,the true driver cell phone use rate (i.e., the proportion of all drivers who are usinga cell phone while operating a motor passengervehicle).
2. Find a 95% confidence interval for p.
3. Give a practical interpretation of the interval, part b.
4. Determine the margin of error in the interval if thenumber of adults in the survey is doubled.

Drivers, who use cell phones while driving, increase their risk to occur an accident. A survey of 2163 adults found that 1298 adults, that is approximately 60% of adults use a cell phone while driving. That is for a repeated confidence interval with same sample size about 95% will contain true proportion of adults use cell phones while driving.

a.

Let’s consider the sample size$\mathrm{n}=2163$and the number of adults who uses cell phones while driving, $\mathrm{x}=1298$.

So, the point estimate of pwhich is the true proportion of all drivers who are using a cell phone operating a motor passenger vehicle is

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{\mathrm{x}}{\mathrm{n}}\\ & =& \frac{1298}{2163}\\ & =& 0.60\end{array}$

b.

The 95% confidence interval for the binomial proportion is defined as,

$\left[\hat{p}-Z_{\alpha }\sqrt{\frac{\hat{p}\hat{q}}{n}}\le z\le \hat{p}+Z_{\alpha }\sqrt{\frac{\hat{p}\hat{q}}{n}}\right]$

Now, the confidence level is data-custom-editor="chemistry" $\mathrm{CL}=0.95$, then,

data-custom-editor="chemistry" $\begin{array}{rcl}\mathrm{\alpha }& =& 1-\mathrm{CL}\\ & =& 1-0.95\\ & =& 0.05\end{array}$

And $\left(\frac{\alpha }{2}\right)=0.025$.

Therefore, by the standard normal probability table, $\begin{array}{rcl}{\mathrm{Z}}_{\left(\frac{\mathrm{\alpha }}{2}\right)}& =& {\mathrm{Z}}_{0.025}\\ & =& 1.96\end{array}$.

Thus, the interval is,

role="math" localid="1658309565277" $\begin{array}{rcl}[\hat{\mathrm{p}}-{\mathrm{Z}}_{\mathrm{\alpha }}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}}& \le & \mathrm{z}\le \hat{\mathrm{p}}+{\mathrm{Z}}_{\mathrm{\alpha }}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}}]=\left[0.60-1.96\sqrt{\frac{0.60\times 0.40}{2163}}\le \mathrm{z}\le 0.60+1.96\sqrt{\frac{0.60\times 0.40}{2163}}\right]\\ & =& \left[0.60-0.020\le \mathrm{z}\le 0.60+0.020\right]\\ & =& \left[0.58\le \mathrm{z}\le 0.62\right]\end{array}$

Therefore, the 95% confidence that approximately between 58% and 62%.

c.

From the confidence interval, one is 95% confident that the true proportion of all drivers who are using a cell phone operating a motor passenger vehicle will contain in the calculated interval.

d.

By the definition, the margin of error is half of the width of the entire confidence interval.

So, the margin of error is $\frac{\left(0.62-0.58\right)}{2}=0.02$.

Now, if the number of adults in the sample is doubled then the margin of error will decrease by the factor $\sqrt{\mathbf{2}}$, as the margin of error is inversely proportional to the square root of the sample size.

Therefore, the margin of error becomes $\frac{0.02}{\sqrt{2}}=0.014$.