Do drivers reduce excessive speed when they encounter police radar? Researchers studied the behavior of a sample of drivers on a rural interstate highway in Maryland where the speed limit was $55$miles per hour. They measured the speed with an electronic device hidden in the pavement and, to eliminate large trucks, considered only vehicles less than $20$feet long. During some time periods (determined at random), police radar was set up at the measurement location. Here are some of the data.

$\begin{array}{lcc}& \text{Number of vehicles}& \text{Number over}65\mathrm{mph}\\ \text{No radar}& 12,931& 5,690\\ \text{Radar}& 3,285& 1,051\\ & & \end{array}$

(a) The researchers chose a rural highway so that cars would be separated rather than in clusters because some cars might slow when they see other cars slowing. Explain why this is important.

(b) Does the proportion of speeding drivers differ significantly when the radar is being used and when it isn’t? Use information from the Minitab computer output below to support your answer.

Given

$x_1=5690$

$n_1=12931$

$x_2=1051$

$n_2=3285$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

This is important for the data of each car to be independent of the other cars.

If a car slows down, because other cars slow down, then this car is not independent of the other cars that slowed down too.

Given

$x_1=5690$

$n_1=12931$

$x_2=1051$

$n_2=3285$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{5690}{12931}\approx 0.440$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{1051}{3285}\approx 0.320$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{5690+1051}{12931+3285}=\frac{6741}{16216}\approx 0.416$

Determine the value of the test statistic:

localid="1650451405878" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.440-0.320}{\sqrt{0.416(1-0.416)}\sqrt{\frac{1}{12931}+\frac{1}{3285}}} \\ \approx 12.46 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650451422257" $$\begin{gathered} P=P(Z<-12.46\text{or}Z>12.46) \\ =2\times P(Z<-12.46) \\ =2\times 0.0001 \\ =0.0002 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P<0.05\Rightarrow \text{Reject}{H}_0$