Question

Seat belt use: The proportion of drivers who use seat belts depends on things like age (young people are more likely to go unbelted) and gender (women are more likely to use belts). It also depends on local law. In New York City, police can stop a driver who is not belted. In Boston at the time of the study, police could cite a driver for not wearing a seat belt only if the driver had been stopped for some other violation. Here are data from observing random samples of female Hispanic drivers in these two cities:

(a) Is this an experiment or an observational study? Why?

(b) Construct and interpret a $95\%$confidence interval for the difference in the proportions of female Hispanic drivers in the two cities who wear seat belts.

(c) Based on the laws in the two cities, we would expect a smaller proportion of drivers to wear seat belts in Boston than in New York. Does the confidence interval in part (b) give good evidence that this is true for female Hispanic drivers? Justify your answer.

Short answer

a). Observational study.

b). $95\%$are sure that the percentage of Hispanic female drivers in New York is between $0.149$and $0.353$higher than the proportion of Hispanic female drivers in Boston.

c). Yes, the confidence interval in part (b) gives good evidence that this is true for female Hispanic drivers.

Step-by-step solution

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

Experiment Study

In order to study their reactions, an experiment intentionally forces certain treatments upon individuals.

Observational Study

Without disrupting the scene, they are studying, an observational study attempts to collect details.

Therefore, an Observational study

Part (b) Step 1: Given Information

Part (b) Step 2: Explanation

For the confidence interval

$\left({\hat{p}}_1-{\hat{p}}_2\right)\pm z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

The sample proportions are

localid="1650369797897" ${\hat{p}}_1=\frac{x_1}{n_1}$

$=\frac{183}{220}$

$=0.832$

localid="1650369848173" ${\hat{p}}_2=\frac{x_2}{n_2}$

$=\frac{68}{117}$

$=0.581$

For the confidence level $1-\alpha =0.95$,

determine $z_{\alpha /2}=z_{0.025}$the z-score is

$z_{\alpha /2}=1.96$

Part (b) Step 3: Explanation

The confidence interval are

$\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

localid="1650391800958" $=(0.832-0.581)-1.96\sqrt{\frac{0.832(1-0.832)}{220}+\frac{0.581(1-0.581)}{117}}$

$=0.149$

$\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$$=(0.832-0.581)+1.96\sqrt{\frac{0.832(1-0.832)}{220}+\frac{0.581(1-0.581)}{117}}$

$=0.353$

Part (c) Step 1: Given Information

Part (c) Step 2: Explanation

According to the information from part (b), the$95\%$ are sure that the percentage of Hispanic female drivers in New York is between $0.149and0.353$higher than the proportion of Hispanic female drivers in Boston.

In the absence of a confidence interval of $0$, the interval suggests that a lower proportion of drivers in Boston are wearing seat belts than in New York and that there is ample evidence supporting the claim.