8.1.1.1.4 - Example: Seatbelt Usage

In the year 2001 Youth Risk Behavior survey done by the U.S. Centers for Disease Control, 747 out of 1168 female 12th graders said they always use a seatbelt when driving. Let’s construct a 95% confidence interval for the proportion of 12th grade females in the population who always use a seatbelt when driving.

\(\widehat{p}=\frac{747}{1168}=.0640\)

First we need to check our assumptions that both \(np \geq 10\) and \(n(1-p) \geq 10\)

\(np=1168 \times 0.640 = 747\) and \(n(1-p)=1168 \times (1-0.640)=421\) Both are greater than 10 so this assumption has been met and we can use the standard normal approximation with this data.

Now we can compute the standard error.

\(SE=\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}=\sqrt{\frac{0.640 (1-0.640)}{1168}}=0.014\)

The \(z^*\) multiplier for a 95% confidence interval is 1.960

Our 95% confidence for interval for \(\widehat{p}\) is \(0.640\pm 1.960(0.014)=0.640\pm0.028=[0.612, \;0.668]\)

We are 95% confident that between 61.2% and 66.8% of all 12th grade females say that they always use a seatbelt when driving.

What if we wanted a 99% confidence interval?

Let’s think about how our interval will change. The 99% confidence interval will be wider than the 95% confidence interval. In order to increase are level of confidence, we will need to expand the interval.

In terms of computing the 99% confidence interval, we will use the same point estimate \(\widehat{p}\) and the same standard error. The multiplier will change though. From the plot below, we see that the \(z^*\) multiplier for a 99% confidence interval is 2.576. The standard error is still 0.14, it has not changed because neither \(n\) nor \(\hat{p}\) have changed.

Standard normal distribution showing the z multipliers for a 99% confidence interval

\(99\%\;C.I.:\;0.640\pm 2.576 (0.014)=0.0640\pm 0.036=[0.604, \; 0.676]\)

We are 99% confidence that between 60.4% and 67.6% of all 12th grade females say that they always use a seatbelt when driving.