Question

In a survey of 802 U.S. adult drivers, 265 state that traffic is getting worse in their community. Construct a 95% confidence interval for the proportion of residents who think traffic is getting worse in their community.

Answer 1

Answer:

The 95% confidence interval for the proportion of residents who think traffic is getting worse in their community is (0.2978, 0.3630).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

In a survey of 802 U.S. adult drivers, 265 state that traffic is getting worse in their community.

This means that $n = 802, \pi = \frac{265}{802} = 0.3304$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3304 - 1.96\sqrt{\frac{0.3304*0.6669}{802}} = 0.2978$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3304 + 1.96\sqrt{\frac{0.3304*0.6669}{802}} = 0.3630$

The 95% confidence interval for the proportion of residents who think traffic is getting worse in their community is (0.2978, 0.3630).