Question

If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People on High-Risk Coast Will Refuse Evacuation Order, Survey of Hurricane Preparedness Finds." This headline was based on a survey of 6138 adults who live within 20 miles of the coast in high hurricane risk counties of eight southern states. In selecting the sample, care was taken to ensure that the sample would be representative of the population of coastal residents in these states.
(a) Use this information to estimate the proportion of coastal residents who would evacuate using a 98% confidence interval. (Round your answers to three decimal places.)

Answer 1

Answer:

The 98% confidence interval would be given by (0.326;0.354)

Step-by-step explanation:

1) Notation and definitions

$X=63$ number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$n=6138$ random sample taken

$\hat p=0.34$ estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$p$ true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by $\alpha=1-0.98=0.02$ and $\alpha/2 =0.01$. And the critical value would be given by:

$z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.34 - 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.326$

$0.34 + 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.354$

The 98% confidence interval would be given by (0.326;0.354)