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Mariulka [41]
3 years ago
10

If a hurricane was headed your way, would you evacuate? The headline of a press release issued January 21, 2009 by the survey re

search company International Communications Research (icrsurvey) states, "Thirty- one Percent of People on High-Risk Coast Will Refuse Evacuation Order, Survey of Hurricane Preparedness Finds." This headline was based on a survey of 5046 adults who live within 20 miles of the coast in high hur- ricane 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 resi- dents in these states.
Use this information to estimate the proportion of coastal residents who would evacuate using a 98% confidence interval.
Write a few sentences interpreting the interval and the confidence level assosiated with the interval.
Mathematics
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

The 98% confidence interval for population proportion of people who refuse evacuation is {0.30, 0.33].

Step-by-step explanation:

The sample drawn is of size, <em>n</em> = 5046.

As the sample size is large, i.e. <em>n</em> > 30, according to the Central limit theorem the sampling distribution of sample proportion will be normally distributed with mean \hat p and standard deviation \sqrt{\frac{\hat p (1-\hat p)}{n} }.

The mean is: \hat p=0.31

The confidence level (CL) = 98%

The confidence interval for single proportion is:

CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]

Here z_{(\alpha /2)} = critical value and <em>α </em>= significance level.

Compute the value of <em>α</em> as follows:

\alpha =1-CL\\=1-0.98\\=0.02

For <em>α</em> = 0.02 the critical value can be computed from the <em>z</em> table.

Then the value of z_{(\alpha /2)} is ± 2.33.

The 98% confidence interval for population proportion is:

CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]\\=[0.31-2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} },\ 0.31+2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} } ]\\=[0.31-0.0152,\ 0.31+0.0152]\\=[0.2948,0.3252]\\\approx[0.30,\ 0.33]

Thus, the 98% confidence interval [0.30, 0.33] implies that there is a 0.98 probability that the population proportion of people who refuse evacuation is between 0.30 and 0.33.  

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