# For a random sample of 1,201 Americans, it was discovered that 1,139 of them lived in neighborhoods with acceptable levels of carbon monoxide....see A) obtain a point estimate for the proportion of Americans who live in neighborhoods with acceptable levels of carbon monoxide. B) construct a 99% confidence interval for the true proportion of Americans who live in neighborhoods with acceptable levels of carbon monoxide. C) you wish do conduct your own study to determine the proportion of Americans who live in neighborhoods with acceptable levels of carbon monoxide. What sample size would you need for the estimate to be within 1.5 percentage points with a 90% confidence if you used the estimate from part(a)? D) again doing your own study what sample size would be needed for the estimate to be within 1.5 percentage points with a 90%confidence if you do NOT have a prior estimate?

We are given that 1139 out of 1201 Americans live in neighborhoods with acceptable carbon monoxide levels.

Thus `n=1201,hat(p)=1139/1201~~.95` , so `hat(q)=1-hat(p)~~.05` .

(1) An acceptable point estimate of the true proprtion is `p=hat(p)=.95`

(2) To produce a confidence interval with 99% confidence level:

`alpha=.01, alpha/2=.005` The interval is given...

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We are given that 1139 out of 1201 Americans live in neighborhoods with acceptable carbon monoxide levels.

Thus `n=1201,hat(p)=1139/1201~~.95` , so `hat(q)=1-hat(p)~~.05` .

(1) An acceptable point estimate of the true proprtion is `p=hat(p)=.95`

(2) To produce a confidence interval with 99% confidence level:

`alpha=.01, alpha/2=.005` The interval is given by:

`hat(p)-z_(alpha/2)sqrt((hat(p)hat(q))/n)<=p<=hat(p)+z_(alpha/2)sqrt((hat(p)hat(q))/n)`

From a standard normal table we have `1-z_(.005)~~2.58` so the confidence interval is:

`.95-2.58sqrt((.95*.05)/1201)<=p<=.95+2.58sqrt((.95*.05)/1201)`

`.934<=p<=.966` with 99% confidence.

(3) We want the sample size with an error less than 1.5% at a 90% confidence level:

Use the formula `n=hat(p)hat(q)((z_(alpha/2))/E)^2` . We have `hat(p)=.95,hat(q)=.05,z_(alpha/2)=z_(.05)~~1.65` (From a standard normal table -- some texts/instructors will use 1.64 or 1.645 or the calculator value; perhaps even interpolating; you will have to check with them) Also we have `E=.015` . Thus:

`n=.95*.05*(1.65/.015)^2~~574.75`

So you need a sample of at least 575 homes.

(4) In (4) we used `hat(p)=.95` which we got from the study. Without the previous study, use `hat(p)=hat(q)=.5` as this yields the maximum required sample size. Thus we would have:

`n=.5*.5*(1.65/.015)^2~~3025` so you would need a sample of at least 3025 homes.

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