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.**