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.