In 2010, the Census Bureau estimated the proportion of all Americans who own their homes to be 0.669. An urban economist wants to estimate the current (2016) proportion of all Americans who own their homes i. Making use of the Census Bureau estimate as a starting point, what sample size does she need to be 90% confident AND have the estimate within 0.02 of the true proportion? SHOW YOUR WORK. If she ignores the Census Bureau estimate, what sample size does she need to be 90% confident AND have the estimate within 0.02 of the true proportion? SHOW YOUR WORK. ii. B. A researcher wants to estimate the mean hours per week that Americans spend watching television. He has read an article about a survey conducted in 2014 that had s = 7.5 hours per week spent watching television. → What sample size does he need to estimate the population mean within 2 hours per week with 95% confidence? SHOW YOUR WORK.

Answer:i) Sample size making use of the Census Bureau: 1,499 American adults.Sample size without making use of the Census Bureau: 1,692 American adultsii)71Step-by-step explanation:i) The sample size n in Simple Random Sampling is given by [tex] \bf n=\frac{z^2p(1-p)}{e^2}[/tex] where  z = 1.645 is the critical value for a 90% confidence level (*) p= 0.669 is the population proportion given by the Census  e = 0.02 is the margin of error so  [tex] \bf n=\frac{(1.645)^2*0.669*0.331}{0.02^2}=1,498.05\approx 1,499[/tex] rounded up to the nearest integer. (*)This is a point z such that the area under the Normal curve N(0,1) 1nside the interval [-z, z] equals 90% = 0.9 It can be obtained with tables or in Excel or OpenOffice Calc with NORMSINV(0.95) If she ignores the Census estimate, the she has to take the largest sample possible that meets the requirements. Let's show it is obtained when p = 0.5 As we said, the sample size n is [tex] \bf n=\frac{z^2p(1-p)}{e^2}[/tex] where  e = 0.02 is the error proportion  z = 1.645 hence [tex] \bf n=\frac{(1.645)^2p(1-p)}{(0.02)^2}=6765.0625p(1-p)=6765.0625p-6765.0625p^2[/tex] taking the derivative with respect to p, we get n'(p)=6765.0625-2*6765.0625p and  n'(p) = 0 when p=0.5 By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.02 Replacing p with 0.5 in the formula for the sample size we get [tex] \bf n=6765.0625*0.5-6765.0625(0.5)^2=1691.27\approx 1,692[/tex] rounded to the nearest integer. ii) When we do not have a proportion but a variable whose approximate standard deviation s is known, then the sample size n in Simple Random Sampling is given by [tex] \bf n=\frac{z^2s^2}{e^2}[/tex] where  z = 2.241 is the critical value for a 95% confidence level (*) s = 7.5 is the estimated population standard deviation e = 2 hours is the margin of error so  [tex] \bf n=\frac{z^2s^2}{e^2}=\frac{(2.241)^2(7.5)^2}{(2)^2}=70.62\approx 71[/tex] (*)This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 95% = 0.95 It can be obtained in Excel or OpenOffice Calc with NORMSINV(0.9875)