Statistics?

From a random sample of 2000 internet users was carried out to test who whould prefer to find out medical info. online or from a library. There were 1400 responses of which 872 preferred to use the internet.





A) what porportion p-bar would use the internet?


B) what is the standard error of the poportion p_bar that would use the internet instead of the library?


C) how large a sample size would be required for a margin of error of at most .01 in a 99% Confidence level to estimate the porportion pbar that prefer the internet instead of the library?

Statistics?
a) 872 / 1400





b) the standard error = sqrt( pbar * (1 - pbar) / n) where n is the sample size.





= sqrt( (872 / 1400) * (1 - 872 / 1400) / 1400)


= 0.01295338





c)





large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval.





For large sample confidence intervals for the proportion in this situation you have:





pHat ± z * sqrt( (pHat * (1-pHat)) / n)





where pHat is the sample proportion


z is the zscore for having α% of the data in the tails, i.e., P( |Z| %26gt; z) = α


n is the sample size





in this case you have a z-score such that:





P( |Z| %26lt; z) = 0.99


P( |Z| %26gt; z) = 0.01


P( Z %26gt; z) = 0.005


P( Z %26gt; 2.575829) = 0.005





to restrict the margin or error we have:





z * sqrt( (pHat * (1-pHat)) / n) %26lt; 0.01


2.575829 * sqrt( (pHat * (1-pHat)) / n) %26lt; 0.01


2.575829 * sqrt( ( (872/1400) * (1- (872/1400))) / n) %26lt; 0.01





we use the proportion from the experiment as a starting point and solve for n





2.575829 * sqrt( ( (872/1400) * (1- (872/1400))) / n) %26lt; 0.01


2.575829 * sqrt( 0.2349061 / n ) %26lt; 0.01


sqrt( 0.2349061 / n ) %26lt; 0.01 / 2.575829


0.2349061 / n %26lt; 1.507183e-05


n %26gt; 0.2349061 / 1.507183e-05


n %26gt; 15585.77





n must be an integer value so n %26gt; 15586
Reply:A) 872/1400 = 0.623


B) The standard error is s/√n, where s = √(np(1-p)) = 18.13


So the standard error is: 0.485


C) Let margin of error, E = 0.01, and we can use p = 0.623 Then,


n = [Zα/2/E]²p(1-p), Where Zα/2 = 2.575 for a 99% confidence interval.


n = [2.575/0.010]²(0.623)(1-0.623) ≈ 15,574

ovary