Quiz5Sols - [(266-k)/(266-1)]^0.5 * SD(pop) / k^0.5 <=

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Quiz #5 1) a) z-value = 1.55 Correct SE = 0.491% , B = [0.595*0.405]^0.5*100% = 49.1% SE = B/n^0.5 = [0.491/(10,000)^0.5] * 100% = 0.491/100 *100% = 0.491% Correct Form 59.5% +- 1.55 * 0.491% = [58.7%, 60.3%] (b) False, Something like, “either the Pop% is in the interval or not, there’s no chance involved”. Or a correction to the statement using the word “confident” instead of chance. 2) a) ) Sample % = 30/90 *100% = 33.3% b) SE WITH finite pop. Correction factor SE = [(N-n)/(N-1)]^0.5 * [frac(students working)*frac(students not working)]^.5 / n^0.5 *100% = [(266-90)/(266-1)]^0.5 * [1/3*2/3]^0.5 / 90^0.5 *100% = [176/265]^0.5 * [0.22]^0.5 / 90^0.5 *100% = [0.81*0.47/9.49] * 100% = 4.0% c) SE(k) <= 1/4*SE(100)
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Unformatted text preview: [(266-k)/(266-1)]^0.5 * SD(pop) / k^0.5 <= [(266-100)/(266-1)]^0.5 * SD(pop) / 100^0.5 (266-k)^0.5 / k^0.5 <= 1/4 * (166)^0.5 / 100^0.5 Squaring both sides we get, (266-k)/k <= 1/16 * 166/100 = 83/800 266 / (1 + 83/800) = 241.00 <= k 3) a) yes, since chance is involved in the sampling procedure b) multi-stage cluster selection bias: students with more classes are more likely to be chosen non-response: none since all subjects picked willingly participate response: none since all subjects picked willingly participate You werent required to mention kinds of bias that dont exist, so you could get full credit by just saying the selection bias part carefully....
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This note was uploaded on 02/08/2011 for the course STAT 21 taught by Professor Anderes during the Fall '08 term at University of California, Berkeley.

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