Question

Students at a certain school were surveyed, and it was estimated that 27% of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.09 with probability 0.99, if before conducting the study (a) you are unwilling to predict the proportion value at your school and (b) you use the results from the surveyed school as a guideline. Click here to view page 1 of the standard normal table. Click here to view page 2 of the standard normal table. a. n= (Round up to the nearest integer.)

Answer 1

Ans:- & Students at a centain school weere Survey, and it was estimated that 27% of college students abstain from drinking alcohol. = Margin of Endle (E)= 0.09 & Significance level=d=ta confidence 21-01992 ore 9-1-P) = (as) cors The Coritical value Zd/2= 20.05= 258. Cesing z table] I 2) If proportion value you are 1 then checling to predict the we teeke p=0.5 = E= Zool 0104 2158 0151005) => 1 (2.58)*70.534 droge 6.6564 60125 0.0081 1.6641 010081 205.444 ~ 206

- Sample Size should be at least n= 206. b) ay 3 we use the result from the Survey 23 P = 0.27 ->) 9- 1-p -1-027 20:13 92 0273 » Ex zoos 0.092-4580127093) 23 Squaring both sides 2 ne (21581776 27) (0-73) 2-58200q? 6.6584 * 0.1971 010081 131237064 01681 1619720667 162 * n = 165*(162) If the above information feels good for you, then let the know by givings alike. "Thank you"