Quiz 3 Solutions, 2013

1.1

ep = .01 + .3*.05 + .2*.1
cat('ep: ',ep,'\n')
ep:  0.045

1.2

c12 = .1*.15*.8
cat('c12: ',c12,'\n')
c12:  0.012

1.3

vp = (.3^2)*(.1^2) + (.2^2)*(.15^2) + 2*.3*.2*.012
cat('vp: ',vp,'\n')
vp:  0.00324

1.4

sp = sqrt(vp)
cat('sp: ',sp,'\n')
sp:  0.056921

2.1

n=1000
y=550
phat = y/n
se = sqrt(phat*(1-phat)/n)

ci = phat +2*se*c(-1,1)
cat('ci:\n')
print(ci)

ci:
[1] 0.5185357 0.5814643
> phat
[1] 0.55
> se
[1] 0.01573213

2.2

#Let's just plug in phat for p.

#.01 = 2*sqrt(phat*(1-phat)/n)

#=>

n = (phat*(1-phat)*4)/(.01^2)
cat('n: ',n,'\n')

#check:
phat=.55
pm = 2*sqrt(phat*(1-phat)/n)
cat('pm: ',pm,'\n')

n:  9900 
pm:  0.01

#another way to do it:

> 0.01573213/.005 #ratio of standard error at n=1000 to se at needed n
[1] 3.146426 #we need to be 3.14 times more accurate
> 3.14^2 #se we need 3.14^2 times more data
[1] 9.8596 # so we need 9.9*1000 observations = 9900.