Question

Which of the following estimates at 95% confidence level most likely comes from a small sample?

A. 62% (+-6%)
B. 65% (+-2%)
C. 60% (+-4%)
D. 71% (+-18%)

Answer 1

smaller samples sizes give us larger errors

Answer 2

E = 1.96 sqrt(pq/n)

we know p and q=1-p if we really want to compare them

sqrt(n) = 1.96 sqrt(pq)/E

n = 1.96^2 pq/E^2

Answer 3

@amistre64 so a smaller sample like 65% with 2%

Answer 4

1.96^2 is constant so we can ignore that ...

for a we get: .68(.38)/(.06)^2 = 65.44

for b we get: .65(.35)/(.02)^2 = 568.75

work it for c and d ... and compare to be sure.

Answer 5

ok I will do that right now

Answer 6

wait for a. where did you get .68 isn't it supposed to be 62 or is it 68 because you add 6 to 62?

Answer 7

I am also confused as to where to get the .38, .35,.06 and .02

Answer 8

just working the error, not the range

we are not concerned with an actual sample size, we just need to detemrine which one would be the smallest

n = 1.96^2 pq/E^2

but 1.96^2 is a constant for all of them so its really just extra baggage and can be ignored

work: n = pq/E^2 and you will get the same basic results for comparison

Answer 9

n = .62(.38)/(.06)^2

n = .65(.35)/(.02)^2

n = .60(.40)/(.04)^2

n = .71(.29)/(.18)^2

Answer 10

p+q = 1, since p = .60, q= 1-.60 = .40

Answer 11

65.44
568.75
150
6.35

Answer 12

good, now the one with the smallest relative number is clear; we could have predicted it since it has the greatest error associated with it.

Answer 13

oh so 71% (+-18%)

Answer 14

thats correct ... to both of you lol

Answer 15

for these type of problems could you just predict the answer because it was largest percentage (71) and the largest error (18)?

Answer 16

you might be able to, but quite frankly i dont like the assumptions about it. i was confident it was D, but i just wasnt sure until the calcs verified it.