Question

Help statistics question!!

Answer 1

Are you referring to np > 5 and n(1-p) > 5 ?

Answer 2

I also found this

Answer 3

oh N = 10n

N = population size
n = sample size

that sounds familiar in a way

Answer 4

sorry N > 10n

Answer 5

alright

Answer 6

i found this too

http://www2.fiu.edu/~tardanic/size.pdf

it says `According to Moore/McCabe, this is true, strictly speaking, as long as the population is at least 100 times larger than the sample.`

Answer 7

ok I guess the rule isn't 100% solid

Answer 8

I would go with your book since that's what the teacher will use

Answer 9

sure

Answer 10

sigma = standard deviation

sigma = sqrt(n*p*(1-p))

we don't know the value of p, but we do know n and sigma
n = 50
sigma = 0.07

are you able to solve for p?

Answer 11

hmm there may be another way to do this

Answer 12

ok I figured out a shorter way

Answer 13

\[\Large \sigma = \sqrt{\frac{p*(1-p)}{n}}\]
\[\Large \sigma = \sqrt{\frac{p*(1-p)}{100}}\]
\[\Large \sigma = \sqrt{\frac{p*(1-p)}{2*50}}\]
\[\Large \sigma = \sqrt{\frac{1}{2}}*\sqrt{\frac{p*(1-p)}{50}}\]
\[\Large \sigma = \sqrt{\frac{1}{2}}*0.07\]
\[\Large \sigma \approx 0.0495 \approx 4.95\%\]

Answer 14

The quantity \[\Large \sqrt{\frac{p*(1-p)}{50}}\] is the given standard deviation of 7% = 0.07