## anonymous one year ago Help statistics question!!

1. jim_thompson5910

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

2. jim_thompson5910

I also found this

3. jim_thompson5910

oh N = 10n N = population size n = sample size that sounds familiar in a way

4. jim_thompson5910

sorry N > 10n

5. jim_thompson5910

alright

6. jim_thompson5910

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.

7. jim_thompson5910

ok I guess the rule isn't 100% solid

8. jim_thompson5910

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

9. jim_thompson5910

sure

10. jim_thompson5910

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?

11. jim_thompson5910

hmm there may be another way to do this

12. jim_thompson5910

ok I figured out a shorter way

13. jim_thompson5910

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

14. jim_thompson5910

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