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anonymous

  • one year ago

Help statistics question!!

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  1. jim_thompson5910
    • one year ago
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    Are you referring to np > 5 and n(1-p) > 5 ?

  2. jim_thompson5910
    • one year ago
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    I also found this

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  3. jim_thompson5910
    • one year ago
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    oh N = 10n N = population size n = sample size that sounds familiar in a way

  4. jim_thompson5910
    • one year ago
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    sorry N > 10n

  5. jim_thompson5910
    • one year ago
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    alright

  6. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    ok I guess the rule isn't 100% solid

  8. jim_thompson5910
    • one year ago
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    I would go with your book since that's what the teacher will use

  9. jim_thompson5910
    • one year ago
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    sure

  10. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    hmm there may be another way to do this

  12. jim_thompson5910
    • one year ago
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    ok I figured out a shorter way

  13. jim_thompson5910
    • one year ago
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    \[\Large \sigma = \sqrt{\frac{p*(1-p)}{n}}\] \[\Large \sigma = \sqrt{\frac{p*(1-p)}{100}}\] \[\Large \sigma = \sqrt{\frac{p*(1-p)}{2*50}}\] \[\Large \sigma = \sqrt{\frac{1}{2}}*\sqrt{\frac{p*(1-p)}{50}}\] \[\Large \sigma = \sqrt{\frac{1}{2}}*0.07\] \[\Large \sigma \approx 0.0495 \approx 4.95\%\]

  14. jim_thompson5910
    • one year ago
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    The quantity \[\Large \sqrt{\frac{p*(1-p)}{50}}\] is the given standard deviation of 7% = 0.07

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