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anonymous

  • one year ago

In a survey conducted with 600 participants across the United States, 450 were found to have studied science in college. If we were to predict the population proportion with a 99.7% confidence, what would the confidence interval be? 69.7% to 80.3% 73.23% to 76.76% 71.46% to 78.53% 73.5% to 76.5%

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  1. anonymous
    • one year ago
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    I have the steps to this problem but I went wrong somewhere so I don't understand it

  2. dan815
    • one year ago
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    okay, so first 450/600 = ?

  3. anonymous
    • one year ago
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    0.75

  4. dan815
    • one year ago
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    okay now for a 99.7 %, u have to find the Z score for that

  5. dan815
    • one year ago
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    the next step is to find out what the standard deviation of thing should be

  6. anonymous
    • one year ago
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    ohhh see, that was the step I was missing, the standard deviation. Except idk how to do it

  7. dan815
    • one year ago
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    oh um also we know the center is definately 0.75 right

  8. anonymous
    • one year ago
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    yes..

  9. dan815
    • one year ago
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    or 75%, some of the choices u are given arent centered around that, so they can eliminated rightway

  10. anonymous
    • one year ago
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    would it be D?

  11. dan815
    • one year ago
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    nvm, all them are centered around 75% cant elimnate any argh

  12. anonymous
    • one year ago
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    oh okay so what do I do now

  13. dan815
    • one year ago
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    C is definitely wrong 71.46% to 78.53% as it should say 71.46% to 78.54%

  14. anonymous
    • one year ago
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    Well I wrote all the answers correctly so idk, I just checked

  15. dan815
    • one year ago
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    its probably just bad rounding

  16. anonymous
    • one year ago
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    So it's definitely not C?

  17. dan815
    • one year ago
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    What is the Z score u foundÉ

  18. anonymous
    • one year ago
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    the formula for z score is \[z= \frac{ X-\mu }{ \sigma }\] right...

  19. dan815
    • one year ago
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    yes thats another one,. but we cant use that right now

  20. anonymous
    • one year ago
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    oh... then what do we use

  21. dan815
    • one year ago
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    i just mean we have to find Z score, for the 99.7 %

  22. dan815
    • one year ago
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    you have to look at that table and chart

  23. anonymous
    • one year ago
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    okay how do I do that

  24. anonymous
    • one year ago
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    http://www.statisticshowto.com/tables/t-distribution-table/ this one right

  25. anonymous
    • one year ago
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    cause this is where I was stuck, I wasn't sure what numbers or whatever to look at on the chart

  26. dan815
    • one year ago
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    click on the 2 tail one

  27. dan815
    • one year ago
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    https://www.youtube.com/watch?v=_x6j-rH33Og you can watch this, i know it says typically not used for sample size greater than 30.. but the standard deviation is unknown so, what else can we do

  28. anonymous
    • one year ago
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    wait is it B?

  29. dan815
    • one year ago
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    no clue it looks like info is missing, what can we do with sample deviation is unknown and the pop deviation is unknown @hartnn

  30. anonymous
    • one year ago
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    here is the entire question

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  31. dan815
    • one year ago
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    ya hm not sure

  32. anonymous
    • one year ago
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    ok I just need help cause I need to answer it soon

  33. dan815
    • one year ago
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    @ganeshie8 is the question missing info??

  34. anonymous
    • one year ago
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    so do you know anything..

  35. anonymous
    • one year ago
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    @ParthKohli

  36. ganeshie8
    • one year ago
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    I am eating at the moment but I found this https://answers.yahoo.com/question/index?qid=20140522184044AAlIk3W

  37. anonymous
    • one year ago
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    I'm on my school computer so yahoo is blocked, could you tell me what it says

  38. ganeshie8
    • one year ago
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    how about this http://gyazo.com/d103059ce0cd9eadfd631e125a3f3bc9

  39. anonymous
    • one year ago
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    no it's still blocked..

  40. ganeshie8
    • one year ago
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    try this

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  41. anonymous
    • one year ago
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    thanks so much :)

  42. ganeshie8
    • one year ago
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    np

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