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AP Statistics question

  1. May 26, 2013 #1
    Sorry, no subforum for statistics so I posted it here..

    1. The problem statement, all variables and given/known data

    So, if I have a given list of proportions

    n = 64

    and I want to compare it to another group of percentages

    n = North American Average

    What type of test would I use?
  2. jcsd
  3. May 26, 2013 #2
    [I condensed your question for clarity and to save space]

    What kind of comparison are you looking for? I'm thinking it's just a two-sample z-test for proportions.
  4. May 26, 2013 #3
    Just to see if there is a significant difference between the proportions or not.

    I am comparing car colors from my school against the National Average. The first group (n=64) is 64 car colors I got from my school parking lot. The second group of proportions, is that of the North American Average.
    Last edited: May 26, 2013
  5. May 26, 2013 #4

    I like Serena

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    Try a goodness-of-fit chi-square test.
  6. May 26, 2013 #5

    Ray Vickson

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    Validity, etc., depends on the nature of your data.

    For instance, how do you condense information about car colors into a proportion? Typically, a proportion like 0.11 could be thought of as the number of 'yes' vs. 'no' answers to some type of question. How does a car's color fit into that type of scheme? The point is that if the numbers you show represent some type of highly 'massaged' figures, their distribution may be an artifact of your data-aggregation method and not reflective of reality. So: show us how you obtained those figures.
  7. May 26, 2013 #6
    Well, I collected 64 car color samples (Chose 64 so it is a large enough number so I could generalize for my entire student parking lot) then had for example 10 black cars and did 10/64 to get the proportion of students who drive black color cars to school on that day.
  8. May 26, 2013 #7
    How formal do you want your test to be? If you really want a formal test, state between what wavelengths of light you consider to be blue, etc. There are other things to consider, but that might be better if you want a boolean ("yes/no") answer. For example, "red" can be very subjective. You could argue that all pink cars are red.

    Excuse my haste. You can use a two-sample z-test, but it would be weird since you are working with more than one proportion. Use a chi-squared based test for goodness of fit. That should work better.
  9. May 26, 2013 #8
    It is pretty informal.

    Okay, yeah, I was thinking of using a chi-squared test, thanks for your thoughts.
  10. May 26, 2013 #9
    hmm.. quick addition..

    Would I do a chi squared GOF for proportions the same as I would for non-proportions?

    just a summation of ((Observed proportion - Expected proportion)^2 / Expected Proportion) to get my chi squared value? And my degrees of freedom would just be number of cars i used to get my data -1 ?
  11. May 26, 2013 #10

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    Not quite.

    The summation should be of ((Observed frequency - Expected frequency)^2 / Expected frequency).
    You converted the frequencies to proportions, but you really need the frequencies.

    The degrees of freedom is the number of colors - 1.
  12. May 26, 2013 #11
    Oh gotcha, so I should convert the average car color proportions to frequencies out of 64?
  13. May 26, 2013 #12

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  14. May 26, 2013 #13
    Alright thanks, really a big help.. one final question, I should round all my expected car colors to whole numbers correct? Because you can't have 5.3 cars..
  15. May 26, 2013 #14

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    No. The expected cars should remain fractional.
  16. May 26, 2013 #15
    Okay, only reason i thought it would be the other way was because my ti-84 would only accept whole numbers or else it would error out... time to do it by hand.

    Really a big help, thanks a lot man.
  17. May 26, 2013 #16

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    Huh? I'd expect your ti-84 to required whole numbers for the observed frequencies, which should indeed be whole, but not for the expected frequencies.
  18. May 26, 2013 #17
    /facepalm.. thanks put it in wrong lol.
  19. May 26, 2013 #18

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