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Math Help - "Frequency distribution of the variances"

  1. #1
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    "Frequency distribution of the variances"

    I am working on a homework assignment from a very vague, unhelpful professor.

    I flipped 20 coins, measured the amount of heads and so forth, and repeated this 100 times.

    One of the questions involves breaking the 100 measurements of "heads" into 25 groups, where group one is the first 4 measurements, group two is the next 4, and so on. The mean and standard deviation of each group was then found.

    I am now asked to "calculate the frequency distribution of the variances of the groups", which I do not understand how to do. I believe they follow a chi-squared distribution, and upon asking the professor I received an unhelpful reply.

    I already made a histogram of the experimental frequency distribution of the variances. I now need to find a calculated, theoretical frequency distribution of these variances.

    How would I go about analyzing this? I am thoroughly lost on this question.

    Thank you for any help in advance!
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  2. #2
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    Your 4 coinflips (denoted here by X) will yield an expected value of 2 times head with a very, very approximated normal distribution of the values 0-4 for this event X. However, the sample sice is very small so that I would not use this approximation. However, if you take the variance of this random variable you get the sum of (X -2)^2 which is indeed chi-squared distributed if you claim your approximation to be valid. Considering your degrees of freedom you should be able to compute your theoretical distribution by chi-squared.

    PS: Did he really make you flip 20 coins 100 times = 2000 coinflips?
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  3. #3
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    Yes. We flipped all 20 coins at once, but had to do it 100 times.

    Thank you for your help, but I am still confused... the only equation I have from lecture and my book is:

    chi^2 = [(degrees of freedom)*(sample variance)] / (variance of whole population)

    A I correct in saying that the degrees of freedom is 24, since degrees of freedom = n-1 and I had 25 samples?

    Also, I computed my sample variance, s^2, of the 25 samples to be 1.2975 and the overall population variance to be 2.492609.

    I honestly do not understand what's going on here. How did you arrive at (X-2)^2, and what does it mean?

    Sorry for asking so many questions... This course is recommended to be taught in 3 semesters but we have to complete it in 1, so I am very lost.
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  4. #4
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    Also I think I was unclear. In each "test", we flipped 20 coins and had to count the number of heads. Then, to make the 25 groups, we made them in this fashion:

    Group 1 = # heads on tosses 1,2,3,and 4
    Group 2 = # heads on tosses 5,6,7, and 8
    etc.

    For my experiment, My group one consists of "9,13,10,9".
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  5. #5
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    You are right about your degrees of freedom.

    The Chi-squared distribution is defined as the sum of squared standard normal distributions. To standardize a random variable, you substract the expected value of it and then divide by the s.d. Thus: ((X -2)/s.d.)^2 (forgot the s.d. before, sorry) These standardized normal distributions describe each group of coinflips. However, after computing the standard normal deviation, the squared sums of these standard normal deviations should reflect the distribution of all your coinflips (a distribution of the particular groups that you looked at. However, it is only a very rough approximation.
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