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Math Help - The following fig show the distribution of digits ....Test the hypothesis

  1. #1
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    The following fig show the distribution of digits ....Test the hypothesis

    Question:
    The following figures show the distribution of digits in numbers chosen at random from a telephone directory

    Digits        : \qquad \ 0 \qquad      1 \qquad        2 \quad \     3 \quad \     4 \qquad      5 \quad \     6 \quad \       7 \quad \     8 \quad \     9
    Frequency : \ 1026 \ 1107 \  997 \  966 \ 1075 \  933 \  1107 \  972 \  964 \  853<br />

    Test the hypothesis that the digits are uniformly distributed at 5% level of significance.
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question:
    The following figures show the distribution of digits in numbers chosen at random from a telephone directory

    Digits        : \qquad \ 0 \qquad      1 \qquad        2 \quad \     3 \quad \     4 \qquad      5 \quad \     6 \quad \       7 \quad \     8 \quad \     9
    Frequency : \ 1026 \ 1107 \  997 \  966 \ 1075 \  933 \  1107 \  972 \  964 \  853<br />

    Test the hypothesis that the digits are uniformly distributed at 5% level of significance.
    \chi^2 or chi squared test, look it up try to do it then let us know what problems you are having

    CB
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  3. #3
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    what shouldbe the Null hypothesis

    Quote Originally Posted by CaptainBlack View Post
    \chi^2 or chi squared test, look it up try to do it then let us know what problems you are having

    CB


    What should be the null hypothesis for this ?
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    Quote Originally Posted by zorro View Post
    What should be the null hypothesis for this ?
    That each digit occurs with equal probability (so the expected frequencies are all equal).

    CB
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    Am i doing it correctly

    Quote Originally Posted by CaptainBlack View Post
    That each digit occurs with equal probability (so the expected frequencies are all equal).

    CB

    From the last post what i have come to understand is that all the frequencies are equal

    H_0 : \sigma^2 = 0
    H_a : \sigma^2 \neq 0
    \alpha : 0.05

    Am i correct???
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  6. #6
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    I am stuck

    Quote Originally Posted by CaptainBlack View Post
    That each digit occurs with equal probability (so the expected frequencies are all equal).

    CB

    I have googled and have lots of documentation but dont know how to solve this question . I am studying for my exam and need to find some document that is close to the above question so that i could understand the chi square test and the hypothesis a little better.....
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  7. #7
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    Is this correct?

    Quote Originally Posted by CaptainBlack View Post
    \chi^2 or chi squared test, look it up try to do it then let us know what problems you are having

    CB


    Null Hypothesis : Digits are uniformly distributed
    <br />
\begin{array}{c|c|c|c|c|c|c|c|c|c}<br />
Digits & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\<br />
Freqs & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933  <br />
\end{array}

    x^2 = \sum \frac{f_o - f_e}{f_e}

    x^2 = 108.1 ...........Is my answer right??

    no of degree of freedom = 9 - 1 = 8

    x_{0.05} ^2 for 8 d.f. = 2.73

    Since the calculated > tabulated

    we reject the null hypothesis
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  8. #8
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    Quote Originally Posted by zorro View Post
    Null Hypothesis : Digits are uniformly distributed
    <br />
\begin{array}{c|c|c|c|c|c|c|c|c|c}<br />
Digits & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\<br />
Freqs & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933  <br />
\end{array}

    x^2 = \sum \frac{f_o - f_e}{f_e}

    x^2 = 108.1 ...........Is my answer right??

    no of degree of freedom = 9 - 1 = 8

    x_{0.05} ^2 for 8 d.f. = 2.73

    Since the calculated > tabulated

    we reject the null hypothesis
    No.

    First I make the expected frequencies 1000 (this is the sum of the observed frequencies divided by the number of bins which here is 10). There were 10 bins corresponding to digits 0 through 9 not 9)

    Second the formula for calculating the value of \chi^2 is:

    \chi^2=\sum_{i=1}^N\frac{(O_i-E_i)^2}{E_i}

    The value I get for this is \chi^2=58.5 with \nu=9. This exceeds both the 5 and 1 \% critical values for the \chi squared distribution with this number of degrees of freedom and so the observed frequencies differ significantly (at which of these levels you choose to report the results)

    CB
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    No.

    First I make the expected frequencies 1000 (this is the sum of the observed frequencies divided by the number of bins which here is 10). There were 10 bins corresponding to digits 0 through 9 not 9)

    Second the formula for calculating the value of \chi^2 is:

    \chi^2=\sum_{i=1}^N\frac{(O_i-E_i)^2}{E_i}

    The value I get for this is \chi^2=58.5 with \nu=9. This exceeds both the 5 and 1 \% critical values for the \chi squared distribution with this number of degrees of freedom and so the observed frequencies differ significantly (at which of these levels you choose to report the results)

    CB
    Thanks mite for ur reply
    cheers
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