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A statistical question

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Can someone tell me when testing for independence using chi square tests, why is the expected frequency of a cell is denoted by the formula
    $$ \frac{\sum row * \sum column}{\sum total } \ $$
     
  2. jcsd
  3. Feb 17, 2017 #2

    BvU

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    Is this a human cell or a plant cell ? Or perhaps an excell ? :smile: Some more description might make your question a bit clearer, perhaps ?
     
  4. Feb 17, 2017 #3

    Ray Vickson

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    In a two-way table, if ##R_i## is the total number in row ##i## and ##C_j## the total number in column ##j## then ##f_i = R_i/N## is the estimated probability of the event for row ##i## and ##g_j = C_j/N## is the estimated probability of the event for column ##j##. Here, ##N = \sum_i R_i = \sum_j C_j## is the total number of observations. Under the hypothesis of independence between rows and columns, the estimated probabilty of the cell ##(i,j)## is ##\bar{p}_{ij} = f_i \,g_j = R_i C_j/N^2.## Thus, the expected frequency of cell ##(i,j)## is ##E_{ij} = N \bar{p}_{ij} = R_i C_j/N.##
     
  5. Feb 17, 2017 #4

    haruspex

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    The null hypothesis is that the two attributes are independent. Call the attributes A (rows representing A and not A) and B (columns representing B and not B). If they are independent then the fraction having attribute A multiplied by the fraction having attribute B should approximately equal the fraction having attributes A and B. I.e. #(A & B) / total = ( #A / total)*( #B /total), so #(A & B) = #A * #B / total.
     
  6. Feb 17, 2017 #5
    Shouldn't it be ##N = \sum_i R_i + \sum_j C_j##
     
  7. Feb 17, 2017 #6

    Ray Vickson

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    No. Try it for yourself on a simple example:
    $$\begin{array}{ccc|l}
    & & &\text{tot.} \\ \hline
    1 & 2 & 3 &6\\
    4 & 5 & 6 & 15 \\
    7 & 8 & 9 & 24\\ \hline
    12 & 15 & 18 & 45\; \leftarrow \text{totals}
    \end{array}
    $$
     
    Last edited: Feb 17, 2017
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