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I have a data set of occurrences per day and I want to see if the day of the week has an influence. However the data is biased: of all the days on which we measured, 22 are Mondays, but only 18 are Sundays.

    | M| T| W| T| F| S| S|
 h_i|60|46|47|57|52|37|33| //total number of occurrences per given day of the week
 w_i|22|22|19|22|20|20|18| 

I have ultimately decided to leave the $h_i$ as they are and formulate the expectation as $E_i = \frac{w_i}{\sum_i w_i}N$ where $N=\sum h_i$

But my first idea was to normalize the $h_i$ to $\frac{h_i}{w_i} \cdot \frac{N}{\sum_i \frac{h_i}{w_i}}$ (normalize by day, then normalization constant so that it adds up to $N$ again)

This yields different statistics and when inserting them in the formula I can see that they differ. But why? It seems so intuitive. I could have been given normalized data, what then?

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    $\begingroup$ If these are per day why do you need to normalise them? $\endgroup$
    – mdewey
    Oct 29 '17 at 15:14
  • $\begingroup$ @mdewey the data I have is per day of one year, which I accumulated to $h_i$ which represent all occurrences on all i-days. We are biased towards days of the week that make up a higher proportion of the total number of days on which we measured. $\endgroup$
    – user2740
    Oct 29 '17 at 16:53
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    $\begingroup$ (1) Your first idea ("normalization") changes the data. For any test that is based on counts, like the chi-squared tests, that's usually a bad idea, because a count in itself carries information about its variability. (2) I believe you intended to define $N=\sum h_i$. (3) If you were given normalized data then the chi-squared test does not apply directly: you would need to recover the original counts and proceed from there. (4) The mechanism that caused differences in the $w_i$ could itself influence the results, so you ought to consider that. $\endgroup$
    – whuber
    Oct 29 '17 at 17:27

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