-1

Units for Chi-Square test contingency table

Let' say we have the following example

Visitors to some place:

representation(1)
              Sunday | Monday
    
    expected    80%     20%
    observed    90%     10%


representation(2)
              Sunday | Monday
    
    expected    0.8     0.2
    observed    0.9     0.1

For case (1):

$\chi_2^2 = \frac{(80-90)^2}{80} + \frac{(20-10)^2}{10} = 11.25 $

For case (2):

$\chi_2^2 = \frac{(.8-.9)^2}{.8} + \frac{(0.2-0.1)^2}{0.1} = 0.1225 $

This could be asked in a different words, when we use kilometers instead of meters ant etc..

Which representation should be choosed?

Also we have some assumption that our "our sum of square" follows the chi-squrare distribution. Why it's correct?

Community
  • 1
Michael D
  • 583
  • 1
  • 3
  • 23
  • maybe this can help https://stats.stackexchange.com/questions/171202/ambiguity-of-chi-squared-test-of-association-how-can-the-conclusion-change-usin/171207#171207 –  Aug 27 '17 at 11:02

1 Answers1

2

The values in the contingency table have to be counts. They can't be fractions, percentages, or measured quantities.

A kind of exception is with a chi-square goodness-of-fit test where you compare counts to expected proportions.

Also, the expected values aren't part of the contingency table per se. They are calculated from the counts in the table.

This solves 99.9% of the difficulties suggested by your question. I'm sure you can come up with examples where it unclear how to count real objects (schools of fish or individual fish?), but these cases are probably rare in practice.

Sal Mangiafico
  • 7,128
  • 2
  • 10
  • 24