# Degrees of freedom for Chi-squared test

I am facing the following dilemma. I am aware of how to handle the one-sided Chi-distribution, but I am falling victim to how to handle degrees of freedom. Let me clarify with an example what I mean.

I have the following obseverd and expected values:

[Observed Data]

#Periods      CountryI   CountryII     CountryIII
#(1900-1950)     100      150            20
#(1951-2000)     59       160            50

[Expected DATA]

#Periods   country I     Country II       CountryIII
#(1900-1950)  118.4         52                40
#(1951-2000)   80.5         90                25


My question is: Since this is a one sided-Chi square test, are the degrees of freedom counted by the formula: (columns-1)(rows-1), in which case I would have $(6-1)(2-1) = 5$?

Or is that really just country1 country2 country3 that matters, so that d.f. would be 3-1=2?

Because d.f. is usually defined as the terms for the chi squared = 6, where we usually subtract 1 from it.

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Could you please clarify how did you get the expected values, and what hypothesis are you trying to test. The correct degrees of freedom depends on that. Most answers assume that you are interested in independence (is the effect of period the same for each country?). –  Aniko Aug 18 '11 at 14:15

How many variables are present in your cross-classification will determine the degrees of freedom of your $\chi^2$-test. In your case, your are actually cross-classifying two variables (period and country) in a 2-by-3 table.

So the dof are $(2-1)\times (3-1)=2$ (see e.g., Pearson's chi-square test for justification of its computation). I don't see where you got the $6$ in your first formula, and your expected frequencies are not correct, unless I misunderstood your dataset.

A quick check in R gives me:

> my.tab <- matrix(c(100, 59, 150, 160, 20, 50), nc=3)
> my.tab
[,1] [,2] [,3]
[1,]  100  150   20
[2,]   59  160   50
> chisq.test(my.tab)

Pearson's Chi-squared test

data:  my.tab
X-squared = 23.7503, df = 2, p-value = 6.961e-06

> chisq.test(my.tab)$expected [,1] [,2] [,3] [1,] 79.6475 155.2876 35.06494 [2,] 79.3525 154.7124 34.93506  - add comment Degrees of Freedom are (r-1)(c-1). You have 2 rows : 1900-1950 and 1950-1999 3 columns: CountryI CountryII CountryIII Thus (2-1)(3-1) = 2 The actual product of r x c should = n (total # of observations) which is six. However, this is not used in your calculation of the df. Edit: If you were doing an 'Goodness of Fit' then yes, it would be n-1 but you have a contigency table (r x c) where r or c not equal to 1 so you have to use the (r-1)(c-1) Edit #2 for dimbo (I can't comment): Expected values should be calculated by (row total)(column total) / (total # of observations) : Thus the expected for r1,c1 position is (270)(159) / (539) which gives the values chi gave you. Edit #: SAS code confirming Chi data question; do a=1 to 2; do b=1 to 3; input var @@; output; end; end; datalines; 100 150 20 59 160 50 ; run; proc freq data = question; weight var; tables a*b / chisq expected norow nocol; run;  Output  Frequency| Expected | Percent | 1| 2| 3| Total --------+--------+--------+--------+ 1 | 100 | 150 | 20 | 270 | 79.647 | 155.29 | 35.065 | | 18.55 | 27.83 | 3.71 | 50.09 ---------+--------+--------+--------+ 2 | 59 | 160 | 50 | 269 | 79.353 | 154.71 | 34.935 | | 10.95 | 29.68 | 9.28 | 49.91 ---------+--------+--------+--------+ Total 159 310 70 539 29.50 57.51 12.99 100.00 Statistics for Table of a by b Statistic DF Value Prob ------------------------------------------------------ Chi-Square 2 23.7503 <.0001 Likelihood Ratio Chi-Square 2 24.2964 <.0001 Mantel-Haenszel Chi-Square 1 23.3700 <.0001 Chi Coefficient 0.2099 Contingency Coefficient 0.2054 Cramer's V 0.2099 Sample Size = 539  - Nothing in the question is indicative of a GOF test because it would require the OP to supply theoretical proportions rather than observed counts (and we couldn't get such a table of expected counts this way). – chl Aug 18 '11 at 15:29 Yup, I was saying that his way of calcualting the expected by making everything into a table with 1 row or 1 column and then just using n-1 was more like the process for a GOF test. – ProcRun Aug 18 '11 at 15:39 add comment Wait a minute, I think Sandra means 5 rather than 6. Maybe chl can correct me on this ... but I think it should be rite. If we take the definition that$\chi^2$is evaluated as follows, $$\chi^2= \sum_{i=1}^{\#Rows}(observed_i - expected_i)^2/expected_i$$ and arrange the data as follows: Observed[O]| Expected[E] | (O-E)^2/E 100 118.4 150 52 20 40 59 80.5 160 90 50 25  Thus, the total number of terms for calculating$\chi^2\$ is 6 (as we are adding the final column of terms together which has 6 rows. As by definition, we have d.f.= no of rows or expected frequencies - 1. Thus we obtain 5.

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by the way why are u saying that Sandra's expected values are incorrect ? –  dimbo Aug 18 '11 at 13:37
mpiktas, can u explain please how i would get the p values using the definition which i stated above -- step by step ? –  dimbo Aug 18 '11 at 14:12
I still dont really understand the other answers. Someone who actually is in a position, can use the definition i provided above, with a step by step procedure to calculate x^2 value and then the p-value. –  dimbo Aug 18 '11 at 14:16
For testing the independence of the two variables, the summation in your formula should be over the number of observed cells, and not only the number of rows. What you describe corresponds to a GOF test, which is not obvious from the question. As we are about to speculate about that, we could imagine variation around RxC tables whereby one of the the variable might be treated as an ordinal outcome, like in this question, but I believe we have to wait for further clarifications :) –  chl Aug 18 '11 at 15:44
Reflect on how you computed the expected values: you had to use two row means and three column means, which is five numbers. However, there is one (linear) relation among them (because they are both related to the grand mean). That means you used 5-1 = 4 degrees of freedom in computing the expected value. Because there are 6 observations, you only have 6 - 4 = 2 degrees of freedom left. Note how your method of placing all the data into a single column obscures the information about degrees of freedom, whereas the standard tabular layout implicitly gives this information. –  whuber Aug 18 '11 at 16:13