Chi square (c^2) is a very popular form of hypotesis testing, but also one that is subject to substantial abuse. You have to consider that you might reject a hypotesis while it is correct and also might confirm a hypotesis while it is false. Now we shall see that you also might make another mistake, simply ask the wrong question before the analyzing has begun. We shall look at alternatives to Chi square. Results obtained in samples do not always agree exactly with theoretical results according to the rules of probability. It is often good to know whether the observed frequences differ signifi cantly from the expected frequences. But there are several difficulties and also several opportunities to come out with unreliable results, as we shall see. 
Tests for Goodness of fit – or more logical, Badnessoffit Statistic
As we in the following will concentrate on messured observed values and expected values according to the rules of probability, we will use o as an observed value and e as an expected value.
Event 
E1 
E2 
…. 
Ek 
Observed 
o1 
o2 
ok 

Expected 
e1 
e2 
ek 
The Chi square c ^2 is defined:
(o1  e1)^2/e1 + (o2  e2)^2/e2 +…… + (on  en)^2/en = S (oj – ej)^2/ej >=0
j from 1 to k
"By how much does the observed frequency diviate from the expected observed"?
Example:
You throw a die 180 times. You expect the die to show 1, 2, 3, 4, 5 and 6 1/6 of the 180 times that means 30 times 1, 30 times 2… 30 times 6.
But actually observe the following result compared with the expectations:
Event 
1 
2 
3 
4 
5 
6 
Observed 
23 
34 
26 
39 
20 
38 
Expected 
30 
30 
30 
30 
30 
30 
c ^2 = [(23  30)^2 + (34  30)^2 + (26  30)^2 + (20  30)^2 + (38  30)^2]/30
= (49 + 16 + 16 + 100 + 64)/30
= 8.17
If c ^2 = 0 the expected and the observed frequences agree exactly. When c ^2 >0 they do not agree. The larger c ^2 is , the greater is the discrepancy between the observed and the expected frequences.
When S oj = S ej = N (here 180)
c ^2 = (S oj^2/ej)  N, j from 1 to k
Imagine that if you make the experiment with 180 throws of die a lot of times, you will then expect to obtain variable frequences of observed events.
Confidence Intervals for c
^2
As can be done with other distributions, we can define 95%, 99% or other confidence limits and intervals for Chi square and estimate the population standard deviation s
in terms of a sample standard deviation s by using the table showing Chi square distribution.
Degrees of Freedom
The n degrees of freedom in a sample of n observation is reduced when you calculate the average or any other parameter that must be estimated beforehand and you use this/ those to calculate further. This means that you have in reality manipulated the raw data so that ej fit oj a little better. Therefore the reduction of degrees of freedom is justified.
In the example above the number of degrees of freedom is 61=5
You cannot claim that your dia is fair (come il faut): c ^2 0.90 < 8.17<c ^2 0.95
Contingency Tables
Example:
i 
j 
Lice 
1 
2 
3 
4 
Total 
Relative 
010 
1030 
3050 
50 

Region 

1 
South 
281 
251 
260 
288 
1080 
0.540 

2 
North 
221 
240 
251 
208 
920 
0.460 

Total frequency 
502 
491 
511 
496 
2000 
1.000 

Relative frequency 
0.251 
0.246 
0.256 
0.248 
1.000 
Zero hypthesis (H0): Assumming independence
Estimated bivariate probabilities
c ^2  calculations
1.
Bivariate 
0.136 
0.133 
0.138 
0.134 
0.115 
0.113 
0.118 
0.114 
2.
Expected frequences Eij 
272 
266 
276 
268 
230 
226 
236 
228 
3.
(Oij  Eij) 
9 
15 
16 
20 
9 
14 
15 
20 
4.
Relative squared 
0.298 
0.846 
0.926 
1.493 
0.352 
0.867 
0.953 
1.754 
Sc^2 = 7.489
Number of Degrees of Freedom = (r  1)(c 1) (here: (21)(41) =3)
The observed c ^2 =7.489 is backeted by c ^2 0.10 =6.25 and c ^2 0.05 = 7.81
Thus: 0.90< propvalue < 0.95
The zero hypothesis that the number of lice does not differ significant from North to South must be rejected at the customary 5% level (and even at 10% level).
Learn more about tests of hypotheses, levels of significance and PropValue in the link about this.
Critics:
The test completely misses the point of asking the wrong question. The right question perhaps concerns salmon breeded unnaturally in fabrics and salmon breeding free or salmon living close together and salmon living more distributed in nature. You perhaps register that the first group seems to have more lice than the other, but this assumption has perhaps nothing to do with the regions South and North. And perhaps it would be of interest to know how the sample of the 2000 individuals actually took place. Are we sure that North/South indicates or add anything else to the assumption of indepence between the two groups.
If you have found a significant difference in the number of lice given a specific level of significance – perhaps in another corrected analyses – you even do not find out how much this difference actually amounts or othewise means. If c ^2 is close to zero you have to look at the analyses with suspion since it is rare that observed frequences agree to well with expected frequences (how did they collect the sample?). Less than c ^2 0.05 or c ^2 0.01 we would decide ("come il faut", but perhaps not in analyses of recovery by medical treatment) that the agreement is to good.
Confidence Interval as an alternative
To compare the difference in the two means of two regions is often much easier and better.
Another Example of Contingency Tables
Two groups, A and B, consist of 200 people who have a desease. A serum is given to group A but not to group B (which we call the control group; otherwise, the two groups are treated identically). It is found in group A and B that 152 and 131 people, respectively recover from the desease. Let us test the hypotesis that serum helps to cure the disease at a level of significance 0.01, 0.05 and 0.10 that means we willing to run the risk that our conclusion is random false in 1, 5 and 10 out of 100 cases.
If assume that the serum has no effect (H0), we expect 140 people in each group to recover and 60 in each group not to recover.
Frequences Observed
Recover 
No recover 
Total 

Group A 
152 
48 
200 
Group B 
131 
69 
200 
Total 
283 
117 
400 
Frequences Expected under H0
Recover 
No recover 
Total 

Group A 
140 
60 
200 
Group B 
140 
60 
200 
Total 
280 
120 
400 
c ^2 = ((152  140)^2/140)+((131  140)^2/140)+((48  60)^2/60)+((69  60)^2/60)
= 5.358
The number of freedom: (21)(21) = 1
Since c ^2 0.95 for 1 degree of freedom is read to 3.84 < 5.358 we have to conclude that the results are not significant at a 0.05 level.
We have to conclude that the serum is not effective or else decision pending further tests.
When numerical variable shall be analysed, to much manipulation of the raw material just dims the picture, and a lot of c ^2test are made alternatively by me thods of comparing averages, regression (simple or multiple), other multiple comparisons. One of problems with Chi is that expected and observed values are mixed before you account the result. You might ask if you are testing your observation or you are testing your expec tations. Its often a quite open question. The conclu sions are seldom so ambigious when you use the other tests as apparently that of the Chitestsanalyzer but the results are often clear and better. A general interpretation of the misuse of Chi: 
"Because two phenomenons appear at the same time at best,
coherence do not necessary light from them for this simple reason"