A sample x₁ , x₂.....is taken from a population and it is necessary to test the hypothesis that F(x) is the distribution function of the population from which the sample has been taken.  The sample distribution function F_s (x) is an approximation of F(x) and if it approximates it sufficiently well the hypothesis can be accepted.  If it deviates the hypothesis is rejected.

The Chi-squared test is a tool which allows us to determine how much the F_s (x) distribution can deviate from F(x) if the hypothesis is true.   A distribution of the deviation is created under the assumption that the hypothesis is true and the number c is determined such that if the hypothesis is true then a deviation greater than c has a preassigned probability called the significance level.

The χ² statistic (pronounced ki) is defined as

The χ² distributions is a family of density functions each one dependent on the number of freedoms denoted by ν.   The exact definition is too complicated to identify on this level website.  This distribution is used to measure how well theoretical data fits the observed data.

The chi-square distribution has the following properties:

The shapes of the χ² distributions for various degrees of freedom are shown in the figure below
The mean of the distribution is equal to the number of degrees of freedom: μ = ν.
The variance is equal to two times the number of degrees of freedom: s _x ² = 2 . ν
When the degrees of freedom are greater than or equal to 2, the maximum value for Y occurs when χ² = ν - 2.
As the degrees of freedom increase, the chi-square curve approaches a normal distribution. When O is and observed sample frequency and E is the expected sample frequency the statistic for χ² as shown below is can be used as an approximation for the true χ² value

The notes below are a basic procedure of completing a χ ² test that F(x) is the distribution function from which a sample x _i..x ₂..x_n is taken.

Step 1)   Divide the x axis into K equal intervals J ₁, J ₂..J _K such that each interval contains at least 5 values of the given sample.  A sample value on a boundary is counted as 0,5 to each side of the boundary.  

Step 2)  Count the number of sample values b _j in each interval J _j (j = 1 to K)

Using a table of distribution values for F(x) determine the probability p_i that the random variable X assumes any value in the each interval I _j
calculate e_j = np _j This is the expected number of samples values if the hypothesis is true

Step 3)  Compute the deviation

Step 4)  Choose a significance value α (0,05, 0,01 , 0,001 etc)

Determine the value of c from

P(χ ² ≤ c) = 1 - α

using the table of chi_Square distribution with ν degrees of freedom .
[example if 1 - α = 0,95 and ν = 6 deg. of freedom then c = 12,592
If χ _o ² ≤ c then the hypothesis is accepted . If χ _o ² > c then the hypothesis is rejected.

Note: the procedure described above is illustrated in the examples below

The number of degrees of freedom ν is taken as

ν =the number of classes /Intervals - 1 = K -1

If any parameters of F(x) (say r parameters) have to be estimated then the number of degrees of freedom =

ν = K - r - 1

The chi-square distribution shown above are constructed so that the total area under each curve is equal to 1.   The area under the curve between 0 and a particular value of a chi-square statistic is the cumulative probability associated with that statistic.   For example, in the figure above, the shaded area represents the cumulative probability for a chi-square equal 3,94 with a (say) a sample size n = 11 with ν degrees of freedom = (n-1) = 10.  The shaded area under the curve = 0,05.

The table below provides Ch-square values relating to F(z) against the number of degrees of freedom

A dice is thrown 180 times and the scores are recorded as shown below.  Confirm that the dice is true . A dice is true if there is equal probability of any score 1 to 6.( = 1/6)

Testing the hypothesis that the dice is true with a 5% level of significance.
There is only one constraint i.e. that the Total number of total of the expected frequencies ∑ E = the total observed frequencies. ∑ O

The number of degrees of freedom ν = 6 - 1 = 5.

With a 1 - α value of 0,95 then c is 11,07 from the table above.  If χ ² < c then hypothesis if accepted
For each toss the expected number of observations =(1/6).180 = 30.    Calculation of the χ ² is as shown below.

χ ² =7,533 is less than 11,07 and therefore there is good confidence that the dice is true.

In this case a continuous distribution is used as a model for testing discrete data. This is reasonable because there are a significant number of sample values-observations (180)

Consider 104 tensile tests on twine resulting in the following table of breaking loads (N).
The sample result to be test to confirm that the population is normal.   The population mean μ and variance σ ² are not known.
the

The average value of the breaking loads is x_m = 269,9 and the variance = (103/104) s _x ² =755,70.
The best estimates for μ = 269,9 and for σ = 27,49. ( Sqrt(755,70)
The number of degrees of freedom ν = K - r -1.    [K = 11    r = 2 -as two population parameters have been estimated ]
Therefore  ν = 1 - 2 - 1 = 8
The value of c for P(χ ² ≤ c) = (1 - α =0,95 ) from the table above = 15,507
The calculations in the table below yields a χ ² = 9,003 which is less than 15,507.
Therefore the the hypothesis that the population is normal is accepted.

x j				Φ (z)		e j = 104p j	b j
-	225	-	-1,6333	0	0,0516	5,3664	5	0,025
225	235	-1,6333	-1,2696	0,0516	0,102	5,2416	8	1,4516
235	245	-1,2696	-0,9058	0,102	0,1814	8,2576	13	2,7236
245	255	-0,9058	-0,542	0,1814	0,2946	11,7728	13,5	0,2534
255	265	-0,542	-0,1782	0,2946	0,4286	13,936	17,5	0,9115
265	275	-0,1782	0,1855	0,4286	0,5714	14,8512	13	0,2308
275	285	0,1855	0,5493	0,5714	0,7088	14,2896	9	1,9581
285	295	0,5493	0,9131	0,7088	0,8186	11,4192	9	0,5125
295	305	0,9131	1,2768	0,8186	0,898	8,2576	5,5	0,9209
305	315	1,2768	1,6406	0,898	0,9495	5,356	5,5	0,0039
315	+	1,6406	+	0,9495	1	5,252	5	0,0121
								χ 2 =9,0034