# Statistics Chi Squared distributions

### Chi Squared distributions

Introduction..... Symbols..... Procedure..... Degrees of Freedom..... Chi-Squared   Distribution..... Chi-Squared   Table..... Examples.....

Introduction

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 Fs (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 Fs (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

Symbols
 n = sample number χ2 = statistic for testing hypothesis c = limit of χ2 defining sample significance f(x) = probability function. (values between 0 and 1) F(x) = probability distribution function. ν = number of degrees of freedoms var = sample variance K = number of classes /intervals Jj class designation Φ (z) = Probability distribution function.(Standardised probability ) α = significance value μ = population /random variable mean σ 2 = population /random variable variance σ = population /random variable standard deviation xm = arithmetic mean of sample sx 2 = variance of sample sx = Standard deviation of sample z = (x - μ ) / σ Equation to standardise prob'y dist'n function/

Procedure.

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..xn 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 pi that the random variable X assumes any value in the each interval I j
calculate ej = 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

Number of freedom

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

Chi Distribution with ν degrees of Freedom.

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

Table of Chi-square values
 Degs Of Freedom ν F(z) 0,005 0,01 0,025 0,05 0,95 0,975 0,99 0,995 1 0 0 0,001 0,004 3,841 5,024 6,635 7,879 2 0,01 0,02 0,051 0,103 5,991 7,378 9,21 10,597 3 0,072 0,115 0,216 0,352 7,815 9,348 11,345 12,838 4 0,207 0,297 0,484 0,711 9,488 11,143 13,277 14,86 5 0,412 0,554 0,831 1,145 11,07 12,832 15,086 16,75 6 0,676 0,872 1,237 1,635 12,592 14,449 16,812 18,548 7 0,989 1,239 1,69 2,167 14,067 16,013 18,475 20,278 8 1,344 1,647 2,18 2,733 15,507 17,535 20,09 21,955 9 1,735 2,088 2,7 3,325 16,919 19,023 21,666 23,589 10 2,156 2,558 3,247 3,94 18,307 20,483 23,209 25,188 11 2,603 3,053 3,816 4,575 19,675 21,92 24,725 26,757 12 3,074 3,571 4,404 5,226 21,026 23,337 26,217 28,3 13 3,565 4,107 5,009 5,892 22,362 24,736 27,688 29,819 14 4,075 4,66 5,629 6,571 23,685 26,119 29,141 31,319 15 4,601 5,229 6,262 7,261 24,996 27,488 30,578 32,801 16 5,142 5,812 6,908 7,962 26,296 28,845 32 34,267 17 5,697 6,408 7,564 8,672 27,587 30,191 33,409 35,718 18 6,265 7,015 8,231 9,39 28,869 31,526 34,805 37,156 19 6,844 7,633 8,907 10,117 30,144 32,852 36,191 38,582 20 7,434 8,26 9,591 10,851 31,41 34,17 37,566 39,997 21 8,034 8,897 10,283 11,591 32,671 35,479 38,932 41,401 22 8,643 9,542 10,982 12,338 33,924 36,781 40,289 42,796 23 9,26 10,196 11,689 13,091 35,172 38,076 41,638 44,181 24 9,886 10,856 12,401 13,848 36,415 39,364 42,98 45,558 25 10,52 11,524 13,12 14,611 37,652 40,646 44,314 46,928 26 11,16 12,198 13,844 15,379 38,885 41,923 45,642 48,29 27 11,808 12,878 14,573 16,151 40,113 43,195 46,963 49,645 28 12,461 13,565 15,308 16,928 41,337 44,461 48,278 50,994 29 13,121 14,256 16,047 17,708 42,557 45,722 49,588 52,335 30 13,787 14,953 16,791 18,493 43,773 46,979 50,892 53,672 31 14,458 15,655 17,539 19,281 44,985 48,232 52,191 55,002 32 15,134 16,362 18,291 20,072 46,194 49,48 53,486 56,328 33 15,815 17,073 19,047 20,867 47,4 50,725 54,775 57,648 34 16,501 17,789 19,806 21,664 48,602 51,966 56,061 58,964 35 17,192 18,509 20,569 22,465 49,802 53,203 57,342 60,275 36 17,887 19,233 21,336 23,269 50,998 54,437 58,619 61,581 37 18,586 19,96 22,106 24,075 52,192 55,668 59,893 62,883 38 19,289 20,691 22,878 24,884 53,384 56,895 61,162 64,181 39 19,996 21,426 23,654 25,695 54,572 58,12 62,428 65,475 40 20,707 22,164 24,433 26,509 55,758 59,342 63,691 66,766 41 21,421 22,906 25,215 27,326 56,942 60,561 64,95 68,053 42 22,138 23,65 25,999 28,144 58,124 61,777 66,206 69,336 43 22,86 24,398 26,785 28,965 59,304 62,99 67,459 70,616 44 23,584 25,148 27,575 29,787 60,481 64,201 68,71 71,892 45 24,311 25,901 28,366 30,612 61,656 65,41 69,957 73,166 46 25,041 26,657 29,16 31,439 62,83 66,616 71,201 74,437 47 25,775 27,416 29,956 32,268 64,001 67,821 72,443 75,704 48 26,511 28,177 30,754 33,098 65,171 69,023 73,683 76,969 49 27,249 28,941 31,555 33,93 66,339 70,222 74,919 78,231 50 27,991 29,707 32,357 34,764 67,505 71,42 76,154 79,49 60 35,534 37,485 40,482 43,188 79,082 83,298 88,379 91,952 70 43,275 45,442 48,758 51,739 90,531 95,023 100,425 104,215 80 51,172 53,54 57,153 60,391 101,879 106,629 112,329 116,321 90 59,196 61,754 65,647 69,126 113,145 118,136 124,116 128,299 100 67,328 70,065 74,222 77,929 124,342 129,561 135,807 140,17

Example 1

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)

 Score 1 2 3 4 5 6 Score 36 35 25 29 38 24

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.

 Observed (O) Expected (E) O-E (O-E)2 (O-E)2 /E 36 30 6 36 1,2 34 30 4 16 0,533 25 30 5 25 0,833 23 30 7 49 1,633 38 30 8 64 2,133 24 30 6 36 1,2 ∑ O = 180 ∑ E = 180 χ 2 =7,533

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)

Example 2

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

 Breaking load ( Newtons) 201 234 242 250 256 261 267 271 277 282 292 300 310 203 234 243 250 256 262 267 271 277 283 293 302 312 221 237 246 252 257 264 268 272 278 284 293 302 315 224 238 246 252 258 264 268 272 278 286 294 304 316 224 239 247 252 259 265 268 273 279 287 296 304 321 229 239 247 253 259 266 269 273 279 289 297 306 326 231 241 249 254 260 266 270 276 281 291 298 307 341 231 241 249 256 261 266 271 276 282 291 299 309 342

The average value of the breaking loads is xm = 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