1. CHI SQUARE DISTRIBUTION

2. The Chi-Square (  2 ) Distribution The chi-square distribution is the probability distribution of the sum of several independent, squared standard normal random variables.

3. The Chi-Square (  2 ) Distribution The chi-square random variable cannot be negative, so it is bound by zero on the left. The chi-square distribution is skewed to the right. The chi-square distribution approaches a normal as the degrees of freedom increase.

4. Area in Right Tail.995.990.975.950.900.100.050.025.010.005 Area in Left Tail df.005.010.025.050.100.900.950.975.990.995 10.00003930.0001570.0009820.0003930.01582.713.845.026.637.88 20.01000.02010.05060.1030.2114.615.997.389.2110.60 30.07170.1150.2160.3520.5846.257.819.3511.3412.84 40.2070.2970.4840.7111.067.789.4911.1413.2814.86 50.4120.5540.8311.151.619.2411.0712.8315.0916.75 60.6760.8721.241.642.2010.6412.5914.4516.8118.55 70.9891.241.692.172.8312.0214.0716.0118.4820.28 81.341.652.182.733.4913.3615.5117.5320.0921.95 91.732.092.703.334.1714.6816.9219.0221.6723.59 102.162.563.253.944.8715.9918.3120.4823.2125.19 112.603.053.824.575.5817.2819.6821.9224.7226.76 123.073.574.405.236.3018.5521.0323.3426.2228.30 133.574.115.015.897.0419.8122.3624.7427.6929.82 144.074.665.636.577.7921.0623.6826.1229.1431.32 154.605.236.267.268.5522.3125.0027.4930.5832.80 165.145.816.917.969.3123.5426.3028.8532.0034.27 175.706.417.568.6710.0924.7727.5930.1933.4135.72 186.267.018.239.3910.8625.9928.8731.5334.8137.16 196.847.638.9110.1211.6527.2030.1432.8536.1938.58 207.438.269.5910.8512.4428.4131.4134.1737.5740.00 218.038.9010.2811.5913.2429.6232.6735.4838.9341.40 228.649.5410.9812.3414.0430.8133.9236.7840.2942.80 239.2610.2011.6913.0914.8532.0135.1738.0841.6444.18 249.8910.8612.4013.8515.6633.2036.4239.3642.9845.56 2510.5211.5213.1214.6116.4734.3837.6540.6544.3146.93 2611.1612.2013.8415.3817.2935.5638.8941.9245.6448.29 2711.8112.8814.5716.1518.1136.7440.1143.1946.9649.65 2812.4613.5615.3116.9318.9437.9241.3444.4648.2850.99 2913.1214.2616.0517.7119.7739.0942.5645.7249.5952.34 3013.7914.9516.7918.4920.6040.2643.7746.9850.8953.67 Values and Probabilities of Chi-Square Distributions

5. Test of Goodness of Fit It enable us to determine how good a fit is between observed and expected frequencies. Where the former is the outcome of samples, the later are determined from the hypothesized population from which the sample is taken. Conventionally the Null hypothesis is formed that says there is no difference between the obtained and expected frequencies, i.e there is a good between observed and expected.

6. Test of Independence Here we try to test whether two different types of attribute of same population are statistically independent or not. Conventionally the Null hypothesis is formed that says there is no difference between the obtained and expected frequencies. In other words attributes under study are independent of each other.

7. The Chi-Square Distribution  The chi-square  ² distribution depends on the number of degrees of freedom  A chi-square point  ² α is the point under a chi-square distribution that gives right-hand tail area 

8. 14-9 A Chi-Square Test for Goodness of Fit Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi-square distribution (with k-1 degrees of freedom) to test the null hypothesis Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi-square distribution (with k-1 degrees of freedom) to test the null hypothesis

9. The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis Assuming equal probabilities, p 1 = p 2 = p 3 = p 4 =0.25 and n=80 PreferenceTanBrownMaroonBlackTotal Observed1240 82080 Expected(np)2020 202080 (O-E)-820-12 00 Goodness-of-Fit Test

10. 50-5 0.4 0.3 0.2 0.1 0.0 z f ( z ) Partitioning the Standard Normal Distribution 1 -0.440.44 0.1700 0.1713 0.1587 0.1700 0.1713 1. Use the table of the standard normal distribution to determine an appropriate partition of the standard normal distribution which gives ranges with approximately equal percentages. p(z<-1) = 0.1587 p(-1<z<-0.44)= 0.1713 p(-0.44<z<0)= 0.1700 p(0<z<0.44)= 0.1700 p(0.44<z<14)= 0.1713 p(z>1) = 0.1587 2. Given z boundaries, x boundaries can be determined from the inverse standard normal transformation: x =  +  z = 125 + 40z. 3. Compare with the critical value of the  2 distribution with k-3 degrees of freedom. Goodness-of-Fit for the Normal Distribution

11. iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755 iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755  2 (0.10,k-3) = 6.5139 > 1.11755  H 0 is not rejected at the 0.10 level Solution

12. Test for Independence

13. Null and alternative hypotheses: H 0 : The two classification variables are independent of each other H 1 : The two classification variables are not independent Chi-square test statistic for independence: Degrees of freedom: df=(r-1)(c-1) Expected cell count: A and B are independent if:P(A  B) = P(A)  P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n A and B are independent if:P(A  B) = P(A)  P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n Contingency Table Analysis: A Chi-Square Test for Independence

14. ijOEO-E(O-E) 2 (O-E) 2 /E 114228.813.2174.246.0500 121831.2-13.2174.245.5846 21619.2-13.2174.249.0750 223420.813.2174.248.3769  2 : 29.0865  2 (0.01,(2-1)(2-1)) =6.63490 H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Contingency Table Analysis:

15. F - Distribution When independent samples of size n1 and n2 are drawn from two normal population, the ratio Follows F- distribution with n1-1 and n2-1 df. Where s1 and s2 are sample variance and

16. Test of Independence Here we try to test whether two different types of attribute of same population are statistically independent or not. Conventionally the Null hypothesis is formed that says there is no difference between the obtained and expected frequencies. In other words attributes under study are independent of each other.

17. Test of Independence Here we try to test whether two different types of attribute of same population are statistically independent or not. Conventionally the Null hypothesis is formed that says there is no difference between the obtained and expected frequencies. In other words attributes under study are independent of each other.

18. Test of Independence Here we try to test whether two different types of attribute of same population are statistically independent or not. Conventionally the Null hypothesis is formed that says there is no difference between the obtained and expected frequencies. In other words attributes under study are independent of each other.