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Chi square Test

Chi square Test, Non Parametric Test, Applications of Chi square Test,

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Chi square Test

  1. 1. Chi Square Test RAHUL BABAR
  2. 2. Chi Square Test Simplest & the most widely used non-parametric test in statistical work
  3. 3. NON PARAMETRIC TEST The test don’t rely on assumptions about population parameters: Population Mean Population Standard Deviation
  4. 4. Chi Square Sum Observed frequency Expected (theoretical) frequency, asserted by the null hypothesis
  5. 5. Null Hypothesis There is no significant difference between the observed and expected frequencies
  6. 6. Chi-square distribution values Degree of freedom : number of elements that can be chosen freely Critical value
  7. 7. Degree of freedom Goodness of Fit Independent Variables & Homogeneity of proportion Number of outcomes – 1 ( No of rows – 1)( No of Columns -1) ( r - 1 )( c - 1 )
  8. 8. Applications of Chi-square test Goodness of fit Independent variables Homogeneity of proportion
  9. 9. Goodness of fit Test of the consistency between a hypothetical and a sample distribution. Example : Coin is tossed 50 times Null hypothesis is that it’s 25 times heads and 25 times tails ( expected observation E )
  10. 10. Event Frequency Head 28 Tail 22 Total 50 Observed Values is, Event O E O - E ( O – E ) ² ( O – E ) ² E Head 28 25 3 9 0.36 Tail 22 25 -3 9 0.36 Total 0.72 Solution : = 0.72 Degree of freedom is = (outcomes – 1) = 2 – 1 = 1 Critical value is 3.841 for df 1 at 0.05 Since 0.72 < 3.841 , our null hypothesis is acceptable which means our coin is fair
  11. 11. Test of Independence To ascertain whether there is any dependency relationship between the two attributes. Example: A company introduced new drug B to cure malaria. It is being compared with existing drug A. Data is shown in next slide. We need to find whether the new drug B is more effective in curing malaria.
  12. 12. Helped (C1) Harmed (C2) No effect (C3) Total Drug A (R1) 44 10 26 80 Drug B (R2) 52 10 18 80 Total 96 20 44 160 Solution : O E O - E ( O-E ) ² ( O – E ) ² E R1C1 44 48 -4 16 0.333 R1C2 10 10 0 0 0 R1C3 26 22 4 16 0.727 R2C1 52 48 4 16 0.333 R2C2 10 10 0 0 0 R2C3 18 22 -4 16 0.727 Total = 2.12
  13. 13. We setup two hypothesis, H0 :There is no difference in the effectiveness of the two drugs H1 : There is difference in the effectiveness of the two drugs Degree of freedom = (r-1)(c-1) = (2-1)(3-1) = 2 Critical value at 0.05 for df 2 is 5.991, and 2.12 < 5.991 So, there is no difference in the effectiveness of the two drugs
  14. 14. Alternative method: Helped Harmed No effect Total Drug A 44 (a) 10 (b) 26 (c) 80 (a+b+c) Drug B 52 (d) 10 (e) 18 (f) 80 (d+e+f) Total 96 (a+d) 20 (b+e) 44 (c+f) 160 (N) Its 2x3 table so for calculating chi-square we use formula: = N a² + b² + c² + N d² + e² + f² - N a+b+c a+d b+e c+f d+e+f a+d b+e c+f = 160 44² + 10² + 26² + 160 52² + 10² + 18² - 160 80 96 20 44 80 96 20 44 = 2.12 Both methods give same value for which is 2.12
  15. 15. Test of homogeneity Test indicates whether the proportions of elements belonging to different groups in two or more populations are similar or not. Example : A company has two factories in Delhi and Mumbai. It is interested to know whether its workers are satisfied with their jobs or not at both places.
  16. 16. Delhi Mumbai Total Fully satisfied 50 70 120 Moderately satisfied 90 110 200 Moderately dissatisfied 160 130 290 Fully dissatisfied 200 190 390 Total 500 500 1000 We setup two hypothesis, H0 :The proportions of workers who belong to the four job satisfaction categories are the same in both Delhi and Mumbai H1 : The proportions of workers who belong to the four job satisfaction categories are not the same in both Delhi and Mumbai Degrees of freedom = (r-1)(c-1) = (4-1)(2-1) = 3 Critical value for 3 df at 0.05 is 7.815
  17. 17. Delhi Mumbai O E O E Fully satisfied 50 60 70 60 Moderately satisfied 90 100 110 100 Moderately dissatisfied 160 145 130 145 Fully dissatisfied 200 195 190 195 = (50-60) ² + (90-100) ² + (160-145) ² + (200-195) ² + (70-60) ² + (110-100) ² + (130-145) ² + (190-195) ² 60 100 145 195 60 100 145 195 = 1.667 + 1.000 + 1.552 + 0.128 + 1.667 + 1.000 + 1.522 + 0.128 = 8.694 Since the value 8.694 > 7.815 we therefore, reject the null hypothesis and conclude that the distribution of job satisfaction for workers in Delhi and Mumbai is not homogeneous
  18. 18. IMPORTANT CHARACTERISTICS OF A CHI SQUARE TEST  This test (as a non-parametric test) is based on frequencies and not on the parameters like mean and standard deviation.  The test is used for testing the hypothesis and is not useful for estimation.  This test can also be applied to a complex contingency table with several classes and as such is a very useful test in research work.  This test is an important non-parametric test as no rigid assumptions are necessary in regard to the type of population, no need of parameter values and relatively less mathematical details are involved.

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Chi square Test, Non Parametric Test, Applications of Chi square Test,

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