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- 1. CHI SQUARE & ANOVA
- 2. DIFFERENCE BETWEEN CHI SQUARE & ANOVA BirinderSingh,AssistantProfessor,PCTE Ludhiana 2 It enables us to test whether more than two population proportions can be considered equal Analysis of Variance (Anova) enables us to test whether more than two population means can be considered equal. Chi Square (χ2 Test) Anova (F Test)
- 3. CHARACTERISTICS OF CHI SQUARE Every Chi square distribution extends indefinitely to right from zero. It is skewed to right As df increases, Chi square curve become more bell shaped and approaches normal distribution. Its mean is degree of freedom Its variance is twice degree of freedom 3 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 4. CHI SQUARE (Χ2 TEST) Chi Square Test deals with analysis of categorical data in terms of frequencies / proportions / percentages. It is primarily of three types: Test of Homogeneity: To determine whether different population are similar w.r.t some characteristics. Test of Independence: Tests whether the characteristics of the elements of the same population are related or independent. Test of Goodness of Fit: To determine whether there is a significant difference between an observed frequency distribution and theoretical probability distribution. 4 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 5. COMPUTATIONAL PROCEDURE – CHI SQUARE TEST Formulate Null & Alternative Hypothesis State type of test Select LOS Compute expected frequencies assuming H0 to be true. Compute χ2 calculated value using 𝜒2 cal = (𝑓𝑜 −𝑓𝑒)2 𝑓𝑒 Extract 𝜒2 crit value from table Compare 𝜒2 cal & 𝜒2 crit and make decision. 5 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 6. CHI SQUARE – TEST OF HOMOGENEITY Based on a study, it is expected that 50% of the students opt for marketing, 30% for finance and 20% for HR. In a sample for 100, it was observed that 61, 24 and 15 opt for these subjects respectively. Do you agree with study findings at 10% LOS? (𝜒2 cal = 4.87) (𝜒2 crit = 4.605) (Rejected) 6 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 7. CHI SQUARE – TEST OF HOMOGENEITY A shoe seller has received the consignment of the order that he had placed for 10000 pair of different sizes. Without physically segregating the sizes and counting no. of pair of shoes of each size, he wants to ascertain that consignment received is as per order . Check at 5% LOS. (𝜒2 cal = 6.87) (𝜒2 crit = 12.592) (Accepted) 7 BirinderSingh,AssistantProfessor,PCTE Ludhiana Size 4 5 6 7 8 9 10 Order qty 500 1500 2000 2000 2000 1500 500 Rec. qty 700 1800 2200 2000 2000 1300 0
- 8. CHI SQUARE – TEST OF INDEPENDENCE FORMULAE TO BE USED Computation of expected frequency Fe = (RT x CT) / GT where RT = Row Total CT = Column Total GT = Grand Total Computation of degree of freedom Df = (r – 1) (c – 1) 8 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 9. CHI SQUARE – TEST OF INDEPENDENCE Following data was collected when a survey was carried out on preference for formal wear in work place: Is there difference in preference due to sex? Check at 20% LOS. (𝜒2 cal = 0.2522) (𝜒2 crit = 1.642) (Accepted) 9 BirinderSingh,AssistantProfessor,PCTE Ludhiana Gents Ladies Yes 520 60 No 80 11
- 10. CHI SQUARE – TEST OF INDEPENDENCE Sample data in respect of viewership of a TV Program for various age groups was collected and is as follows: Is viewership of program independent of age ? Check at 5% LOS. (𝜒2 cal = 26.01) (𝜒2 crit = 9.488) (Rejected) 10 BirinderSingh,AssistantProfessor,PCTE Ludhiana 15-25 26-40 41-50 Always 75 180 105 Sometimes 50 60 40 Never 25 20 5
- 11. CHI SQUARE – TEST OF GOODNESS OF FIT Gordon Company requires that college seniors who are seeking positions will be interviewed. For staffing purposes, the director of recruitment thinks that the interview process can be approximated by a binomial distribution with p = 0.40 i.e. Can he conclude that BD at p = 0.4 provides a good description of observed frequencies. Check at 20% LOS. (𝜒2 cal = 5.041) (𝜒2 crit = 4.642) (Rejected) 11 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 12. ANALYSIS OF VARIANCE (ANOVA) It enables us to test for the significance of the differences among more than two sample means. Using Anova, we will be able to make inferences about whether our samples are drawn from population having the same mean. Examples: Comparing the mileage of five different brands of cars Testing which of the four different training methods produces the fastest learning record Comparing the average salary of three different companies In each of these cases, we would compare the means of more than two sample means. F-Distribution is used to analyze certain situations 12 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 13. ASSUMPTIONS Populations are normally distributed Samples are random and independent Population Variances are equal. 13 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 14. COMPUTATIONAL PROCEDURE IN ANOVA (ONE WAY) Define Null & Alternative Hypothesis Select estimator & determine its distribution Select Significance Level Calculate Sum of all observations: T = Ʃxi Calculate correction factor: CF = T2 / nT where nT = sample size Calculate Sum of squares total, SST = Σ(Σ𝑥𝑖 2) − CF Calculate Sum of squares between columns, SSB = Σ((Σ𝑥𝑖)2 /𝑛𝑖) − CF Calculate Sum of squares within columns, SSW = SST – SSB Calculate Mean of squares between groups, MSB = SSB / (k – 1) where k = no. of samples Calculate Mean of squares within groups, MSW = SSW / (nT – k) Calculate Fcal = MSB / MSW Calculate Fcrit = F(dfnum, dfden, α) where dfnum = k – 1, dfden = nT – k Compare Fcal & Fcrit and make your statistical & managerial decisions 14 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 15. PRACTICE PROBLEM – ONE WAY ANOVA Three group of students are taught a statistical technique by three different methods. When tested on one problem, sample scores of 3 students selected at random from each of the group as under: At 0.05 LOS, do the means of populations taught by three methods differ? (Fcal = 1.5, Fcrit = 5.14) (Accepted) 15 BirinderSingh,AssistantProfessor,PCTE Ludhiana Group 1 Group 2 Group 3 3 5 3 4 7 7 5 6 5
- 16. 16 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 17. PRACTICE PROBLEM – ONE WAY ANOVA IAA wanted to find out if average sale of small cars namely Swift, Jazz & Figo is same in Tier II cities. It obtained quarterly sales data from 5 such cities A,B,C,D,E as shown. What conclusion can be drawn at 0.05 LOS? (Fcal = 1.5, Fcrit = 5.14) (Accepted) 17 BirinderSingh,AssistantProfessor,PCTE Ludhiana City Swift Jazz Figo A 32 26 30 B 28 34 - C 25 - 28 D 34 33 32 E 31 31 26
- 18. COMPUTATIONAL PROCEDURE IN ANOVA (TWO WAY) Define Null & Alternative Hypothesis Select estimator & determine its distribution Select Significance Level Calculate Sum of all observations: T = Ʃxi Calculate correction factor: CF = T2 / nT where nT = sample size Calculate Sum of squares columns, SSC = Σ((Σ𝑥𝑗)2 /𝑛𝑖) − CF Calculate Sum of squares rows, SSR = Σ((Σ𝑥𝑖)2/𝑛𝑗) − CF Calculate Sum of squares total, SST = Σ(Σ𝑥𝑖 2 ) − CF Calculate Sum of square error, SSE = SST – (SSC + SSR) Calculate Mean of squares column, MSC = SSC / (c – 1) Calculate Mean of squares rows, MSR = SSR / (r – 1) Calculate Mean of squares error, MSE = SSE / (c – 1) (r – 1) Calculate Fcal = MSC/MSE & MSR/MSE Calculate Fcrit = F(dfnum, dfden, α) where dfnum = c-1 or r-1, dfden = (c-1)(r-1) Compare Fcal & Fcrit and make your statistical & managerial decisions 18 BirinderSingh,AssistantProfessor,PCTE Ludhiana
- 19. PRACTICE PROBLEM – TWO WAY ANOVA Three salesmen Kallu, Lallu & Mallu were assigned three cities A,B & C. The data on sales for quarter ending June 2017 achieved by them is: Is there any significant difference in sales made by 3 of them? Is there any significant difference in sales made in 3 cities? Check at 0.05 LOS? (F1cal = 9.23, F1crit = 6.94) (F2cal = 3.25, F2crit = 6.94) (Rejected) & (Accepted) 19 BirinderSingh,AssistantProfessor,PCTE Ludhiana City Kallu Mallu Lallu A 4 3 4 B 3 2 5 C 5 3 6
- 20. DECISION FLOW DIAGRAM - ESTIMATION 20 BirinderSingh,AssistantProfessor,PCTE Ludhiana Start Is n≥30 Is pop. Known to be normally distributed Use ‘Z’ table Stop Use a Statistician Is SD known ? Use ‘Z’ table Stop Use ‘t’ table Stop