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Chi-Square Goodness of Fit and Independence Testing Tutorial Steps

This tutorial covers the chi-square goodness of fit and independence tests as part of statistical analysis. It provides detailed steps for conducting these tests using sample data, including defining hypotheses, locating critical regions, calculating chi-square statistics, and making conclusions in APA format. Both examples and formulas are included to ensure comprehensive understanding, covering scenarios such as automobile accident distributions and preferences among photographs. The importance of critical chi-square values and decision making in hypothesis testing is emphasized for accurate data interpretation.

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Chi-Square Goodness of Fit and Independence Testing Tutorial Steps

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  1. Stats 2020 Tutorial

  2. Chi-Square Goodness of Fit

  3. fo pe fe Steps What we know: n = 300, α= .05 and... The observed number (fo) and percentage of drivers in each category:

  4. Steps (cont.) • State the hypotheses:Ho: The distribution of auto accidents is the same as the distribution of registered drivers.H1: The distribution of auto accidents is different/dependent/related to age.

  5. “C” is the number of columns Steps (cont.) • Locate the critical regiondf = C - 1 = 3 - 1 = 2For df = 2 and α= .05, the critical 𝝌2= 5.99

  6. fo pe fe Steps (cont.) • Calculate the chi-square statisticfe = pnAge < 20:.16(300) = 48Age 20-29:.28(300) = 84Age ≥ 30:.56(300) = 168 Notice that for both the observed (fo) and expected (fe) frequency, that the sum of the frequencies should equal n.

  7. fo pe fe Steps (cont.) + + (140-168)2/168 𝝌2 = (68-48)2/48 (92-84)2/84 = 8.3333 + .7619 + 4.6667 = 13.76

  8. Steps (cont.) • State a decision and conclusionDecision:Critical𝝌2 = 5.99Obtained 𝝌2 = 13.76Therefore, reject HoConclusion (in APA format)The distribution of automobile accidents is not identical to the distribution of registered drivers, 𝝌2 (2, n = 300) = 13.76, p < .05. df = 2

  9. Chi-Square Goodness of Fit

  10. fo pe fe Steps What we know: n = 150, α= .05 and... Assuming all groups are equal, we divide our proportions equally into 3: 1/3 = .3333 for each proportion

  11. Steps (cont.) • State the hypotheses:Ho: There is no preference among the three photographs.H1: There is a preference among the three photographs.

  12. Steps (cont.) • Locate the critical regiondf = C - 1 = 3 - 1 = 2For df = 2 and α= .05, the critical 𝝌2= 5.99

  13. fo pe fe Steps (cont.) • Calculate the chi-square statisticfe = pnOriginal:.3333(150) = 50Eyes farther:.3333(150) = 50Eyes closer:.3333(150) = 50

  14. fo pe fe Steps (cont.) + + (27-50)2/50 𝝌2 = (51-50)2/50 (72-50)2/50 = .02 + 9.68 + 10.58 = 20.28

  15. Steps (cont.) • State a decision and conclusionDecision:Critical𝝌2 = 5.99Obtained 𝝌2 = 20.28Therefore, reject HoConclusion (in APA format)Participants showed significant preferences among the three photograph types, 𝝌2 (2, n = 150) = 20.28, p < .05.

  16. Chi-Square Test for Independence

  17. Opinion Row totals Residence 100 200 Column totals 154 146 Steps What we know: n = 300, α= .05 and... Of the 300 participants, 100 are from the city, and200 are from the suburbs That is, 68+86 = 154 That is, 86+114 = 200

  18. Steps (cont.) • State the hypotheses:Ho: Opinion is independent of residence. That is, the frequency distribution of opinions has the same form for residents of the city and the suburbs.H1: Opinion is related to residence.

  19. Steps (cont.) • Locate the critical regiondf = (# of columns - 1) (# of rows -1) = (2 - 1) (2 - 1) = 1 x 1 = 1For df = 1 and α= .05, the critical 𝝌2= 3.84

  20. Opinion Row totals Residence 100 200 Column totals 154 146 Steps (cont.)

  21. Opinion Row totals Residence 100 200 Column totals 154 146 Steps (cont.) City frequencies fefavour = 154(100) / 300 = 51.33feoppose = 146(100) / 300 = 48.67 Suburb frequencies fefavour = 154(200) / 300 = 102.67feoppose = 146(200) / 300 = 97.33

  22. Steps (cont.) • Calculate chi-square statisic 𝝌2 = 5.4138 + 5.7097 + 2.7066 + 2.8551 = 16.69

  23. Steps (cont.) • State a decision and conclusionDecision:Critical𝝌2 = 3.84Obtained 𝝌2 = 16.69Therefore, reject HoConclusion (in APA format)Opinions in the city are different from those in the suburbs, 𝝌2 (1, n = 300) = 16.69, p < .05.

  24. Steps (cont.) • Part b) Phi-coefficient (effect size)? ɸ = √(𝝌2 / N) = √(.0556) = .236 Therefore, it is a small effect.

  25. Spearman Correlation What we know: n = 5 (that is, there are five X-Y pairs)

  26. XRANK YRANK 2 1 3 4 5 2 1 4 3 5 Step 1. Rank the X and Y Values The order of your X and Y values by increasing value

  27. XRANK YRANK 2 1 3 4 5 2 1 4 3 5 Step 2. Compute the correlation D D2 0 0 -1 1 0 0 0 1 1 0 2 = ΣD2

  28. rs = 1 - 6(2) = 1 - 12 = 1 - 12 D2 5(52-1) 5(24) 120 Step 2. Cont. Using the Spearman formula, we obtain = 1 - .1 = + 0.90

  29. Mann-Whitney U A B

  30. Steps What we know: nA = 6, nB = 6, α= .05 A B

  31. Steps (cont.) • State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.

  32. Steps (cont.) • Locate the critical regionFor a non-directional test with α = .05, andnA = 6, and nB = 6, the critical U = 5.

  33. 1 2 3 4 5 6 7 8 9 10 11 12 Rank Score Sample Points for Treatment A 9 10 12 14 17 37 39 40 41 44 45 104 B B B B B A A A A A A B 1 1 1 1 1 1 Steps (cont.) Step 3: First: Identify the scores for treatment A Second: For each treatment A score, count how many scores in treatment B have a higher rank. Third: UA = the sum of the above points for Treatment A, therefore, UA = 6.

  34. 1 2 3 4 5 6 7 8 9 10 11 12 Rank Score Sample Points for Treatment A 9 10 12 14 17 37 39 40 41 44 45 104 B B B B B A A A A A A B 1 1 1 1 1 1 Steps (cont.) Alternatively, UA can be computed based on the sum of the Treatment A ranks. This is a less tedious option for large samples. 𝚺 RA= 6 + 7 + 8 + 9 + 10 + 11 = 51 Computation continued on the next slide

  35. nB(nA+1) 6(6+1) 1 2 3 4 5 6 7 8 9 10 11 12 Rank Score Sample Points for Treatment A - 𝚺 RA - 51 UA= nAnB + = 6(6) + 9 10 12 14 17 37 39 40 41 44 45 104 2 2 B B B B B A A A A A A B 1 1 1 1 1 1 Steps (cont.) = 36 + 21 - 51 = 6

  36. Steps (cont.) Since UA + UB = nAnB and we know UA = 6 UB can be derived accordingly… UB = nAnB - UA = 6(6) - 6 = 36 - 6 = 30 The smallerU value is the Mann-Whitney U statistic, so U = 6.

  37. Steps (cont.) Step 4: Decision and Conclusion U = 6 is greater than the critical value of U = 5, therefore we fail to reject Ho. The treatment A and B scores were rank-ordered and a Mann-Whitney U-test was used to compare the ranks for Treatment A (n=6) and B (n=6). The results show no significant difference between the two treatments, U = 6, p > .05, with the sum of the ranks equal to 51 for treatment A and 27 for treatment B.

  38. Wilcoxon Signed-Ranks Test

  39. Steps Differences ranked from smallest to largest (in relation to 0) FINAL RANK RANK POSITION DIFF. 8 3 10 6 4 2 9 7 5 1 8 2.5 10 6 4 2.5 9 7 5 1 -11 -2 -18 -7 4 -2 -14 -9 -5 1 Use average of the ranks for the final rank (2+3)/2 = 2.5 TiedDiff.

  40. FINAL RANK DIFF. 11 2 18 7 -4 2 14 9 5 -1 8 2.5 10 6 4 2.5 9 7 5 1 Steps

  41. Steps (cont.) • State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.

  42. Steps (cont.) • Locate the critical regionFor a non-directional test with α = .05, andn = 10, the critical T = 8. • Compute the sum of the ranks for the positive and negative difference scores:𝚺R+ = 8+2.5+10+6+2.5+9+7+5 = 50𝚺R- = 4+1 = 5The Wilcoxon T is the smaller of these sums, therefore, T = 5.

  43. Steps (cont.) • Decision and ConclusionT = 5 is less than the critical value of T = 8, therefore we reject Ho.The treatment I and II scores were rank-ordered by the magnitude in difference scores, and the data was evaluated using the Wilcoxon T. The results show a significant difference in scores, T = 5, p < .05, with the ranks for increases totalling 50, and for decreases totalling 5.

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