Download presentation

Presentation is loading. Please wait.

Published byBraiden Hockin Modified over 2 years ago

1
Stats 2020 Tutorial

2
Chi-Square Goodness of Fit

3
Steps Age < 20Age 20-29Age ≥ 30 6892140 0.160.280.56 fofo pepe fefe What we know: n = 300, α =.05 and... The observed number (f o ) and percentage of drivers in each category:

4
Steps (cont.) 1. State the hypotheses: H o : The distribution of auto accidents is the same as the distribution of registered drivers. H 1 : The distribution of auto accidents is different/dependent/related to age.

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

6
Steps (cont.) 3. Calculate the chi-square statistic fe = pn Age < 20:.16(300) = 48 Age 20-29:.28(300) = 84 Age ≥ 30:.56(300) = 168 Age < 20Age 20-29Age ≥ 30 6892140 0.160.280.56 4884168 fofo pepe fefe Notice that for both the observed (fo) and expected (fe) frequency, that the sum of the frequencies should equal n.

7
Steps (cont.) Age < 20Age 20-29Age ≥ 30 6892140 0.160.280.56 4884168 fofo pepe fefe 2 = (68-48) 2 /48 (92-84) 2 /84 (140-168) 2 /168+ + = 8.3333 +.7619 + 4.6667 = 13.76

8
Steps (cont.) 4. State a decision and conclusion Decision: Critical 2 = 5.99 Obtained 2 = 13.76 Therefore, reject Ho Conclusion (in APA format) The distribution of automobile accidents is not identical to the distribution of registered drivers, 2 (2, n = 300) = 13.76, p <.05. df = 2

9
Chi-Square Goodness of Fit

10
Steps OriginalEyes fartherEyes closer 517227 0.33 fofo pepe fefe 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.) 1. State the hypotheses: H o : There is no preference among the three photographs. H 1 : There is a preference among the three photographs.

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

13
Steps (cont.) 3. Calculate the chi-square statistic fe = pn Original:.3333(150) = 50 Eyes farther:.3333(150) = 50 Eyes closer:.3333(150) = 50 OriginalEyes fartherEyes closer 517227 0.33 50 fofo pepe fefe

14
Steps (cont.) 2 = (51-50) 2 /50 (72-50) 2 /50 (27-50) 2 /50+ + =.02 + 9.68 + 10.58 = 20.28 OriginalEyes fartherEyes closer 517227 0.33 50 fofo pepe fefe

15
Steps (cont.) 4. State a decision and conclusion Decision: Critical 2 = 5.99 Obtained 2 = 20.28 Therefore, reject Ho Conclusion (in APA format) Participants showed significant preferences among the three photograph types, 2 (2, n = 150) = 20.28, p <.05.

16
Chi-Square Test for Independence

17
Steps What we know: n = 300, α =.05 and... Of the 300 participants, 100 are from the city, and 200 are from the suburbs FavourOppose City6832 Suburb86114 Opinion Residence Row totals Column totals154146 100 200 That is, 68+86 = 154 That is, 86+114 = 200

18
Steps (cont.) 1. State the hypotheses: H o : Opinion is independent of residence. That is, the frequency distribution of opinions has the same form for residents of the city and the suburbs. H 1 : Opinion is related to residence.

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

20
Steps (cont.) FavourOppose City 6832 Suburb 86114 Opinion Residence Row totals Column totals 154146 100 200 Cellfofe(fo-fe)(fo-fe) 2 (fo-fe) 2 /fe City favour 68 City oppose 32 Suburb favour 86 Suburb oppose 11 4

21
Steps (cont.) FavourOppose City 6832 Suburb 86114 Opinion Residence Row totals Column totals 154146 100 200 City frequencies fe favour = 154(100) / 300 = 51.33 fe oppose = 146(100) / 300 = 48.67 Suburb frequencies fe favour = 154(200) / 300 = 102.67 fe oppose = 146(200) / 300 = 97.33

22
Steps (cont.) Cellfofe(fo-fe)(fo-fe) 2 (fo-fe) 2 /fe City favour 6851.3316.67277.88895.4138 City oppose 3248.67-16.67277.88895.7097 Suburb favour 86102.67-16.67277.88892.7066 Suburb oppose 11497.3316.67277.88892.8551 2 = 5.4138 + 5.7097 + 2.7066 + 2.8551 = 16.69 3.Calculate chi-square statisic

23
Steps (cont.) 4. State a decision and conclusion Decision: Critical 2 = 3.84 Obtained 2 = 16.69 Therefore, reject Ho Conclusion (in APA format) Opinions in the city are different from those in the suburbs, 2 (1, n = 300) = 16.69, p <.05.

24
Steps (cont.) Part b) Phi-coefficient (effect size)? ɸ = √( 2 / 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
Step 1. Rank the X and Y Values X RANK Y RANK 2134521345 2143521435 The order of your X and Y values by increasing value

27
Step 2. Compute the correlation DD2D2 0 1 0 0011000110 X RANK Y RANK 2134521345 2143521435 2 = ΣD 2

28
Step 2. Cont. Using the Spearman formula, we obtain D2D2 r s = 1 - 6(2) 5(5 2 -1) = 1 - 12 5(24) = 1 - 12 120 = 1 -.1 = + 0.90

29
Mann-Whitney U AB

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

31
Steps (cont.) 1. State the hypotheses: H o : There is no difference between the two treatments. H 1 : There is a difference between the two treatments.

32
Steps (cont.) 2. Locate the critical region For a non-directional test with α =.05, and n A = 6, and n B = 6, the critical U = 5.

33
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: U A = the sum of the above points for Treatment A, therefore, U A = 6. Rank Score Sample Points for Treatment A 1 2 3 4 5 6 7 8 9 10 11 12 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

34
Steps (cont.) Alternatively, U A can be computed based on the sum of the Treatment A ranks. This is a less tedious option for large samples. R A = 6 + 7 + 8 + 9 + 10 + 11 = 51 Computation continued on the next slide Rank Score Sample Points for Treatment A 1 2 3 4 5 6 7 8 9 10 11 12 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

35
Steps (cont.) U A = n A n B + n B (n A +1) - R A 2 = 6(6) + 6(6+1) - 51 2 = 36 + 21 - 51 = 6 Rank Score Sample Points for Treatment A 1 2 3 4 5 6 7 8 9 10 11 12 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

36
Steps (cont.) Since U A + U B = n A n B and we know U A = 6 U B can be derived accordingly… U B = n A n B - U A = 6(6) - 6 = 36 - 6 = 30 The smaller U 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 DIFF. RANK POSITION -11 -2 -18 -7 4 -2 -14 -9 -5 1 Differences ranked from smallest to largest (in relation to 0) 8 3 10 6 4 2 9 7 5 1 FINAL RANK 8 2.5 10 6 4 2.5 9 7 5 1 Tied Diff. Use average of the ranks for the final rank (2+3)/2 = 2.5

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

41
Steps (cont.) 1. State the hypotheses: H o : There is no difference between the two treatments. H 1 : There is a difference between the two treatments.

42
Steps (cont.) 2. Locate the critical region For a non-directional test with α =.05, and n = 10, the critical T = 8. 3. Compute the sum of the ranks for the positive and negative difference scores: R + = 8+2.5+10+6+2.5+9+7+5 = 50 R - = 4+1 = 5 The Wilcoxon T is the smaller of these sums, therefore, T = 5.

43
Steps (cont.) 4. Decision and Conclusion T = 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.

Similar presentations

OK

Chapter 13. The Chi Square Test ( ) : is a nonparametric test of significance - used with nominal data -it makes no assumptions about the shape of the.

Chapter 13. The Chi Square Test ( ) : is a nonparametric test of significance - used with nominal data -it makes no assumptions about the shape of the.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on earth movements and major landforms in africa Ppt on art of war by sun tzu Slides for ppt on pollution Ppt on autonomous car project Ppt on jewellery management system Ppt on social media and nurses Ppt on geothermal energy in india Ppt on limitation act alberta Ppt on operation research in linear programming Download ppt on limits and derivatives