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Stats 2020 Tutorial. Chi-Square Goodness of Fit. f o. p e. f e. Steps. What we know: n = 300, α = .05 and... The observed number (f o ) and percentage of drivers in each category:. Steps (cont.).

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## Stats 2020 Tutorial

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**fo**pe fe Steps What we know: n = 300, α= .05 and... The observed number (fo) and percentage of drivers in each category:**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.**“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**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.**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**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**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**Steps (cont.)**• State the hypotheses:Ho: There is no preference among the three photographs.H1: There is a preference among the three photographs.**Steps (cont.)**• Locate the critical regiondf = C - 1 = 3 - 1 = 2For df = 2 and α= .05, the critical 𝝌2= 5.99**fo**pe fe Steps (cont.) • Calculate the chi-square statisticfe = pnOriginal:.3333(150) = 50Eyes farther:.3333(150) = 50Eyes closer:.3333(150) = 50**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**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.**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**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.**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**Opinion**Row totals Residence 100 200 Column totals 154 146 Steps (cont.)**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**Steps (cont.)**• Calculate chi-square statisic 𝝌2 = 5.4138 + 5.7097 + 2.7066 + 2.8551 = 16.69**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.**Steps (cont.)**• Part b) Phi-coefficient (effect size)? ɸ = √(𝝌2 / N) = √(.0556) = .236 Therefore, it is a small effect.**Spearman Correlation**What we know: n = 5 (that is, there are five X-Y pairs)**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**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**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**Mann-Whitney U**A B**Steps**What we know: nA = 6, nB = 6, α= .05 A B**Steps (cont.)**• State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.**Steps (cont.)**• Locate the critical regionFor a non-directional test with α = .05, andnA = 6, and nB = 6, the critical U = 5.**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.**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**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**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.**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.**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.**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**Steps (cont.)**• State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.**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.**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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