Download - Stats 2020 Tutorial
![Page 1: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/1.jpg)
Stats 2020 Tutorial
![Page 2: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/2.jpg)
Chi-Square Goodness of Fit
![Page 3: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/3.jpg)
Steps
Age < 20 Age 20-29 Age ≥ 30
68 92 140
0.16 0.28 0.56
fo
pe
fe
What we know:n = 300, α = .05
and...The observed
number (fo) and percentage of drivers in each
category:
![Page 4: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/4.jpg)
Steps (cont.)
1.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.
![Page 5: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/5.jpg)
Steps (cont.)
2.Locate the critical region
df = C - 1 = 3 - 1 = 2
For df = 2 and α = .05, the critical 𝝌2 = 5.99
“C” is the number of columns
![Page 6: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/6.jpg)
Steps (cont.)
3.Calculate the chi-square statisticfe = pn
Age < 20:.16(300) = 48
Age 20-29:.28(300) = 84
Age ≥ 30:.56(300) = 168
Age < 20 Age 20-29 Age ≥ 30
68 92 140
0.16 0.28 0.56
48 84 168
fo
pe
fe
Notice that for both the observed (fo) and expected (fe) frequency, that the sum of the frequencies should equal
n.
![Page 7: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/7.jpg)
Steps (cont.)
Age < 20 Age 20-29 Age ≥ 30
68 92 140
0.16 0.28 0.56
48 84 168
fo
pe
fe
𝝌2 = (68-48)2/48 (92-84)2/84(140-
168)2/168+ +
= 8.3333 + .7619 + 4.6667= 13.76
![Page 8: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/8.jpg)
Steps (cont.)4.State a decision and conclusion
Decision:Critical 𝝌2 = 5.99Obtained 𝝌2 = 13.76Therefore, reject Ho
Conclusion (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
![Page 9: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/9.jpg)
Chi-Square Goodness of Fit
![Page 10: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/10.jpg)
Steps
Original Eyes farther Eyes closer
51 72 27
0.33 0.33 0.33
fo
pe
fe
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
![Page 11: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/11.jpg)
Steps (cont.)
1.State the hypotheses:
Ho: There is no preference among the three photographs.H1: There is a preference among the three photographs.
![Page 12: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/12.jpg)
Steps (cont.)
2.Locate the critical region
df = C - 1 = 3 - 1 = 2
For df = 2 and α = .05, the critical 𝝌2 = 5.99
![Page 13: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/13.jpg)
Steps (cont.)
3.Calculate the chi-square statisticfe = pn
Original:.3333(150) = 50
Eyes farther:.3333(150) = 50
Eyes closer:.3333(150) = 50
Original Eyes farther Eyes closer
51 72 27
0.33 0.33 0.33
50 50 50
fo
pe
fe
![Page 14: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/14.jpg)
Steps (cont.)
𝝌2 = (51-50)2/50 (72-50)2/50 (27-50)2/50+ += .02 + 9.68 + 10.58= 20.28
Original Eyes farther Eyes closer
51 72 27
0.33 0.33 0.33
50 50 50
fo
pe
fe
![Page 15: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/15.jpg)
Steps (cont.)4.State a decision and conclusion
Decision:Critical 𝝌2 = 5.99Obtained 𝝌2 = 20.28Therefore, reject Ho
Conclusion (in APA format)Participants showed significant preferences among the three photograph types, 𝝌2 (2, n = 150) = 20.28, p < .05.
![Page 16: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/16.jpg)
Chi-Square Test for Independence
![Page 17: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/17.jpg)
Steps
What we know:n = 300, α = .05
and...Of the 300 participants,
100 are from the city, and200 are from the suburbs
Favour Oppose
City 68 32
Suburb 86 114
Opinion
Residence
Row totals
Column totals 154 146
100
200
That is, 68+86 = 154 That is, 86+114 = 200
![Page 18: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/18.jpg)
Steps (cont.)
1.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.
![Page 19: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/19.jpg)
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 𝝌2 = 3.84
![Page 20: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/20.jpg)
Steps (cont.)
Favour Oppose
City 68 32
Suburb 86 114
Opinion
Residence
Row totals
Column totals
154 146
100
200
Cell fo fe(fo-fe)
(fo-fe)2 (fo-fe)2/fe
City favour
68
City oppos
e32
Suburb
favour86
Suburb
oppose
114
![Page 21: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/21.jpg)
Steps (cont.)
Favour Oppose
City 68 32
Suburb 86 114
Opinion
Residence
Row totals
Column totals
154 146
100
200
City frequenciesfefavour = 154(100) / 300 = 51.33feoppose = 146(100) / 300 = 48.67
Suburb frequenciesfefavour = 154(200) / 300 = 102.67feoppose = 146(200) / 300 = 97.33
![Page 22: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/22.jpg)
Steps (cont.)Cell fo fe (fo-fe) (fo-fe)2 (fo-fe)2/fe
City favour
68 51.33 16.67277.888
95.4138
City oppose
32 48.67 -16.67277.888
95.7097
Suburb favour
86102.6
7-16.67
277.8889
2.7066
Suburb oppose
114 97.33 16.67277.888
92.8551
𝝌2 = 5.4138 + 5.7097 + 2.7066 + 2.8551
= 16.69
3.Calculate chi-square statisic
![Page 23: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/23.jpg)
Steps (cont.)
4.State a decision and conclusion
Decision:Critical 𝝌2 = 3.84Obtained 𝝌2 = 16.69Therefore, reject Ho
Conclusion (in APA format)Opinions in the city are different from those in the suburbs, 𝝌2 (1, n = 300) = 16.69, p < .05.
![Page 24: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/24.jpg)
Steps (cont.)• Part b) Phi-coefficient
(effect size)?
ɸ = √(𝝌2 / N)
= √(.0556) = .236
Therefore, it is a small effect.
![Page 25: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/25.jpg)
Spearman Correlation
What we know:n = 5
(that is, there are five X-Y pairs)
![Page 26: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/26.jpg)
Step 1. Rank the X and Y Values
XRANK YRANK
21345
21435
The order of your X and Y values by increasing value
![Page 27: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/27.jpg)
Step 2. Compute the correlation
D D2
00-110
00110
XRANK YRANK
21345
21435
2 = ΣD2
![Page 28: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/28.jpg)
Step 2. Cont.Using the Spearman formula, we obtain
D2
rs = 1 - 6(2)5(52-1)
= 1 - 125(24)
= 1 - 12120
= 1 - .1 = + 0.90
![Page 29: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/29.jpg)
Mann-Whitney U
A B
![Page 30: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/30.jpg)
StepsWhat we know:
nA = 6, nB = 6, α = .05
A B
![Page 31: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/31.jpg)
Steps (cont.)
1.State the hypotheses:
Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.
![Page 32: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/32.jpg)
Steps (cont.)2.Locate the critical region
For a non-directional test with α = .05, andnA = 6, and nB = 6, the critical U = 5.
![Page 33: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/33.jpg)
Steps (cont.)
Step 3:First: Identify the scores for treatment ASecond: 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.
RankScoreSamplePoints for Treatment A
1 2 3 4 5 6 7 8 9 10 11 129 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
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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
RankScoreSamplePoints for Treatment A
1 2 3 4 5 6 7 8 9 10 11 129 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
![Page 35: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/35.jpg)
Steps (cont.)
UA = nAnB +nB(nA+1)- 𝚺 RA
2
= 6(6) + 6(6+1) - 512
= 36 + 21 - 51
= 6
RankScoreSamplePoints for Treatment A
1 2 3 4 5 6 7 8 9 10 11 129 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
![Page 36: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/36.jpg)
Steps (cont.)Since
UA + UB = nAnB
and we know UA = 6UB can be derived
accordingly…
UB = nAnB - UA
= 6(6) - 6= 36 - 6
= 30
The smaller U value is the Mann-Whitney U statistic, so U = 6.
![Page 37: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/37.jpg)
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.
![Page 38: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/38.jpg)
Wilcoxon Signed-Ranks Test
![Page 39: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/39.jpg)
Steps
DIFF.RANK
POSITION
-11-2
-18-74-2
-14-9-51
Differences ranked from smallest to
largest (in relation to 0)
83
106429751
FINAL RANK
82.51064
2.59751
TiedDiff.
Use average of the ranks for the final
rank(2+3)/2 = 2.5
![Page 40: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/40.jpg)
Steps
DIFF.
112
187-42
1495-1
FINAL RANK
82.51064
2.59751
![Page 41: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/41.jpg)
Steps (cont.)
1.State the hypotheses:
Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.
![Page 42: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/42.jpg)
Steps (cont.)2.Locate the critical region
For a non-directional test with α = .05, andn = 10, the critical T = 8.
3.Compute the sum of the ranks for the positive and negative difference scores:
𝚺R+ = 8+2.5+10+6+2.5+9+7+5 = 50 𝚺R- = 4+1 = 5
The Wilcoxon T is the smaller of these sums, therefore, T = 5.
![Page 43: Stats 2020 Tutorial](https://reader035.vdocuments.pub/reader035/viewer/2022062321/568139ba550346895da159e5/html5/thumbnails/43.jpg)
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.