Stats 2020 Tutorial
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Transcript of Stats 2020 Tutorial
Stats 2020 Tutorial
Chi-Square Goodness of Fit
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:
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.
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
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.
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
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
Chi-Square Goodness of Fit
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
Steps (cont.)
1.State the hypotheses:
Ho: There is no preference among the three photographs.H1: There is a preference among the three photographs.
Steps (cont.)
2.Locate the critical region
df = C - 1 = 3 - 1 = 2
For df = 2 and α = .05, the critical 𝝌2 = 5.99
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
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
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.
Chi-Square Test for Independence
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
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.
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
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
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
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
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.
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)
Step 1. Rank the X and Y Values
XRANK YRANK
21345
21435
The order of your X and Y values by increasing value
Step 2. Compute the correlation
D D2
00-110
00110
XRANK YRANK
21345
21435
2 = ΣD2
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
Mann-Whitney U
A B
StepsWhat we know:
nA = 6, nB = 6, α = .05
A B
Steps (cont.)
1.State the hypotheses:
Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.
Steps (cont.)2.Locate the critical region
For a non-directional test with α = .05, andnA = 6, and nB = 6, the critical U = 5.
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
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
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
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.
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.
Wilcoxon Signed-Ranks Test
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
Steps
DIFF.
112
187-42
1495-1
FINAL RANK
82.51064
2.59751
Steps (cont.)
1.State the hypotheses:
Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.
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.
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.