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Chi-Square Goodness of Fit

fo

pe

fe

StepsWhat 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.

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

Chi-Square Goodness of Fit

fo

pe

fe

StepsWhat 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

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.

Chi-Square Test for Independence

Row totals

Residence

100

200

Column totals

154

146

StepsWhat 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

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.)

- 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.

XRANK

YRANK

2

1

3

4

5

2

1

4

3

5

Step 1. Rank the X and Y ValuesThe order of your X and Y values by increasing value

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

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.

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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