Stats 2020 tutorial
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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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Stats 2020 tutorial

Stats 2020 Tutorial


Chi square goodness of fit

Chi-Square Goodness of Fit


Steps

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

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 cont1

“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


Steps cont2

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.


Steps cont3

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 cont4

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 fit1

Chi-Square Goodness of Fit


Steps1

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 cont5

Steps (cont.)

  • State the hypotheses:Ho: There is no preference among the three photographs.H1: There is a preference among the three photographs.


Steps cont6

Steps (cont.)

  • Locate the critical regiondf = C - 1 = 3 - 1 = 2For df = 2 and α= .05, the critical 𝝌2= 5.99


Steps cont7

fo

pe

fe

Steps (cont.)

  • Calculate the chi-square statisticfe = pnOriginal:.3333(150) = 50Eyes farther:.3333(150) = 50Eyes closer:.3333(150) = 50


Steps cont8

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 cont9

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

Chi-Square Test for Independence


Steps2

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 cont10

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 cont11

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


Steps cont12

Opinion

Row totals

Residence

100

200

Column totals

154

146

Steps (cont.)


Steps cont13

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 cont14

Steps (cont.)

  • Calculate chi-square statisic

𝝌2 = 5.4138 + 5.7097 + 2.7066 + 2.8551

= 16.69


Steps cont15

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 cont16

Steps (cont.)

  • Part b) Phi-coefficient (effect size)?

ɸ = √(𝝌2 / N)

= √(.0556)

= .236

Therefore, it is a small effect.


Spearman correlation

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

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


Step 2 compute the correlation

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


Step 2 cont

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

Mann-Whitney U

A

B


Steps3

Steps

What we know:

nA = 6, nB = 6, α= .05

A

B


Steps cont17

Steps (cont.)

  • State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.


Steps cont18

Steps (cont.)

  • Locate the critical regionFor a non-directional test with α = .05, andnA = 6, and nB = 6, the critical U = 5.


Steps cont19

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.


Steps cont20

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


Steps cont21

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 cont22

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 cont23

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

Wilcoxon Signed-Ranks Test


Steps4

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.


Steps5

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 cont24

Steps (cont.)

  • State the hypotheses:Ho: There is no difference between the two treatments.H1: There is a difference between the two treatments.


Steps cont25

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 cont26

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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