mean and variance

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Mean and variance

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Mean and variance. Central tendency. Data:2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13. Mean:2+3+3+3+4+6+6+9+12+13+13. = 6.72. 11. Median:2, 3, 3, 3, 4, 6 , 6, 9, 12, 13, 13. = 6. Mode:2, 3, 3, 3 , 4, 6, 6, 9, 12, 13, 13. = 3. Range. Data:2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13. - PowerPoint PPT Presentation

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Page 1: Mean and  variance

Mean and variance

Page 2: Mean and  variance

Central tendency

Data: 2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13

Mean: 2+3+3+3+4+6+6+9+12+13+13

11= 6.72

Median: 2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13 = 6

Mode: 2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13 = 3

Page 3: Mean and  variance

Range

Data: 2, 3, 3, 3, 4, 6, 6, 9, 12, 13, 13

Range: 2 - 13

Page 4: Mean and  variance

Variance and SD

(x1 – x)2

N- 1

Page 5: Mean and  variance

Variance

S words

12345678

3749129114

Page 6: Mean and  variance

Variance

S words

12345678

3749129114

59 / 8 = 7.4 (mean)

Page 7: Mean and  variance

Variance

S words (=X1 – Xmean)

12345678

3749129114

3 – 7.4 7 – 7.44 – 7.49 – 7.412 – 7.49 – 7.411 – 7.44 – 7.4

59 / 8 = 7.4 (mean)

Page 8: Mean and  variance

Variance

S words (=X1 – Xmean) d1

12345678

3749129114

3 – 7.4 7 – 7.44 – 7.49 – 7.412 – 7.49 – 7.411 – 7.44 – 7.4

–4.4–0.4–3.41.64.61.63.6–3.4

59 / 8 = 7.4 (mean)

Page 9: Mean and  variance

Variance

S words (=X1 – Xmean) d1

12345678

3749129114

3 – 7.4 7 – 7.44 – 7.49 – 7.412 – 7.49 – 7.411 – 7.44 – 7.4

–4.4–0.4–3.41.64.61.63.6–3.4

59 / 8 = 7.4 (mean)

0 / 8 = 0

Page 10: Mean and  variance

Variance

S words (=X1 – Xmean) d1 d12 (residuals)

12345678

3749129114

3 – 7.4 7 – 7.44 – 7.49 – 7.412 – 7.49 – 7.411 – 7.44 – 7.4

–4.4–0.4–3.41.64.61.63.6–3.4

19.360.1611.562.5621.162.5612.9611.56

59 / 8 = 7.4 (mean)

0 / 8 = 0 81.87

Page 11: Mean and  variance

Variance

81.87

8 - 1= 11.7

Page 12: Mean and  variance

Standard Deviation

81.87

8 - 1= 3.42

Page 13: Mean and  variance

Standard Deviation

70% of all data points fall within 1 SD.

Mean +/- SD = range of 70% of the data

7.4 +/- 3.42 = 3.98 – 10.82 words

Page 14: Mean and  variance

z scores

Test 1 – candidate A Test 2 – candidate B

Scenario Score Mean SD Score Mean SD

1 41 49 53 49

Page 15: Mean and  variance

z scores

Test 1 – candidate A Test 2 – candidate B

Scenario Score Mean SD Score Mean SD

1 41 49 53 49

2 41 49 53 58

Page 16: Mean and  variance

z scores

Test 1 – candidate A Test 2 – candidate B

Scenario Score Mean SD Score Mean SD

1 41 49 53 49

2 41 49 6 53 58 6

Page 17: Mean and  variance

z scores

Test 1 – candidate A Test 2 – candidate B

Scenario Score Mean SD Score Mean SD

1 41 49 53 49

2 41 49 53 58

3 41 49 8 53 58 5

Page 18: Mean and  variance

z scores

x1 – x

SD

Page 19: Mean and  variance

z scores

S words

12345

7342365163

Page 20: Mean and  variance

z scores

S words

12345

7342365163

M = 265 / 5 = 53SD = 15.12

Page 21: Mean and  variance

z scores

S words (=X1 – Xmean) d1

12345

7342365163

73 – 53 42 – 5336 – 5351 – 5363 – 53

20–11–17–210

M = 265 / 5 = 53SD = 15.12

0

Page 22: Mean and  variance

z scores

S Number of words (=X1 –

Xmean)

d1 z = (d1 / SD)

12345

7342365163

73 – 53 42 – 5336 – 5351 – 5363 – 53

20–11–17–210

1.32–0.73–1.12–0.130.66

265 / 5 = 53 (m)SD = 15.12

0

Page 23: Mean and  variance

z scores

x1 – x

SD

Page 24: Mean and  variance

Exercise

Zwei Kandidaten haben an zwei unterschiedlichen Sprachtests teilgenommen. Kandidat A hat 121 Punkte erzielt, Kandidat B hat 177 Punkte erzielt. Im ersten Test (an dem Kandidat A teilgenommen hat) lag der Mittelwert bei 92 und die Standardabweichung bei 14; im zweiten Test (an dem Kandidat B teilgenommen hat) lag der Mittelwert bei 143 und die Standardabweichung bei 21. Welcher der beiden Kandidaten hat besser abschlossen (im Vergleich zu allen übrigen Kandidaten)?

Page 25: Mean and  variance

Exercise

ZA = 121 – 92 / 14 = 2.07

ZB = 177 – 143 / 21 = 1.62

Page 26: Mean and  variance

Coefficient of variance

SD

MeanCV =

Page 27: Mean and  variance

Coefficient of variance

Over a 4 months period a mean number of 90 parking tickets was issued. The standard deviation was 5. The tickets yielded an average of $5400 per day and the SD was $775. Where do you have more variability, in the number of parking tickets that were issued each day or in the amount of money that was generate each day?

Page 28: Mean and  variance

Coefficient of variance

Parking tickets: Mean = 90SD = 5

Fines: Mean = 5400SD = 775

CV1 = 5/90 = 0.06

CV2 = 775/5400 = 0.14

Page 29: Mean and  variance

SPSS

A linguist has conducted a study comparing

memory for adjectives with that for nouns. She

randomly allocates 20 participants to two

conditions. She then presents to one of the groups

of 10 participants a list of 20 adjectives and to the

other group a list of 20 nouns. Following this, she

asks each group to try to remember as many of the

words they were presented with as possible. Here

are the data.

Page 30: Mean and  variance

SPSS

Adjectives: 10, 6, 7, 9, 11, 9, 8, 6, 9, 8

Nouns: 12, 13, 16, 15, 9, 7, 14, 12, 11, 13

1. What is the independent variable?

2. What is the dependent variable?

3. Is this a correlational analysis or an experiment?

4. Is this a bewteen subjects or a within subjects

design?

5. Enter the data into SPSS.