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 Range: 2 - 13

Variance and SD (x1 – x)2 N- 1

Variance S words 1 2 3 4 5 6 7 8 9 12 11

Variance S words 1 2 3 4 5 6 7 8 9 12 11  59 / 8 = 7.4 (mean)

Variance S words (=X1 – Xmean) 1 2 3 4 5 6 7 8 9 12 11 3 – 7.4 7 – 7.4 4 – 7.4 9 – 7.4 12 – 7.4 11 – 7.4  59 / 8 = 7.4 (mean)

Variance S words (=X1 – Xmean) d1 1 2 3 4 5 6 7 8 9 12 11 3 – 7.4 7 – 7.4 4 – 7.4 9 – 7.4 12 – 7.4 11 – 7.4 –4.4 –0.4 –3.4 1.6 4.6 3.6  59 / 8 = 7.4 (mean)

Variance S words (=X1 – Xmean) d1 1 2 3 4 5 6 7 8 9 12 11 3 – 7.4 7 – 7.4 4 – 7.4 9 – 7.4 12 – 7.4 11 – 7.4 –4.4 –0.4 –3.4 1.6 4.6 3.6  59 / 8 = 7.4 (mean)  0 / 8 = 0

Variance S words (=X1 – Xmean) d1 d12 (residuals) 1 2 3 4 5 6 7 8 9 12 11 3 – 7.4 7 – 7.4 4 – 7.4 9 – 7.4 12 – 7.4 11 – 7.4 –4.4 –0.4 –3.4 1.6 4.6 3.6 19.36 0.16 11.56 2.56 21.16 12.96  59 / 8 = 7.4 (mean)  0 / 8 = 0  81.87

Variance 81.87 8 - 1 = 11.7

Standard Deviation 81.87 8 - 1 = 3.42

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

z scores Test 1 – candidate A Test 2 – candidate B Scenario Score Mean SD 1 41 49 53

z scores Test 1 – candidate A Test 2 – candidate B Scenario Score Mean SD 1 41 49 53 2 58

z scores Test 1 – candidate A Test 2 – candidate B Scenario Score Mean SD 1 41 49 53 2 6 58

z scores Test 1 – candidate A Test 2 – candidate B Scenario Score Mean SD 1 41 49 53 2 58 3 8 5

z scores x1 – x SD

z scores S words 1 2 3 4 5 73 42 36 51 63

z scores S words 1 2 3 4 5 73 42 36 51 63 M = 265 / 5 = 53 SD = 15.12

z scores S words (=X1 – Xmean) d1 1 2 3 4 5 73 42 36 51 63 73 – 53 42 – 53 36 – 53 51 – 53 63 – 53 20 –11 –17 –2 10 M = 265 / 5 = 53 SD = 15.12  0

z scores S Number of words (=X1 – Xmean) d1 z = (d1 / SD) 1 2 3 4 5 73 42 36 51 63 73 – 53 42 – 53 36 – 53 51 – 53 63 – 53 20 –11 –17 –2 10 1.32 –0.73 –1.12 –0.13 0.66  265 / 5 = 53 (m) SD = 15.12  0

z scores x1 – x SD

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)?

Exercise ZA = 121 – 92 / 14 = 2.07 ZB = 177 – 143 / 21 = 1.62

Coefficient of variance SD Mean CV =

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?

Coefficient of variance Parking tickets: Mean = 90 SD = 5 Fines: Mean = 5400 SD = 775 CV1 = 5/90 = 0.06 CV2 = 775/5400 = 0.14

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.

SPSS Adjectives: 10, 6, 7, 9, 11, 9, 8, 6, 9, 8 Nouns: 12, 13, 16, 15, 9, 7, 14, 12, 11, 13 What is the independent variable? What is the dependent variable? Is this a correlational analysis or an experiment? Is this a bewteen subjects or a within subjects design? Enter the data into SPSS.