Competence and a new baby, M

1. Competence and a new baby, M | mastery baby | 0 1 | Total -----------+----------------------+---------- 0 | 67 84 | 151 | 44.37 55.63 | 100.00 | 67.00 80.77 | 74.02 -----------+----------------------+---------- 1 | 33 20 | 53 | 62.26 37.74 | 100.00 | 33.00 19.23 | 25.98 -----------+----------------------+---------- Total | 100 104 | 204 | 49.02 50.98 | 100.00 | 100.00 100.00 | 100.00

2. Competence and a new baby, F | mastery baby | 0 1 | Total -----------+----------------------+---------- 0 | 74 77 | 151 | 49.01 50.99 | 100.00 | 73.27 74.76 | 74.02 -----------+----------------------+---------- 1 | 27 26 | 53 | 50.94 49.06 | 100.00 | 26.73 25.24 | 25.98 -----------+----------------------+---------- Total | 101 103 | 204 | 49.51 50.49 | 100.00 | 100.00 100.00 | 100.00

3. Depression and a new baby, M | depress baby | 0 1 | Total -----------+----------------------+---------- 0 | 92 59 | 151 | 60.93 39.07 | 100.00 | 75.41 71.95 | 74.02 -----------+----------------------+---------- 1 | 30 23 | 53 | 56.60 43.40 | 100.00 | 24.59 28.05 | 25.98 -----------+----------------------+---------- Total | 122 82 | 204 | 59.80 40.20 | 100.00 | 100.00 100.00 | 100.00

4. Depression and a new baby, F | depress baby | 0 1 | Total -----------+----------------------+---------- 0 | 91 60 | 151 | 60.26 39.74 | 100.00 | 76.47 70.59 | 74.02 -----------+----------------------+---------- 1 | 28 25 | 53 | 52.83 47.17 | 100.00 | 23.53 29.41 | 25.98 -----------+----------------------+---------- Total | 119 85 | 204 | 58.33 41.67 | 100.00 | 100.00 100.00 | 100.00

5. Central Limit Theorem • Know the sampling distribution of means from properties of the population • Mean of sampling distribution of means is mean of the population • Standard deviation of sampling distribution of means is times sd of population • Sampling distribution is normal

6. .4 .3 .2 norm .1 0 -3 -2 -1 0 1 2 3 x Normal Distribution 95% Mean Standard deviation

7. Normal Distribution

8. Normal Probabilities z p 1. -3 .0013499 2. -2 .0227501 3. -1 .1586553 4. 0 .5 5. 1 .8413448 6. 2 .9772499 7. 3 .9986501 Z score Why we have a normal table!!!

9. Translation Mean = 0 sd = 1 0 1 2 3 100 116 132 148 Mean = 100 sd = 16

10. Translation (part 1) • Original metric of variable, IQ: mean is 100, std. dev. is 16 • New metric (z score), mean is 0, std. dev. is 1. • To make the new variable, z, have a mean of zero, subtract the mean of the old variable from all observations

11. Translation (part 2) • To make the new variable, z, have a standard deviation of 1, divide each deviation by the sd of X

12. Translation (Part 3) 100 116 132 148 Mean = 100, sd = 16 Subtract mean of 100 16 32 48 Mean = 0, sd = 16 0 Divide by sd of 16 Mean = 0, sd = 1 0 1 2 3 The order in which you do these steps is important!

13. Not all distributions are normal • We have already looked at the binomial • The same principles hold in looking at the binomial distribution. The formulae are different.

14. Binomial, p = .3

15. Binomial example • Let p = .3 and n = 7 • In this case f can be 0,1,2,3,4,5,6, or 7 • This creates a discrete distribution with a probability for each separate outcome .082

16. Binomial example (cont.) 7 .3 .118 = .247 21 .09 .168 = .318

17. Binomial Results f comb pf qnf p 0 1 1 .0823543 .082 1 7 .3 .117649 .247 2 21 .09 .16807 .317 3 35 .027 .2401 .226 4 35 .0081 .343 .097 5 21 .00243 .49 .025 6 7 .000729 .7 .004

18. Binomial • Normal approximation with large sample • A large sample? np/q > 9 & nq/p > 9 • If p=.3 then n > 21

19. Binomial with p = .32, n = 21 Note normal curve Mean = np = 6.3 Sd = sqrt(npq) = 2.1