1. Hypothesis testing applied to means

2. Characteristics of the Sampling Distribution of the mean The sampling distribution of means will have the same mean as the population : 0 = :

3. Characteristics of the Sampling Distribution of the mean The sampling distribution of means has a smaller variance.   x N 2 2    x N   x = standard deviation of the mean = standard error This is because the means of samples are less likely to be extreme compared to individual scores.

4. Characteristics of the Sampling Distribution of the mean The shape of the sampling distribution approximates a normal curve if either: the population of individual cases is normally distributed the sample size being considered is 30 or more

6. z X        115100 15 10. Probability is approximately.16

7. Example Jill now has to choose 25 intelligent people. 0 = 106 IQ test: := 100 F = 15. Hypotheses: H 1 : : 1 … : 2 H 0 : : 1 = : 2 = 100

8. Sampling Distribution : 1 = : 2 = 100   x N  15 25 15 5 3 z X x    z X x        106100 3 6 3 200. Look up area in the tail of the curve =.0228 (one tailed) or.0456 (two tailed) If signficance level (") =.05, reject the Null Hypothesis

9. t Distribution $What happens if we do not have the population standard deviation? <We can use the sample standard deviation and an estimate. $Problem <Cannot use z and the normal distribution to estimate probability. BSample variance tends to underestimate the population variance  Have to use a slightly different distribution - Student=s t Distribution z X x    t X s x  

10. df = n - 1 Different t distribution for each degree of freedom Degrees of freedom

11. Sample:N = 56 0 = 104.3 S = 12.58 Norms: : = 100 Hypotheses: H 1 : : 1 … : 2 H 0 : : 1 = : 2 = 100

12. t X s x     X s N    10413100 1258 56..  4125 1682.. t  245. df = 56 - 1 = 55 t crit or t.025 = " 2.009 t obs > t crit Reject H 0

13. Percentage Points of the t Distribution See Howell page 247

14. Matched-Sample t-test Use related samples In SPSS called the >Paired-Samples t-test' Data: Howell p 193

15. H0:H0:  D  12 0 t D s D s D s N DD D       00

16. D s N D       04290 1604 31 429 288 149..... df = N - 1 (N is number of pairs of observations) df = 31 - 1 = 30 t.025 (30) = " 2.042 t obs < t crit Fail to reject H 0

17. Two Independent Samples Distribution of Differences between Means

18. Variance of the Distributions of the Means  1 2 1 2 2 2 NN & Standard Error of the Distributions of the Means  1 1 2 2 NN & Variance of the Distribution of Mean Difference   XX NN 12 2 1 2 1 2 2 2   Standard Error of the Distribution of Mean Difference   XX NN 12 1 2 1 2 2 2   Mean of the Distribution of Mean Differences  12 0 

19. t - test for two independent samples t X s x   t XX s XX    ()() 1212 12     ()()XX s N s N 1212 1 2 1 2 2 2     ()XX s N s N 12 1 2 1 2 2 2

20. Pooled Variance estimate s NsNs NN p 2 11 2 22 2 12 11 2    ()() t XX s N s N    () 12 1 2 1 2 2 2    ()XX s N s N pp 12 2 1 2 2 t XX s NN p          () 12 2 12 11 Degrees of Freedom: df = (N 1 - 1) + (N 2 - 1) = N 1 + N 2 -2