Individual values of X

1. Distribution of a population Frequency How many individuals   Individual values of X

2. Distribution of a sample Frequency How many individuals s Individual values of X

3. Distribution of a sample: The way in which the individual scores within a sample are spread out in relation to the mean. Sampling Distribution: The way in which sample means, drawn from the same population, spread out from the population mean.

4. Sampling Distribution Frequency How many samples  Mean values of all possible samples.

5. Measure of Variability: Standard error - the average amount that sample means differ from the population mean.

7. If: a) randomly select enough samples b) all samples were the same size

8. Then: The distribution of those sample means would form a normal distribution.

9. We can use the normal distribution to predict the likelihood of randomly selecting samples with particular means.

11.  = 45  = 10 N = 16 What is the likelihood that a sample, selected at random from this population, would have a mean of 43 or less?

12. 1) compute the standard error 2) compute Z 3) Use the Z table to determine the probability

13.  = 45  = 10 N = 16 1) compute the standard error

14. = 2.5

15. 2) compute Z using the standard error Z = - .8

16. Reminders about Z table: a column: Z score b column: prob. between mean and Z c column: prob. beyond z into the tail

17. b column + c column = .5000

18. b b c c

19. Z table: z score a b c .79 .2852 .2148 .80 .2881 .2119 .81 .2910 .2090

20. Z table: z score a b c .79 .2852 .2148 .80 .2881 .2119 .81 .2910 .2090

21. The prob. that a sample mean would below 43 is .2119.

22. .5000 or 50.0 % .2881 or 28.81% .2119 or 21.19%  Raw means 43 45 Z scores - .8 0

23. With the probability info we can: - Determine how often we expect a particular sample to be drawn.

24. - Establish the probability that a particular sample that we have is a random sample.

25. You develop an expensive training program for athletes. You say that your program reduce event times for runners.

26. Typical running time: 50 sec. After your program the group’s average running time is 49 sec.

27. Is the program effective, or is this random variability?

28. Assume that :  = 50 sec. N = 15  = 4 sec.

29. standard error = 1.03

30. Z = - .97

31. Probability that a sample would have a mean of 49 or a more extreme score? P = .1660

32. 33.40% 16.60% Raw score Z score 49 50 - .97 0

33. Sampling Distribution Frequency How many samples  Mean values of all possible samples.

34. The convention is that we will assume that a sample is different from the population mean for reasons other than chance when the probability of randomly selecting the sample is less than or equal to .05 (5%)

35. Statistical Significance: The outcome could occur by chance alone less than 5% of the time.

36. Null hypothesis: Mathematical statement that describes the population in the absence of any treatment. Ho :  = 50

37. Alternative hypothesis: The mathematical statement that describes the population if the treatment given would be expected to have an effect and therefore change the population. H1 :   50

38. If an outcome is statistically significant we reject the null hypothesis and accept the alternative.

39. 2.5% Ho 2.5% H1

40. Why is the 5% split into two parts? H1 :   50 This doesn’t suggest the direction of an unusual outcome.

41. Samples means much less than the population mean of 50 as well as samples means much greater than 50 would be unusual.

42. Ho :  = 50 2.5% 2.5% H1 :   50 50

43. This is know as a two tailed test. Both tails of the distribution are included as possible outcomes in the hypothesis statement.

44. Assume a teacher knows that students generally have an average score on a test of 85. The standard deviation of all of the students has been 7 points. Ho :  = 85 H1 :   85

45. The teacher uses a new curriculum this year and finds that the class (N = 20) has an average on the test of 81. Has this curriculum made a difference in the performance of the class?

47.  = 85  = 7 n = 20

49. 81 - 85 Z = 1.56 Z = --2.56

50. From the Z table: Prob. of a Z score between 0 and 2.56 is .4948 (col. B) The prob. of a Z score beyond 2.56 is .0052 (col. C)