Using Statistics To Make Inferences 3

1. Using Statistics To Make Inferences 3 Summary Review the normal distribution Z test Z test for the sample mean t test for the sample mean 1 Wednesday, 08 October 20145:07 PM

2. Goals To perform and interpret a Z test. To perform and interpret tests on the sample mean. To produce a confidence interval for the population mean. Know when to employ Z and when t. Practical Perform a t test. Perform a two sample t test, in preparation for next week. 2

3. Normal Distribution Tables present results for the standard normal distribution (μ=0, σ=1). 3

4. Use of Tables Prob(1≤z≤∞) = 68% of the observations Prob(-∞≤z≤-1) = lie within 1 standard deviation 0.16 of the mean Prob(1.96≤z≤∞) = 95% of the observations lie Prob(-∞≤z≤-1.96) = within 1.96 standard 0.025 deviations of the mean Prob(2.58≤z≤∞) = 99% of the observations lie Prob(-∞≤z≤-2.58) = within 2.58 standard 0.005 deviations of the mean 4

5. Use of Tables Prob(2.58≤z≤∞) = Prob(-∞≤z≤-2.58) = 0.005 Prob(1.96≤z≤∞) = Prob(-∞≤z≤-1.96) = 0.025 Prob(1≤z≤∞) = Prob(-∞≤z≤-1) = 0.16 5

6. Testing Hypothesis 6

7. Z Test For a value x taken from a population with mean μ and standard deviation σ, the Z-score is 7

8. The Central Limit Theorem When taking repeated samples of size n from the same population.   • The distribution of the sample means is centred around the true population mean • The spread of the distribution of the sample means is smaller than that of the original observations. • The distribution of the sample means approximates a Normal curve. 8

9. Central Limit Theorem If the standard deviation of the individual observations is σ then the standard error of the sample mean value is For a sample mean, , with mean μ and standard deviation the Z-score is 9

10. Example 1 mean score 100 standard deviation 16 What is the probability a score is higher than 108? Prob(x≥108) = Prob(z≥0.5) = 0.309 10

11. Example 2 mean score 100 standard deviation 16 The sample mean of 25 individuals is found to be 110. The null hypothesis, no real effect present, is that μ = 100. Wish to test if the mean significantly exceeds this value. 11

12. Solution 2 Prob( ≥ 100) = Prob(z≥3.125) = 0.0009, beyond our basic table Since the p-value is less than 0.001 the result is highly significant, the null hypothesis is rejected. The sample average is significantly higher. 12

13. Estimating The Population Mean Confidence interval for the population mean Don’t forget to multiply or divide before you add or subtract 13 This test is not available in SPSS

14. Estimating The Population Mean Confidence interval for the population mean We can be 100(1-2α)% certain the population mean lies in the interval 14

15. Normal Values Notation commonly used to denote Z values for confidence interval is Zα where 100(1 - 2α) is the desired confidence level in percent. 15

16. Example 3 standard deviation 16 mean of a sample of 25 individuals is found to be 110   Require 95% confidence interval for the population mean 16

17. Solution 3 95% sure the population mean lies in the interval [103.7,116.3] 17

18. Is there a snag? 18

19. If The Population Standard Deviation Is Not Available? t values with ν = n – 1 degrees of freedom (ν the Greek letter nu) Note in this module, typically, the sample variance is required. Divide by n-1 Don’t forget to multiply or divide before you add or subtract 19

20. If The Population Standard Deviation Is Not Available? t values 20

21. Two Tail t To obtain confidence limits a two tail probability is employed since it refers to the proportion of values of the population mean, both above and below the sample mean. 21

22. Example 4 An experiment results in the following estimates. Obtain a 90% confidence interval for the population mean. 22

23. Example 4 Given We can be 90% (α=0.05) sure that the population mean lies in this interval [68.6,74.2]. 23

24. One Sample t-Test The basic test statistic is 24

25. Interpreting t-values If tcalc<tν(α) then we cannot reject the null hypothesis that μ=m. If tcalc>tν(α) the null hypothesis is rejected, the true mean μ differs significantly at the 2α level from m. 25

26. Example 5 Claimed mean is 75 seconds, the times taken for 20 volunteers are 72 64 69 82 76 70 58 64 81 75 71 76 60 78 64 65 73 69 84 77 H0: there is no effect so μ = 75 H1: μ ≠ 75 (two tail test) 26

27. Solution 5 72 64 69 82 76 70 58 64 81 75 71 76 60 78 64 65 73 69 84 77 n = 20 Σx = 72 + 64 + … + 84 + 77 = 1428 Σx2 = 722 + 642 + … + 842 + 772 = 102984 n = 20 Σx = 1428 Σx2 = 102984 27

28. Solution 5 n = 20 Σx = 1428 Σx2 = 102984 28

29. Solution 5 n = 20 Σx = 1428 Σx2 = 102984 Note in this module, typically, the sample variance is required. Divide by n-1 s = 7.34 29

30. Solution 5 30

31. Conclusion 5 t = 2.193 Since 2.093<2.193<2.861 0.01<p-value<0.05 (note 2α since two tail) There is sufficient evidence to reject H0 at the 5% level. The experiment is not consistent with a mean of 75. In fact the 95% confidence interval is [68.0,74.8] which, as expected, excludes 75. The precise p value may be found from software. 31

32. SPSS 5 Analyze > Compare Means > One Sample t Test Note insertion of test value 32

33. SPSS 5 Basic descriptive statistics for a manual test 33

34. SPSS 5 As predicted 0.01 < p-value < 0.05 The confidence interval is 75-7.04 to 75-0.16 that is [67.96, 74.84]. 34

35. Graph?Graph > Legacy Dialogs > Error Bar 35

36. Graph?Graph > Legacy Dialogs > Error Bar 36

37. Example 6 Experimental data 0.235 0.252 0.312 0.264 0.323 0.241 0.284 0.306 0.248 0.284 0.298 0.320 Test whether these data are consistent with a population mean of 0.250. H0 is that μ = 0.250 37

38. Solution 6 t11(0.005)=3.106 t11(0.0025)=3.497 38

39. Conclusion 6 t = 3.333 Since 3.106 < 3.333 < 3.497 0.005 < p-value < 0.01 There is sufficient evidence to reject H0 at the 1% level. The experimental mean would not appear to be consistent with 0.250 39

40. SPSS 6 As predicted p-value < 0.01 The confidence interval is 0.250+0.010 to 0.250+0.050 that is [0.26, 0.30]. 40

41. Read Read Howitt and Cramer pages 40-50 Read Russo (e-text) pages 134-145 Read Davis and Smith pages 133-134, 139-143, 200-205, 237-264 41

42. Practical 3 This material is available from the module web page. http://www.staff.ncl.ac.uk/mike.cox Module Web Page 42

43. Practical 3 This material for the practical is available. Instructions for the practical Practical 3 Material for the practical Practical 3 43

44. Whoops! From testimony by Michael Gove, British Secretary of State for Education, before their Education Committee: "Q98 Chair: [I]f 'good' requires pupil performance to exceed the national average, and if all schools must be good, how is this mathematically possible? "Michael Gove: By getting better all the time. "Q99 Chair: So it is possible, is it? "Michael Gove: It is possible to get better all the time. "Q100 Chair: Were you better at literacy than numeracy, Secretary of State? "Michael Gove: I cannot remember." Oral Evidence, British House of Commons, January 31, 2012, p. 28 44

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