1. 3.11 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 Thursday, 11 June 201511:38 PM

2. 3.22 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.

3. 3.33 Normal Distribution Series123 μ001 σ1½1 Tables present results for the standard normal distribution (μ=0, σ=1).

4. 3.44 Use of Tables Prob(1≤z≤∞) = 68% of the observations Prob(-∞≤z≤-1) = lie within 1 standard deviation 0.16of 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

5. 3.55 Use of Tables Z0.00-0.01-0.02-0.03-0.04-0.05-0.06-0.07-0.08-0.09 0.1590.1560.1540.1520.1490.1470.1450.1420.1400.138 -1.90.0290.0280.027 0.026 0.0250.024 0.023 -2.50.006 0.005 Prob(1≤z≤∞) = Prob(-∞≤z≤-1) = 0.16 Prob(1.96≤z≤∞) = Prob(-∞≤z≤-1.96) = 0.025 Prob(2.58≤z≤∞) = Prob(-∞≤z≤-2.58) = 0.005

6. 3.66 Testing Hypothesis Null hypothesis H0H0 assumes that there is no real effect present Alternate hypothesis H1H1 assumes that there is some effect

7. 3.77 Z Test For a value x taken from a population with mean μ and standard deviation σ, the Z-score is

8. 3.88 The Central Limit Theorem When taking repeated samples of size n from the same population. 3.The distribution of the sample means approximates a Normal curve. 2.The spread of the distribution of the sample means is smaller than that of the original observations. 1.The distribution of the sample means is centred around the true population mean

9. 3.99 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

10. 3.1010 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 Z0.00-0.01-0.02-0.03-0.04-0.05-0.06-0.07-0.08-0.09 -0.50.3090.3050.3020.2980.2950.2910.2880.2840.2810.278

11. 3.1111 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.

12. 3.1212 Solution 2 Prob( ≥ 100) = Prob(z≥3.125) = 0.0009, beyond our basic table Z0.00-0.01-0.02-0.03-0.04-0.05-0.06-0.07-0.08-0.09 -3.00.001 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.

13. 3.1313 Estimating The Population Mean Confidence interval for the population mean Sample mean nSample size σPopulation standard deviation (known) Tabulated value of the z-score that achieves a significance level of α in a two tail test Don’t forget to multiply or divide before you add or subtract This test is not available in SPSS

14. 3.1414 Estimating The Population Mean Confidence interval for the population mean Sample mean nSample size σPopulation standard deviation (known) Tabulated value of the z-score that achieves a significance level of α in a two tail test We can be 100(1-2α)% certain the population mean lies in the interval

15. 3.1515 Normal Values Conf. level Prob. α One Tail ZαZα 90%0.051.645 95%0.0251.960 99%0.0052.576 Z0.00-0.01-0.02-0.03-0.04-0.05-0.06-0.07-0.08-0.09 -1.60.0550.0540.0530.0520.0510.0490.0480.0470.046 -1.90.0290.0280.027 0.026 0.0250.024 0.023 -2.50.006 0.005 Notation commonly used to denote Z values for confidence interval is Z α where 100(1 - 2 α ) is the desired confidence level in percent.

16. 3.1616 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

17. 3.1717 Solution 3 95% sure the population mean lies in the interval [103.7,116.3]

18. 3.1818 Is there a snag?

19. 3.1919 If The Population Standard Deviation Is Not Available? t values with ν = n – 1 degrees of freedom ( ν the Greek letter nu) Sample mean n Sample size ν Degrees of freedom, n-1 in this case s Sample standard deviation α Proportion of occasions that the true mean lies outside the range tνtν Critical value of t from tables Don’t forget to multiply or divide before you add or subtract Note in this module, typically, the sample variance is required. Divide by n-1

20. 3.2020 If The Population Standard Deviation Is Not Available? t values Sample mean n Sample size ν Degrees of freedom, n-1 in this case s Sample standard deviation α Proportion of occasions that the true mean lies outside the range tνtν Critical value of t from tables

21. 3.2121 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.

22. 3.2222 Example 4 An experiment results in the following estimates. Obtain a 90% confidence interval for the population mean.

23. 3.2323 Example 4 Given ν p=0.05p=0.025p=0.005p=0.0025 191.7292.0932.8613.174 We can be 90% (α=0.05) sure that the population mean lies in this interval [68.6,74.2].

24. 3.2424 One Sample t-Test The basic test statistic is

25. 3.2525 Interpreting t-values If t calc < t ν (α) then we cannot reject the null hypothesis that μ=m. If t calc > t ν (α) the null hypothesis is rejected, the true mean μ differs significantly at the 2α level from m.

26. 3.2626 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 H 0 : there is no effect so μ = 75 H 1 : μ ≠ 75 (two tail test)

27. 3.2727 Solution 5 n = 20 Σx = 1428 Σx 2 = 102984 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 Σx 2 = 72 2 + 64 2 + … + 84 2 + 77 2 = 102984

28. 3.2828 Solution 5 n = 20 Σx = 1428 Σx 2 = 102984

29. 3.2929 Solution 5 n = 20 Σx = 1428 Σx 2 = 102984 s = 7.34 Note in this module, typically, the sample variance is required. Divide by n-1

30. 3.3030 Solution 5 νp=0.05p=0.025p=0.005p=0.0025p=0.0010 191.7292.0932.8613.1743.579 In an attempt to “estimate” p.

31. 3.3131 Conclusion 5 Since 2.093<2.193<2.861 0.01<p-value<0.05 (note 2α since two tail) t = 2.193 There is sufficient evidence to reject H 0 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.

32. 3.3232 SPSS 5 Analyze > Compare Means > One Sample t Test Note insertion of test value

33. 3.3333 SPSS 5 Basic descriptive statistics for a manual test

34. 3.3434 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].

35. 3.3535 Graph? Graph > Legacy Dialogs > Error Bar

36. 3.3636 Graph? Graph > Legacy Dialogs > Error Bar

37. 3.3737 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. H 0 is that μ = 0.250

38. 3.3838 Solution 6 νp=0.05p=0.025p=0.005p=0.0025p=0.0010 111.7962.2013.1063.4974.025 t 11 (0.005)=3.106 t 11 (0.0025)=3.497 In an attempt to “estimate” p.

39. 3.3939 Conclusion 6 t = 3.333 Since 3.106 < 3.333 < 3.497 0.005 < p-value < 0.01 There is sufficient evidence to reject H 0 at the 1% level. The experimental mean would not appear to be consistent with 0.250

40. 3.4040 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].

41. 3.4141 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

42. 3.4242 Practical 3 This material is available from the module web page. http://www.staff.ncl.ac.uk/mike.cox Module Web Page

43. 3.4343 Practical 3 This material for the practical is available. Instructions for the practical Practical 3 Material for the practical Practical 3

44. 3.4444 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

45. 3.4545 Whoops!

46. 3.4646 Whoops!