1. Testing means, part II The paired t-test

2. Outline of lecture Options in statistics –sometimes there is more than one option One-sample t-test: review –testing the sample mean The paired t-test –testing the mean difference

3. A digression: Options in statistics

4. Example A student wants to check the fairness of the loonie She flips the coin 1,000,000 times, and gets heads 501,823 times. Is this a fair coin?

5. H o : The coin is fair (p heads = 0.5). H a : The coin is not fair (p heads ≠ 0.5). n = 1,000,000 trials x = 501,823 successes Under the null hypothesis, the number of successes should follow a binomial distribution with n=1,000,000 and p=0.5

6. Test statistic

7. Binomial test P = 2*Pr[X ≥ 501,823] P = 2*(Pr[X = 501,823] + Pr[X = 501,824] + Pr[X = 501,825] + Pr[X = 501,826] +... + Pr[X = 999,999] + Pr[X = 1,000,000]

8. Central limit theorem The sum or mean of a large number of measurements randomly sampled from any population is approximately normally distributed

9. Binomial Distribution

10. Normal approximation to the binomial distribution

11. Example A student wants to check the fairness of the loonie She flips the coin 1,000,000 times, and gets heads 501,823 times. Is this a fair coin?

12. Normal approximation Under the null hypothesis, data are approximately normally distributed Mean: np = 1,000,000 * 0.5 = 500,000 Standard deviation: s = 500

13. Normal distributions Any normal distribution can be converted to a standard normal distribution, by Z-score

14. From standard normal table: P = 0.0001

15. Conclusion P = 0.0001, so we reject the null hypothesis This is much easier than the binomial test Can use as long as p is not close to 0 or 1 and n is large

16. Example A student wants to check the fairness of the loonie She flips the coin 1,000,000 times, and gets heads 500,823 times. Is this a fair coin?

17. A Third Option! Chi-squared goodness of fit test Null expectation: equal number of successes and failures Compare to chi-squared distribution with 1 d.f.

18. Test statistic: 13.3 Critical value: 3.84

19. Coin toss example Binomial test Normal approximation Chi-squared goodness of fit test Most accurate Hard to calculate Assumes: Random sample Approximate Easier to calculate Assumes: Random sample Large n p far from 0, 1 Approximate Easier to calculate Assumes: Random sample No expected <1 Not more than 20% less than 5

20. Coin toss example Binomial test Normal approximation Chi-squared goodness of fit test in this case, n very large (1,000,000) all P < 0.05, reject null hypothesis

21. Normal distributions Any normal distribution can be converted to a standard normal distribution, by Z-score

22. t distribution We carry out a similar transformation on the sample mean mean under H o estimated standard error

23. How do we use this? t has a Student's t distribution Find confidence limits for the mean Carry out one-sample t-test

24. t has a Student’s t distribution*

25. * Under the null hypothesis Uncertainty makes the null distribution FATTER

26. Confidence interval for a mean  (2) = 2-tailed significance level df = degrees of freedom, n-1 SE Y = standard error of the mean

27. Confidence interval for a mean 95 % Confidence interval: Use α(2) = 0.05

28. Confidence interval for a mean c % Confidence interval: Use α(2) = 1-c/100

29. Sample Null hypothesis The population mean is equal to  o One-sample t-test Test statistic Null distribution t with n-1 df compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

30. The following are equivalent: Test statistic > critical value P < alpha Reject the null hypothesis Statistically significant

31. Quick reference summary: One-sample t-test What is it for? Compares the mean of a numerical variable to a hypothesized value, μ o What does it assume? Individuals are randomly sampled from a population that is normally distributed Test statistic: t Distribution under H o : t-distribution with n-1 degrees of freedom Formulae:Y = sample mean, s = sample standard deviation

32. 32 Comparing means Goal: to compare the mean of a numerical variable for different groups. Tests one categorical vs. one numerical variable Example: gender (M, F) vs. height

33. 33 Paired vs. 2 sample comparisons

34. 34 Paired designs Data from the two groups are paired There is a one-to-one correspondence between the individuals in the two groups

35. 35 More on pairs Each member of the pair shares much in common with the other, except for the tested categorical variable Example: identical twins raised in different environments Can use the same individual at different points in time Example: before, after medical treatment

36. 36 Paired design: Examples Same river, upstream and downstream of a power plant Tattoos on both arms: how to get them off? Compare lasers to dermabrasion

37. 37 Paired comparisons - setup We have many pairs In each pair, there is one member that has one treatment and another who has another treatment “Treatment” can mean “group”

38. 38 Paired comparisons To compare two groups, we use the mean of the difference between the two members of each pair

39. 39 Example: National No Smoking Day Data compares injuries at work on National No Smoking Day (in Britain) to the same day the week before Each data point is a year

40. 40 data

41. 41 Calculate differences

42. 42 Paired t test Compares the mean of the differences to a value given in the null hypothesis For each pair, calculate the difference. The paired t-test is a one-sample t-test on the differences.

43. 43 Hypotheses Ho: Work related injuries do not change during No Smoking Days ( μ =0) Ha: Work related injuries change during No Smoking Days ( μ≠ 0)

44. 44 Calculate differences

45. 45 Calculate t using d’s

46. 46 Caution! The number of data points in a paired t test is the number of pairs. -- Not the number of individuals Degrees of freedom = Number of pairs - 1 Here, df = 10-1 = 9

47. 47 Critical value of t So we can reject the null hypothesis: Stopping smoking increases job-related accidents in the short term. Test statistic: t = 2.45

48. 48 Assumptions of paired t test Pairs are chosen at random The differences have a normal distribution It does not assume that the individual values are normally distributed, only the differences.

49. Quick reference summary: Paired t-test What is it for? To test whether the mean difference in a population equals a null hypothesized value, μ do What does it assume? Pairs are randomly sampled from a population. The differences are normally distributed Test statistic: t Distribution under H o : t-distribution with n-1 degrees of freedom, where n is the number of pairs Formula: