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Testing means, part II The paired t-test 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 A digression: Options in statistics 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? Ho: The coin is fair (pheads = 0.5). Ha: The coin is not fair (pheads ≠ 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 8 e-04 6 e-04 Probability 4 e-04 2 e-04 0 e+00 498000 499000 500000 x 501000 502000 Test statistic 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] Central limit theorem The sum or mean of a large number of measurements randomly sampled from any population is approximately normally distributed Binomial Distribution Normal approximation to the binomial distribution The binomial distribution, when number of trials n is large and probability of success p is not close to 0 or 1, is approximated by a normal distribution having mean np and standard deviation np( 1-p). 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? Normal approximation • Under the null hypothesis, data are approximately normally distributed • Mean: np = 1,000,000 * 0.5 = 500,000 • Standard deviation: s= n p 1− p = 1,000,000∗ 0.5∗ 1− 0.5 • s = 500 Normal distributions • Any normal distribution can be converted to a standard normal distribution, by Z= Y− m s Z-score Z= Y− m s 501,823− 500,000 Z= = 3.646 500 From standard normal table: P = 0.0001 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 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? 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. Result Heads Tails Observed Expected 501823 500000 498167 500000 Test statistic: 13.3 Critical value: 3.84 Coin toss example Binomial test Normal approximation Most accurate Hard to calculate Assumes: Random sample Approximate Easier to calculate Assumes: Random sample Large n p far from 0, 1 Chi-squared goodness of fit test Approximate Easier to calculate Assumes: Random sample No expected <1 Not more than 20% less than 5 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 Normal distributions • Any normal distribution can be converted to a standard normal distribution, by Z= Y− m s Z-score t distribution • We carry out a similar transformation on the sample mean mean under Ho Y− m t= s/ n estimated standard error How do we use this? • t has a Student's t distribution • Find confidence limits for the mean • Carry out one-sample t-test t has a Student’s t distribution* t has a Student’s t distribution* Uncertainty makes the null distribution FATTER * Under the null hypothesis Confidence interval for a mean Y ± SE Y t 2 ,df (2) = 2-tailed significance level df = degrees of freedom, n-1 SEY = standard error of the mean Confidence interval for a mean Y ± SE Y t 2 ,df 95 % Confidence interval: Use α(2) = 0.05 Confidence interval for a mean Y ± SE Y t 2 ,df c % Confidence interval: Use α(2) = 1-c/100 One-sample t-test Null hypothesis The population mean is equal to o Sample Test statistic t= Y− o compare Null distribution t with n-1 df s/ n How unusual is this test statistic? P < 0.05 Reject Ho P > 0.05 Fail to reject Ho The following are equivalent: • • • • Test statistic > critical value P < alpha Reject the null hypothesis Statistically significant Quick reference summary: One-sample t-test • What is it for? Compares the mean of a numerical variable to a hypothesized value, μo • What does it assume? Individuals are randomly sampled from a population that is normally distributed • Test statistic: t • Distribution under Ho: t-distribution with n-1 degrees of freedom • Formulae:Y = sample mean, s = sample standard deviation t= Y− o s/ n 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 32 Paired vs. 2 sample comparisons 33 Paired designs • Data from the two groups are paired • There is a one-to-one correspondence between the individuals in the two groups 34 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 35 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 36 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” 37 Paired comparisons • To compare two groups, we use the mean of the difference between the two members of each pair 38 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 39 data Year 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 Injur ies before No Smoking Day 516 610 581 586 554 632 479 583 445 522 Injur ies on No Smoking Day 540 620 599 639 607 603 519 560 515 556 40 Calculate differences Injur ies before No Smoking Day Injur ies on No Smoking Day Differ ence 516 540 (d) 24 610 620 10 581 586 599 639 18 53 554 632 479 607 603 519 53 -29 40 583 445 560 515 -23 70 522 556 34 41 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. 42 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) 43 Calculate differences Injur ies before No Smoking Day Injur ies on No Smoking Day Differ ence 516 540 (d) 24 610 620 10 581 586 599 639 18 53 554 632 479 607 603 519 53 -29 40 583 445 560 515 -23 70 522 556 34 44 Calculate t using d’s d =25 2 d s =1043 .78 n =10 25 -0 t = =2.45 1043 .78 /10 45 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 46 Critical value of t t 0.05 2 ,9 = 2.26 Test statistic: t = 2.45 So we can reject the null hypothesis: Stopping smoking increases job-related accidents in the short term. 47 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. 48 Quick reference summary: Paired t-test • What is it for? To test whether the mean difference in a population equals a null hypothesized value, μdo • What does it assume? Pairs are randomly sampled from a population. The differences are normally distributed • Test statistic: t • Distribution under Ho: t-distribution with n-1 degrees of freedom, where n is the number of pairs • Formula: d − do t= SE d