Outline of lecture. Central limit theorem

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1 Outline of lecture Testing means, part II The paired t-test 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? H o : The coin is fair (p heads = 0.5). H a : The coin is not fair (p heads 7 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. / $ ( ( ( ) * $ $ ' ( ( ( ) + $!! ( ( ( ) + $!, ( ( ( ) + $! # ( ( ( ) + $!! " # $ $ $! " " $ $ $ % $ $ $ $ $ % $ & $ $ $ % $ ' $ $ $ - Test statistic Binomial test P = 2*Pr[X8501,823] 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

2 Normal approximation to the binomial distribution! "#$%&'() &*+$,&-./&%0.&('1$2"#'$'0) %#/$(3$./&*+-$!$&-$+*/4# *',$5/(%*%&+&.6$(3$-077#--$"$&-$'(.$7+(-#$.($8$(/$91$&- *55/(:&) *.#,$%6$*$'(/) *+$,&-./&%0.&('$"*;&'4$)#*'$!"$*', -.*',*/,$,#;&*.&('$ 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 = 500 Normal distributions!! " "# $ Conclusion Any normal distribution can be converted to a standard normal distribution, by!! " "# $!!!"#$%&'"!""$"""!'()*)!"" From standard normal table: P = P = , so we reject the null hypothesis This is much easier than the binomial test Can use as long as p is not close to 0 or 1 and n is large 9+:;0/) Example A student wants to check the fairness of the loonie She flips the coin 1,000,000 times, and gets heads 500,823 times. Is this a fair coin? A Third Option! Chi-squared goodness of fit test Null expectation: equal number of successes and failures Compare to chi-squared distribution with 1 d.f. <):=45 >1:)/?)@ A-B);5)@ C)2@: %$&#'D %$$$$$ E234:!"#&,F %$$$$$ Test statistic: 13.3 Critical value: 3.84

3 Coin toss example Coin toss example Normal distributions Binomial test Normal approximation Chi-squared goodness of fit test Binomial test Normal approximation Chi-squared goodness of fit test Any normal distribution can be converted to a standard normal distribution, by 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 in this case, n very large (1,000,000) all P < 0.05, reject null hypothesis!! " "# $ 9+:;0/) t distribution How do we use this? t has a Student s t distribution* We carry out a similar transformation on the sample mean t has a Student's t distribution Find confidence limits for the mean Carry out one-sample t-test mean under H o '" "# )! $* #% estimated standard error t has a Student s t distribution* Uncertainty makes the null distribution FATTER Confidence interval for a mean '" ('( '" ) )$&% *+, Confidence interval for a mean '" ('( '" ) )$&% *+, * Under the null hypothesis G'H(I('+5234)@(:3JK3L3;2K;)(4)?)4!"(I(@)J/)):(0L(L/))@0MN(K+& OA P (I(:52K@2/@()//0/(0L(5Q)(M)2K 95 % Confidence interval: Use!+&,-.-"("!

4 >K)+:2MB4)(5+5):5 Confidence interval for a mean O2MB4) R=44(Q6B05Q):3: #$%&'(')*+,-(.&/%+.& -0&%1)+*&,(& - The following are equivalent: '" ('( '" ) )$&% *+, E):5(:5253:53; )! '" "+ - $* #% ;0MB2/) R=44(@3:5/31=530K,&2-,$&.34&!" Test statistic > critical value P < alpha Reject the null hypothesis Statistically significant c % Confidence interval: Use!+&,-.-#/01#"" C0S(=K=:=24(3:(5Q3:(5):5(:5253:53;T.(U($V$%.(W($V$% <)X);5(C 0 Y234(50(/)X);5(C 0 Quick reference summary: One-sample t-test What is it for? Compares the mean of a numerical variable to a hypothesized value, " 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 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 Paired vs. 2 sample comparisons )! '" "+ - $* #% Paired designs Data from the two groups are paired There is a one-to-one correspondence between the individuals in the two groups 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 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

5 Paired comparisons - setup We have many pairs In each pair, there is one member that has one treatment and another who has another treatment Paired comparisons To compare two groups, we use the mean of the difference between the two members of each pair Example: National No Smoking Day Data compares injuries at work on National No (in Britain) to the same day the week before Each data point is a year Treatment can mean group Year data Injuries before No Injuries on No Calculate differences Injuries before No Injuries on No Difference (d) 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 Hypotheses Ho: Work related injuries do not change during No s ("=0) Ha: Work related injuries change during No s (" 70) Calculate differences Injuries before No Injuries on No Difference (d) Calculate t using d s t = d = 25 s d 2 = n = /10 =

6 \2=530K] 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 Critical value of t ) $"("!$&% *2%!&(&) Test statistic: t = 2.45 O0(S)(;2K(/)X);5(5Q)(K=44(Q6B05Q):3:Z(O50BB3KJ(:M0[3KJ( 3K;/)2:):(X01+/)425)@(2;;3@)K5:(3K(5Q)(:Q0/5(5)/MV 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 Quick reference summary: Paired t-test What is it for? To test whether the mean difference in a population equals a null hypothesized value, " 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: )! ' + "+ +- '( '+