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Section 4.4 Creating Randomization Distributions.

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Presentation on theme: "Section 4.4 Creating Randomization Distributions."— Presentation transcript:

1 Section 4.4 Creating Randomization Distributions

2 Randomization Distributions How do we estimate P-values using randomization distributions? Today we’ll discuss ways to simulate randomization samples for a variety of situations. 1.Simulate samples, assuming H 0 is true 2.Calculate the statistic of interest for each sample 3.Find the p-value as the proportion of simulated statistics as extreme as the observed statistic

3 In a randomized experiment on treating cocaine addiction, 48 people were randomly assigned to take either Desipramine (a new drug), or Lithium (an existing drug), and then followed to see who relapsed Question of interest: Is Desipramine better than Lithium at treating cocaine addiction? Cocaine Addiction

4 What are the null and alternative hypotheses? What are the possible conclusions? Cocaine Addiction

5 What are the null and alternative hypotheses? What are the possible conclusions? Cocaine Addiction Reject H 0 : Desipramine is better than Lithium Do not reject H 0 : We cannot determine from these data whether Desipramine is better than Lithium Let p D, p L be the proportion of cocaine addicts who relapse after taking Desipramine or Lithium, respectively. H 0 : p D = p L H a : p D < p L

6 RRRRRR RRRRRR RRRRRR RRRRRR RRRRRR RRRRRR RRRRRR RRRRRR RRRR RRRRRR RRRRRR RRRRRR RRRR RRRRRR RRRRRR RRRRRR Desipramine Lithium 1. Randomly assign units to treatment groups

7 RRRR RRRRRR RRRRRR NNNNNN RRRRRR RRRRNN NNNNNN RR NNNNNN R = Relapse N = No Relapse RRRR RRRRRR RRRRRR NNNNNN RRRRRR RRRRRR RRNNNN RR NNNNNN 2. Conduct experiment 3. Observe relapse counts in each group Lithium Desipramine 10 relapse, 14 no relapse18 relapse, 6 no relapse 1. Randomly assign units to treatment groups

8 To see if a statistic provides evidence against H 0, we need to see what kind of sample statistics we would observe, just by random chance, if H 0 were true Measuring Evidence against H 0

9 “by random chance” means by the random assignment to the two treatment groups “if H 0 were true” means if the two drugs were equally effective at preventing relapses (equivalently: whether a person relapses or not does not depend on which drug is taken) Simulate what would happen just by random chance, if H 0 were true… Cocaine Addiction

10 RRRR RRRRRR RRRRRR NNNNNN RR RRRR RRRRNN NNNNNN RR NNNNNN 10 relapse, 14 no relapse18 relapse, 6 no relapse

11 RRRRRR RRRRNN NNNNNN NNNNNN RRRRRR RRRRRR RRRRRR NNNNNN RNRN RRRRRR RNRRRN RNNNRR NNNR NRRNNN NRNRRN RNRRRR Simulate another randomization Desipramine Lithium 16 relapse, 8 no relapse12 relapse, 12 no relapse

12 RRRR RRRRRR RRRRRR NNNNNN RR RRRR RNRRNN RRNRNR RR RNRNRR Simulate another randomization Desipramine Lithium 17 relapse, 7 no relapse11 relapse, 13 no relapse

13 Simulate Your Own Sample In the experiment, 28 people relapsed and 20 people did not relapse. Create cards or slips of paper with 28 “R” values and 20 “N” values. Pool these response values together, and randomly divide them into two groups (representing Desipramine and Lithium) Calculate your difference in proportions Plot your statistic on the class dotplot To create an entire randomization distribution, we simulate this process many more times with technology: StatKeyStatKey

14 www.lock5stat.com/statkey p-value

15 Randomization Distribution Center A randomization distribution is centered at the value of the parameter given in the null hypothesis. A randomization distribution simulates samples assuming the null hypothesis is true, so

16 Randomization Distribution We need to generate randomization samples assuming the null hypothesis is true.

17 Randomization Distribution a) 10.2 b) 12 c) 45 d) 1.8 Randomization distributions are always centered around the null hypothesized value.

18 Randomization Distribution a) How extreme 10.2 is b) How extreme 12 is c) How extreme 45 is d) What the standard error is e) How many randomization samples we collected We want to see how extreme the observed statistic is.

19 Randomization Distribution In a hypothesis test for H 0 :  1 =  2, H a :  1 >  2 sample mean #1 = 26 and sample mean #2 = 21. What do we require about the method to produce the randomization samples? We need to generate randomization samples assuming the null hypothesis is true.

20 Randomization Distribution a) 0 b) 1 c) 21 d) 26 e) 5 The randomization distribution is centered around the null hypothesized value,  1 -  2 = 0 In a hypothesis test for H 0 :  1 =  2, H a :  1 >  2 sample mean #1 = 26 and sample mean #2 = 21. Where will the randomization distribution be centered?

21 Randomization Distribution a) The standard error b) The center point c) How extreme 26 is d) How extreme 21 is e) How extreme 5 is We want to see how extreme the observed difference in means is. In a hypothesis test for H 0 :  1 =  2, H a :  1 >  2 sample mean #1 = 26 and sample mean #2 = 21. What do we look for in the randomization distribution?

22 Randomization Distribution For a randomization distribution, each simulated sample should… be consistent with the null hypothesis use the data in the observed sample reflect the way the data were collected

23 In randomized experiments the “randomness” is the random allocation to treatment groups If the null hypothesis is true, the response values would be the same, regardless of treatment group assignment To simulate what would happen just by random chance, if H 0 were true: Reallocate cases to treatment groups, keeping the response values the same Randomized Experiments

24 Observational Studies In observational studies, the “randomness” is random sampling from the population To simulate what would happen, just by random chance, if H 0 were true: Simulate drawing samples from a population in which H 0 is true How do we simulate sampling from a population in which H 0 is true when we only have sample data? Adjust the sample to make H 0 true, then bootstrap!

25 Let  the average human body temperature H 0 :  = 98.6 H a :  ≠ 98.6 Adjust the sample by adding 98.6 – 98.26 = 0.34 to each value. The sample mean becomes 98.6, exactly the value given by the null hypothesis. Bootstrapping the adjusted sample allows us to simulate drawing samples as if the null is true! Body Temperatures sample mean = 98.26

26 In StatKey, when we enter the null hypothesis, this shifting is automatically done for us StatKey Body Temperatures p-value = 0.002

27 Exercise and Gender 1. State null and alternative hypotheses 2. Devise a way to generate a randomization sample that Uses the observed sample data Makes the null hypothesis true Reflects the way the data were collected Do males exercise more hours per week than females? sample mean difference x m – x f = 3

28 Exercise and Gender 1.H 0 :  m =  f H a :  m >  f 2.Generating a randomization distribution can be done with the “shift groups” method: To make H 0 true set the sample means equal by adding 3 to every female value. Now bootstrap from this modified sample Note: There are other ways. In StatKey, the default randomization method is “Reallocate Groups”, but “Shift Groups” is also an option.

29 Exercise and Gender p-value = 0.095

30 Exercise and Gender The p-value is 0.095. Using α = 0.05, we conclude…. a) Males exercise more than females, on average b) Males do not exercise more than females, on average c) Nothing Do not reject the null… we can’t conclude anything.

31 Blood Pressure and Heart Rate Is blood pressure negatively correlated with heart rate? 1. State null and alternative hypotheses 2. Devise a way to generate a randomization sample that Uses the observed sample data Makes the null hypothesis true Reflects the way the data were collected sample corre lation r = -0.037

32 Blood Pressure and Heart Rate 1.H 0 :  = 0 H a :  < 0 2.Generating a randomization distribution: Two variables have correlation 0 if they are not associated (null hypothesis). We can “break the association” by randomly shuffling one of the variables. Each time we do this, we get a sample we might observe just by random chance, if there really is no correlation

33 Blood Pressure and Heart Rate p-value = 0.219 Even if blood pressure and heart rate are not correlated, we would see correlations this extreme about 22% of the time, just by random chance.

34 Randomization Distributions: Cocaine Addiction (randomized experiment) Rerandomize cases to treatment groups, keeping response values fixed Body Temperature (single mean) Shift to make H 0 true, then bootstrap Exercise and Gender (observational study) Shift to make H 0 true, then bootstrap Blood Pressure and Heart Rate (correlation) Randomly shuffle one variable

35 As long as the original data is used and the null hypothesis is true for the randomization samples, most methods usually give similar p-values StatKey generates the randomizations for us. We will not be concerned with the details of the process. It is enough to understand the general principles. Generating Randomization Samples

36 Summary Randomization samples should be generated Consistent with the null hypothesis Using the observed data Reflecting the way the data were collected The specific method varies with the situation, but the general idea is always the same


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