How do we estimate P-values using randomization distributions?
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
Today we’ll discuss ways to simulate randomization samples for a variety of situations.

• 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?

• What are the null and alternative hypotheses?
• What are the possible conclusions?

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

R R R R R R
R R R R R R
R R R R R R
R R R R R R
R R R R R R
R R R R R R
R R R R R R
R R R R R R
Desipramine
R R R R
R R R R R R
R R R R R R
R R R R R R
1. Randomly assign units to treatment groups
Lithium
R R R R
R R R R R R
R R R R R R
R R R R R R

2. Conduct experiment
3. Observe relapse counts in each group
R = Relapse
N = No Relapse
Desipramine
1. Randomly assign units to treatment groups
R R R R
R R R R R R
N N N N N N
N N N N N N
10 relapse, 14 no relapse
 p
D
 p
L
10

18
24 24
 
.333
Lithium
R R R R
R R R R R R
R R R R R R
N N N N N N
18 relapse, 6 no relapse

0
0,
0

• “ 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…

R R R R R R
R R R R N N
N N N N N N
N N N N N N
10 relapse, 14 no relapse
R R R R R R
R R R R R R
R R R R R R
N N N N N N
18 relapse, 6 no relapse

R R R R R R
R R R R N N
N N N N N N
N N N N N N
R R R R R R
R R R R R R
R R R R R R
N N N N N N
Desipramine
R N R N
R R R R R R
R N R R R N
R N N N R R
16 relapse, 8 no relapse
Simulate another randomization p
D
 p
L

16

12
24 24

0.167
Lithium
N N N R
N R R N N N
N R N R R N
R N R R R R
12 relapse, 12 no relapse

Desipramine
R R R R R R
R N R R N N
R R N R N R
R N R N R R
17 relapse, 7 no relapse
Simulate another randomization p
D
 p
L

17

11
24 24

0.250
Lithium
R R R R R R
R R R R R R
R R R R R R
N N N N N N
11 relapse, 13 no relapse


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: StatKey

p-value www.lock5stat.com/statkey

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

In a hypothesis test for H
0
:  = 12 vs H a
:  < 12, we have a sample with n = 45 and 𝑥 = 10.2.
What do we require about the method to produce randomization samples?
a) b) c)
 = 12
 < 12 𝑥 = 10.2
We need to generate randomization samples assuming the null hypothesis is true.

In a hypothesis test for H
0
:  = 12 vs H a
:  < 12, we have a sample with n = 45 and 𝑥 = 10.2
.
Where will the randomization distribution be centered?
a) b) c) d)
10.2
12
45
1.8
Randomization distributions are always centered around the null hypothesized value.

In a hypothesis test for H
0
:  = 12 vs H a
:  < 12, we have a sample with n = 45 and 𝑥 = 10.2.
a) b) c) d) e)
What will we look for on the randomization distribution?
How extreme 10.2 is
How extreme 12 is
How extreme 45 is
What the standard error is
We want to see how extreme the observed statistic is.
How many randomization samples we collected

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? a) b) c) d)

1

1 𝑥
1 𝑥
1
= 
2
> 
2
= 26, 𝑥
2 𝑥
2
= 5
= 21
We need to generate randomization samples assuming the null hypothesis is true.

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? a) b) c) d) e)
0
1
21
26
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.
What do we look for in the randomization distribution?
a) b) c) d) e)
The standard error
The center point
How extreme 26 is
How extreme 21 is
How extreme 5 is
We want to see how extreme the observed difference in means is.

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

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

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!

Let   the average human body temperature
H
0
H a
:  = 98.6
:  ≠ 98.6
sample mean = 98.26
• 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!

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

Exercise and Gender
Do males exercise more hours per week than females? sample mean difference x m
– x f
= 3
1.
2.
State null and alternative hypotheses
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

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.

p-value =
0.095

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

Blood Pressure and Heart Rate
Is blood pressure negatively correlated with heart rate?
sample corre lation r = -0.037
1.
2.
State null and alternative hypotheses
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

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

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.

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

• 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.


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