Transcript Chapter 5 Slides

Unit 2: Comparing Two Groups
In Unit 1, we learned the basic process of
statistical inference using tests and
confidence intervals. We did all this by
focusing on a single proportion.
 In Unit 2, we will take these ideas and
extend them to comparing two groups.
We will compare two proportions, two
independent means, and paired data.

Chapter 5:
Comparing Two Proportions
5.1: Descriptive (Two-Way Tables)
5.2: Inference with Simulation-Based Methods
5.3: Inference with Theory-Based Methods
Positive and Negative Perceptions

Consider these two questions:
◦ Are you having a good year?
◦ Are you having a bad year?

Do people answer each question in such
a way that would indicated the same
answer? (e.g. Yes for the first one and
No for the second.)
Positive and Negative Perceptions
Researchers questioned 30 students
(randomly giving them one of the two
questions).
 They then recorded if a positive or
negative response was given.
 Is this an observational study or
randomized experiment?

Positive and Negative Perceptions

Observational units
◦ The 30 students

Variables
◦ Question wording (good year or bad year)
◦ Perception of their year (positive or negative)

Which is the explanatory and which is the
response?
Raw Data in a Spreadsheet
Individual
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Type of
Question
Good Year
Good Year
Bad Year
Good Year
Good Year
Bad Year
Good Year
Good Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Response
Individual
Positive
Negative
Positive
Positive
Negative
Positive
Positive
Positive
Positive
Negative
Negative
Negative
Positive
Negative
Positive
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Type of
Question
Good Year
Bad Year
Good Year
Good Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Bad Year
Response
Positive
Positive
Positive
Positive
Positive
Negative
Positive
Negative
Positive
Negative
Positive
Negative
Positive
Positive
Negative
Two-Way Tables

A two-way table organizes data
◦ Summarizes two categorical variables
◦ Also called contingency table

Are students more likely to give a positive
response if they were given the good year
question?
Positive response
Negative response
Total
Good Year
Bad Year
Total
15
3
18
4
8
12
19
11
30
Two-Way Tables


Conditional proportions will help us better
determine if there is an association between
the question asked and the type of response.
We can see that those given the positive
question were more likely to respond
positively.
Good Year
Positive response
Negative response
Total
Bad Year
15/18 ≈ 0.83 4/12 ≈ 0.33
3
8
18
12
Total
19
11
30
Segmented Bar Graphs


We can use segmented
bar graphs to see this
association.
Remember that variables
are associated if the
conditional proportion of
the outcomes for one
group differ from the
conditional proportion of
outcomes in other
groups.
Perceptions
Those responding to the good year
question were more likely to answer
positively (83% to 33%) than those
responding positively to the bad year
question.
 The statistic we will be using to measure
this is the difference in proportions.

◦ 0.83 - 0.33 = 0.50 higher for the good year
question than the bad year question.
A Sneak Peak at Comparing Two
Proportions: Simulation-Based Approach
In the next section we will conduct tests of
significance to compare two proportions and I
want to give you a preview of that here.
 We will assume there is no association between
the variables (i.e. the two population
proportions are the same) and decide if two
sample proportions differ enough to conclude
this would be very unlikely just by random
chance.

Hypotheses
Null Hypothesis: There is no association
between which question is asked and the
type of response. (The proportion of
positive responses will be the same in
each group. )
 Alternative Hypothesis: There is an
association between which question is
asked and the type of response. (The
proportion of positive responses will be
different in each group. )

Results


The difference in proportions of positive
responses is 0.83 − 0.33 = 0.50.
How likely is a difference this great or
greater if the type of question asked made
no difference in how the student would
respond?
Positive response
Negative response
Total
Good Year
Bad Year
Total
15 (83%)
3
18
4 (33%)
8
12
19
11
30
Random Reassignment
Notice that 19 students gave a positive response. If the
null hypothesis is true, these 19 would have given a
positive response no matter which question was asked.
 Therefore, under a true null hypothesis, we can
randomly place these 19 people into either group and
they will still give a positive response. This replicates
the random assignment that was done in the
experiment.
 We will also keep constant the 18 that receive the
positive question and 12 that receive the negative
question.

Positive response
Negative response
Total
Good Year
Bad Year
Total
random
random
18
random
random
12
19
11
30
You can think about this random reassignment with the
raw data as well. It doesn’t matter which question was
asked, the responses will be the same. Therefore, we
can shuffle the type of question and leave the responses
fixed. This is equivalent to keeping the same column and
row totals and just shuffling the inside of the two-way
table as described earlier.
Individual
1
2
3
4
5
6
7
Type of Question
Good Year
Good Year
Bad Year
Good Year
Good Year
Bad Year
Good Year
Response
Positive
Negative
Positive
Positive
Negative
Positive
Positive
Individual
16
17
18
19
20
21
22
Type of Question
Good Year
Bad Year
Good Year
Good Year
Good Year
Bad Year
Good Year
Response
Positive
Positive
Positive
Positive
Positive
Negative
Positive
8
9
10
11
12
13
14
15
Good Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Positive
Positive
Negative
Negative
Negative
Positive
Negative
Positive
23
24
25
26
27
28
29
30
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Good Year
Bad Year
Bad Year
Negative
Positive
Negative
Positive
Negative
Positive
Positive
Negative
Random Reassignment

I did this once and found a difference in
the proportions of positive responses for
the two questions of 0.50 − 0.83 = −0.33
Positive response
Negative response
Total
Good Year
Bad Year
Total
9 (50%)
9
18
10 (83%)
2
12
19
11
30
Random Reassignment

I did this again and found a difference in
the proportions of positive responses for
the two questions of 0.61 − 0.67 = −0.06
Positive response
Negative response
Total
GoodYear
BadYear
Total
11 (61%)
7
18
8 (67%)
4
12
19
11
30
Random Reassignment

I did this again and found a difference in
the proportions of positive responses for
the two questions of 0.67 − 0.58 = 0.09
Positive response
Negative response
Total
GoodYear
Bad Year
Total
12 (67%)
6
18
7 (58%)
5
12
19
11
30
Random Reassignment
In my three randomizations, I have yet to
see a difference in proportions that is as
far away from zero as the observed
difference of 0.5.
 Let’s do some more randomizations to
develop a null distribution.

Collection 1
-0.6
Dot Plot
-0.4 -0.2
0.0
0.2
0.4
difference_in_proportions
0.6
Random Reassignment

After 1000 randomizations, only 7 were as
far away from zero as our observed
proportion.
Conclusion

Since we have a p-value of 7/1000 or
0.007, we can conclude the alternative
hypothesis and say we have strong
evidence that how the question is
phrased affects the response.
Applets

Let’s look at how this is done in two
applets
◦ Simulation for Two Proportions
◦ Simulation for Multiple Proportions
Exploration 5.1: Murderous Nurse?
Example 5.2: Swimming With Dolphins
Swimming with Dolphins
Is swimming with dolphins therapeutic
for patients suffering from clinical
depression?
 Researchers recruited 30 subjects aged 1865 with a clinical diagnosis of mild to
moderate depression.
 Discontinued antidepressants and
psychotherapy 4 weeks prior to and
throughout the experiment
 30 subjects went to an island near Honduras
 Randomly assigned to two treatment groups
Swimming with Dolphins





Both groups engaged in one hour of
swimming and snorkeling each day.
One group swam in the presence of
dolphins and the other group did not.
Participants in both groups had identical
conditions except for the dolphins
After 2 weeks, each subjects’ level of
depression was evaluated, as it had been at
the beginning of the study
The response variable is if the subject
achieved substantial reduction in depression.
Swimming with Dolphins

Observational units
◦ The 30 subjects with mild to moderate
depression.

Explanatory variable
◦ Swimming with dolphins or not

Response variable
◦ Reduction in depression or not

Are the variables quantitative or
categorical?
Swimming with Dolphins
Is this study an observational study or an
experiment?
 Are the subjects in this study a random
sample from a larger population?

Swimming with Dolphins
Results
Dolphin
group
Control
group
Total
Improved
10 (67%)
3 (20%)
13
Did Not Improve
5
12
17
Total
15
15
30
Swimming with Dolphins
The difference in proportions of
improvers is 0.67 – 0.20 = 0.47.
 There are two possible explanations for
an observed difference of 0.47.

◦ A tendency to be more likely to improve with
dolphins
◦ The 13 subjects were going to show
improvement with or without dolphins and
random chance assigned more improvers to
the dolphins
Swimming with Dolphins
Null hypothesis: Dolphins don’t help
◦ Swimming with dolphins is not associated
with substantial improvement in depression
Alternative hypothesis: Dolphins help
◦ Swimming with dolphins increases the
probability of substantial improvement in
depression symptoms
Swimming with Dolphins
The parameter is the (long-run)
difference in the probability of improving
between receiving dolphin therapy and
the control.
 Our statistic is the observed difference in
sample proportions (𝑝𝑑𝑜𝑙𝑝ℎ𝑖𝑛𝑠 − 𝑝𝑐𝑜𝑛𝑡𝑟𝑜𝑙 )
or 0.67 – 0.20 = 0.47.

Swimming with Dolphins
Null Hypothesis: The probability someone
exhibits substantial improvement after swimming
with dolphins is the same as the probability
someone exhibits substantial improvement after
swimming without dolphins.
Alternative Hypothesis: The probability
someone exhibits substantial improvement after
swimming with dolphins is higher than the
probability someone exhibits substantial
improvement after swimming without dolphins.
Swimming with Dolphins

Since we defined our parameter as the
difference in probability of improving
between the 2 groups, we can write our
hypotheses this way as well:
H0: 𝜋dolphins = 𝜋control
𝜋dolphins  𝜋control = 0
H a:
𝜋dolphins > 𝜋control
𝜋dolphins  𝜋control > 0
Swimming with Dolphins
If the null hypothesis is true (dolphin
therapy is not better) we would have 13
improvers and 17 non-improvers regardless
of the group they were in.
Any differences we see between groups
arise solely from the randomness in the
assignment to the groups.
Swimming with Dolphins

We can perform this simulation with cards.
◦ 13 black cards represent the improvers
◦ 17 red cards represent the non-improvers



We assume these outcomes would happen
no matter which treatment group subjects
were in.
Shuffle the cards and put 15 in one pile
(dolphin therapy) and 15 in another (control
group)
An improver is equally likely to be assigned
to each group
Swimming with Dolphins
In the actual study, there were 10
improvers (diff of 0.47) in the dolphin
group.
 We conducted 3 simulations and got 8, 5,
and 6 improvers in the dolphin therapy
group. (notice the diff in proportions)

Swimming with Dolphins

We did 1000 repetitions to develop a null
distribution.
◦ Why is it centered at about 0?
◦ What does each dot represent?
Swimming with Dolphins

13 out of 1000 results had a difference of
0.47 or higher (p-value = 0.013).
0.47−0
0.178
≈ 2.65 SD above zero.

0.47 is

Using either the p-value or standardized
statistic, we have strong evidence against the
null and can conclude that swimming with
dolphins increases the probability of substantial
improvement in depression symptoms.
Swimming with Dolphins

A 95% confidence interval for the difference in
the probability using the standard deviation
from the null distribution is 0.467 + 2(0.178) =
0.467 + 0.356 or (0.111to 0.823)
We are 95% confident that when allowed
to swim with dolphins, the probability of
improving is between 0.111 and 0.823
higher than when no dolphins are present.
 How does this interval back up our
conclusion from the test of significance?

Swimming with Dolphins

Can we say that the presence of dolphins
caused this improvement?
◦ Since this was a randomized experiment, and
assuming everything was identical between the
groups, we have strong evidence that dolphins
were the cause

Can we generalize to a larger population?
◦ Maybe mild to moderately depressed 18-65 year
old patients willing to volunteer for this study
◦ We have no evidence that random selection was
used to find the 30 subjects.
Exploration 5.2: Contagious Yawns?



MythBusters investigated this.
50 subjects were ushered into a small room
by co-host Kari.
She yawned as she ushered 34 in the room
and for 16 she didn’t yawn. We will assume
she randomly decided who would received
the yawns.
Comparing Two Proportions:
Theory-Based Approach
Section 5.3
Introduction
Just as with a single proportion, we can
often predict results of a simulation using
a theory-based approach.
 The theory-based approach also gives a
simpler way to generate a confidence
intervals.

Smoking and Birth Gender
Smoking and Gender
How does parents’ behavior affect the sex of
their children?
 Fukuda et al., 2002 (Japan) found the following:

◦ 255 of 565 births (45.1%) where both parents
smoked more than a pack a day were boys.
◦ 1975 of 3602 births (54.8%) where both parents did
not smoke were boys.

Other studies have shown a reduced male to
female birth ratio where high concentrations of
other environmental chemicals are present (e.g.
industrial pollution, pesticides)
Smoking and Gender


A segmented bar graph and 2-way table
Let’s compare the proportions to see if the
difference is statistically significantly.
Smoking and Gender
Null Hypothesis:
 There is no association between smoking
status of parents and sex of child.
 The probability of having a boy is the same for
parents who smoke and don’t smoke.
 𝜋smoking - 𝜋nonsmoking = 0
Alternative Hypothesis:
 There is an association between smoking status
of parents and sex of child.
 The probability of having a boy is not the same
for parents who smoke and don’t smoke
 𝜋smoking - 𝜋nonsmoking ≠ 0
Smoking and Gender
What are the observational units in the
study?
 What are the variables in this study?
 Which variable should be considered the
explanatory variable and which the
response variable?
 Can you draw cause-and-effect
conclusions for this study?

Smoking and Gender
OK to shuffle?
 In the last section we “re-randomized”
subjects to treatment groups to simulate
the null distribution.
 In this study the parents weren’t
randomized to the treatment, since it’s
observational, but we can still represent
the null hypothesis of no association
through randomization.

Smoking and Gender
Use the 3S Strategy to asses the strength
1. Statistic:
 The proportion of boys born to
nonsmokers minus boys born to smokers
is 0.548 – 0.451 = 0.097.
Smoking and Gender
2. Simulate:
 Use the Multiple Proportions applet to
simulate
 Many repetitions of shuffling the 2230 boys
and 1937 girls to the 565 smoking and 3602
nonsmoking parents
 Calculate the difference in proportions of
boys between the groups for each repetition.
 Shuffling simulates the null hypothesis of no
association
Smoking and Gender
3. Strength of
evidence:
 Nothing as extreme
as our observed
statistic (≥ 0.097 or
≤ −0.097) occurred
in 5000 repetitions,
 How many SDs is
0.097 above the
mean?
Smoking and Gender
Notice the null distribution is centered at
zero and is bell-shaped.
 This, along with its standard deviation can
be predicted using normal distributions.

Smoking and Gender

We can use either the Multiple
Proportion applet or the Theory-Based
Inference applet to find the p-value
Smoking and Gender
Estimation (Confidence Intervals)
How different are the population proportions
of having a boy between non-smoking and
smoking parents?
 How different are the probabilities of having a
boy between non-smoking and smoking
parents?
 The parameter we are estimating is the
difference in population proportions (or
probabilities) (𝜋ns − 𝜋s).
 It should be centered at 0.097, our sample
difference in proportions


Smoking and Gender

From our test of significance, do we
expect 0 to be in the interval of plausible
values for the difference in the population
proportions?
Smoking and Gender

Again, either applet can be used to
determine a confidence interval.

We are 95% confident that the probability
of a boy baby is 0.053 to 0.141 higher for
families where neither parent smokes
compared to families with two smoking
parents
Smoking and Gender

We can also write the confidence interval in
the form:
◦ statistic ± margin of error.
Our statistic is the observed sample
difference in proportions, 0.097.
 We can find the margin of error by
subtracting the statistic (center) from the
upper endpoint or 0.141 – 0.097 = 0.044.
 0.097 ± 0.044

◦
Is the margin of error about the standard deviation?
Smoking and Gender

How would the interval change if the
confidence level was 99%?
Smoking and Gender
Written as the statistic ± margin of error
0.097 ± 0.058.
 Margin of error

◦ 0.058 for the 99% confidence interval
◦ 0.044 for the 95% confidence interval
Smoking and Gender

How would the 95% confidence interval
change if we were estimating
𝜋smoker – 𝜋nonsmoker
instead of
𝜋nonsmoker – 𝜋smoker ?
Smoking and Gender
(−0.141, −0.053) or −0.097 ± 0.044
instead of
 (0.053, 0.141) or 0.097 ± 0.044


The negative signs indicate the probability
of a boy born to smoking parents is lower
than that for nonsmoking parents.
Smoking and Gender
Validity Conditions of Theory-Based
 Same as with a single proportion.
 Should have at least 10 observations in
each of the cells of the 2 x 2 table.
Male
Female
Total
Smoking
Parents
Nonsmoking
Parents
Total
255
310
565
1975
1627
3602
2230
1937
4167
Smoking and Gender



The strong significant result in this study
yielded quite a bit of press when it came out
Soon other studies came out which found
no relationship between smoking and gender
One article argued that confounding
variables like social factors, diet,
environmental exposure or stress were the
reason for different study’s results. (These
are all possible since it was an observational
study.)
Formulas

How was z = 4.30 found?
𝑧=
𝑝1 − 𝑝2
1
1
𝑝(1 − 𝑝)
+
𝑛1 𝑛2
Formulas

How was the margin of error for the
difference in proportions 0.044 found?
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑟 ⨯

𝑝1 (1 − 𝑝1 ) 𝑝2 (1 − 𝑝2 )
+
𝑛1
𝑛2
The multiplier is dependent upon the
confidence level.
◦ 1.645 for 90% confidence
◦ 1.96 for 95% confidence
◦ 2.576 for 99% confidence
Strength of Evidence
As the proportions move farther away
from each other, the strength of evidence
increases.
 As sample size increases, the strength of
evidence increases.

• Let’s run this
previous test using
both the
Simulation-Based
and the TheoryBased Applets.
• Donating Blood
Exploration 5.3
• Questions 1-14
(skip 2 and 5)