Short-ish example: Handedness and Career Field
Long example: Handedness and Brain Mapping
ANOVA: Jumping from 2 sample t-test to 3+ samples.
Final review session (optional for the class)
Aug 7 10am-noon. West Mall Centre 3260.
Final exam questions recommended so far: Zero.
Don’t forget about the secondary resources on the bottom of
the webpage, they can help.
We took a survey of people in four career fields and found if
they were left or right handed.
These are the observed counts.
Most of the respondents are right handed except for in the
athletics field, where a few more than half are left handed. Is
this a fluke?
We want to know or if career and handedness are somehow
dependent.
We have a 2x4 crosstab, so we should use a chi-squared test.
These are the results:
Degrees of freedom =
2
χ =
There is
evidence against independence.
We have a 2x4 crosstab, so we should use a chi-squared test.
These are the results:
Degrees of freedom = 3
2
χ = 50.434
There is very significant evidence against independence.
The chi-squared test has a very small p-value (less than .001).
Do the results of this test tell us that there are more left
handed people in athletics in general?
The chi-squared test has a very small p-value (less than .001).
Do the results of this test tell us that there are more left
handed people in athletics in general?
No.
Chi-squared only checks whether two variables are
independent, not specific trends within them.
By comparing the expected and observed counts, we can see
that the athletic field is much different from the others.
We can use this information to guide a next step even if we’re
not getting definite answers from just the expected counts.
We could try merging the other three fields into “non-athletic”
and “athletic”, as long as those three fields together fairly
represented everything non-athletic.
In that case, the odds ratio shows that someone in the athletic
field has 7.371 times the odds of being left handed as
someone in a non-athletic profession.
The confidence interval shows that this odds ratio is
significantly more than 1 at the alpha = 0.025 level.
One more cross tab analysis.
Because they belong in pairs.
We randomly selected 150 people from the pool of righthanded people, 70 from the pool of left-handed people, and
10 ambidextrous people (who use both hands equally well).
Then we performed a brain scan and found that some people’s
brains operated in a “mirrored” fashion to most other people.
(The opposite half was activated that what we would expect)
Ordinary
Mirrored
Total
Right
126
24
150
Left
37
33
70
Both
4
6
10
Total
167
63
230
Ordinary
Mirrored
Total
Right
126
24
150
Left
37
33
70
Both
4
6
10
Total
167
63
230
a) What is the probability of someone from this sample
having mirrored brain activity?
Ordinary
Mirrored
Right
126
24
Left
37
33
Both
4
6
Total
150
70
10
Total
167
63
230
a) What is the probability of someone from this sample
having mirrored brain activity?
Probability uses the frequency of the event in question and the
total.
Pr(Mirrored) = # that are mirrored / # total
63 / 270 = 0.273
Ordinary
Right
126
Left
37
Both
4
Total
Mirrored
24
33
6
167
63
Total
150
70
10
230
b) What are the odds of someone from this sample having
mirrored brain activity?
Ordinary
Right
126
Left
37
Both
4
Total
Mirrored
24
33
6
167
63
Total
150
70
10
230
b) What are the odds of someone from this sample having
mirrored brain activity?
Odds use the frequency of the event and the frequency of
everything that wasn’t the event.
Odds of Mirrored = # mirrored to # not mirrored
= 63 to 167 = 0.377
Or: Pr / (1-Pr) = .273/(1-.273) = 0.377
c) Why can’t we use this data to estimate the probability
and odds that someone has mirrored brain activity in the
population as a whole?*
We selected 150 righties, 70 lefties, and 10 ambidextrous
people on purpose. They weren’t randomly selected.
We may have done this to increase the number of left-handed
people in the sample to improve the quality of methods like
the chi-squared test.
If we used this data to talk about the whole population (if we
inferred something about the population) we would be
assuming that about 70/230 or 30% of people are left handed.
(In reality, it’s more like 12%)
Each left-handed person had a greater chance of being in our
sample than each right-handed person, so the chances of
being in the sample weren’t equal for everyone.
That makes the sample, with regards to handed-ness, not
random.
Ordinary
Mirrored
Total
Right
126
24
150
Left
37
33
70
Both
4
6
10
Total
167
63
230
e) What are the odds of someone in this sample having
mirrored brain activity conditional on them being
right-handed?
Right
Ordinary
Mirrored
126
24
Total
150
Left
37
Both
4
Total
33
6
63
70
10
230
167
e) What are the odds of someone in this sample having
mirrored brain activity conditional on them being
right-handed?
Only use the data from the right-handed cases because of the
condition.
Odds = 24 to 126 = 0.190
e) Can we estimate the odds of a right-handed person in the
population having mirrored brain activity?
YES.
The distortion from selection a certain proportion of
left/right/both handed people doesn’t affect our estimate
when we’re only considering one type of handedness to begin
with.
For this question, we’re only interested in right-handed
people, so that’s our population. Every right-handed
person had an equal chance of being in the same, therefore
sampling was random.
What’s the difference between parts C and E?
Why is it okay to infer for only right handed people and not
every one?
When considering all people
Sample
Population
Right
65%
83%
Left
30%
12%
Both
5%
5%
When only considering right-handed people
Sample
Population
Right
100%
100%
Left
0%
0%
Both
0%
0%
When considering all people
Sample
Population
Right
65%
83%
Left
30%
12%
Both
5%
5%
If we had picked the number of left/right handed people to
match the population, this would have been a stratified
sample.
Each of the types of handedness would be a stratum, and
the brain activity would be the value of interest.
We want to test if handedness/brain activity are independent.
Ordinary
Mirrored
Right
126
24
Left
37
33
Both
4
6
Total
167
63
Total
150
70
10
230
We would use the chi-squared test.
However, there is a potential problem. We have very few
people in the both-handed cells. Chi-squared generally needs
an expected frequency of 5 in each cell to be accurate.
Ordinary
Mirrored
Right
126
24
Left
37
33
Both
4
6
Total
167
63
Total
150
70
10
230
The common solution:
Merge two categories.
Which two should we merge? For any benefit, it has to be
Both-handed with something?
In this case, it makes more sense to merge Left-handed and
Both-handed. The frequencies indicate they are the two most
similar categories.
Ordinary
Right
126
Mirrored
24
Total
150
Other
41
39
80
Total
167
63
230
Now we have a 2x2 cross tab. In this case, we could use either
the chi-square or the odds ratio.
The odds ratio is easier to compute, and it lets us do one-tailed
tests, so we’ll do that.
Ordinary
Mirrored
Right
126
24
Other
41
39
Total
167
63
Total
150
80
230
Odds ratio by quick formula: AD/BC
OR = (126)(39) / (24)(41) = 4.99
The odds of having mirrored brain activity if you something
other than right handed is 4.99 times as much as it is when you
are right handed.
Ordinary
Mirrored
Right
126 A
24 C
Other
41 B
39 D
The interpretation of the odds ratio is found directly in the
quick formula: AD/BC
The odds are (OR) times as high if (A or D) instead of (B or C)
Here, that’s the odds of (Ordinary Righthand or Mirrored
Left/Bothhand) are 4.99 times higher than (Ordinary
Left/Bothhand or Mirrored Righthand)
The 95% confidence interval of the odds ratio is
(2.656 to 9.389)
That means at alpha = 0.05, the handed-ness and having
mirrored brain activity or not are dependent on each other.
Why? The odds ratio of 1 is not in the confidence interval.
An odds ratio of 1 implies that the odds of one thing are just as
much as the odds of another thing. In our case that would be
the odds of having a mirrored brain are the same regardless of
handedness.
This slide for interest, skip in lecture:
*Technically,
we could use our sample to infer to the population (for all
handedness), but we would need a method called weighting.
Since we oversampled the left-handed people, we could use a
weighted probability that counts each sampled left-handed
person for less so that their total contribution to the
probability, their WEIGHT in other words, matched the
proportion of people that are left handed in the population.
Here, lefties are 30% of the sample, but 12% of the population,
so each lefty would count for a 12/30 of a sampled person.
19 slide example problem, woo! All that leaves is AnOVa
A simple ANOVA in a familiar form.
Around Week 7-9, we looked at how to find a difference
between the means of two groups.
We did this by taking a sample mean from each group and
comparing the difference to the standard error.
Our method of testing whether this difference was significant
(in other words, testing the null hypothesis that the difference
between the true means was zero) was the t-score.
It was always t = (difference) / Standard Error.
The definition of standard error depended on the details
(paired/independent, pooled/non-pooled standard deviation)
If the difference was bigger, the t-score was bigger and we
more often rejected the null hypothesis.
It’s easier to say a difference is real when the sample mean
difference is larger. (Easier to detect larger effects)
If there was more scatter between the points within a group,
the standard error got bigger, and we more often failed to
reject the null hypothesis.
Standard error also gets smaller when there are more data
points.
In every case that the t-test is used, you’re ultimately just
answering one question over and over again:
Are the differences between the two groups large compared
to the differences within each group?
Can we use the t-test to determine if there are differences
between any of the three means from three samples?
We can’t do this all as a single t-test, because the t-test is only
a comparison between two sample means. We have three
We could test each pair of groups and look for differences.
If we found a significant difference between two means, that
would imply that not all the means are the same.
We’d need to test:
Mean of group 1 vs group 2
Mean of group 2 vs group 3
Mean of group 1 vs group 3
Doing multiple t-tests takes time, and what’s worse: It opens
up the issue of multiple testing (the more tests you to, more
likely you are to commit an error like falsely rejecting the null)
A much cleaner solution is the F-statistic of ANOVA.
MS stands for Mean Squared, and MSwithin is the average
squared difference from a data point to the average for the
that group. It’s the mean squared WITHIN a group
If we were just looking at a single group, this average squared
distance would be the standard deviation squared, or the
variance.
MSwithin is large when the spread within the samples is large.
Spread/variance within a sample makes it hard to detect
differences between the samples, and so the F-statistics gets
smaller, just like the t-statistic.
MSbetween is large when there are large differences between the
sample means. MSbetween stands for the differences between
means, instead of within them.
Here, the average (squared) difference from a group mean to
the grand mean, the average of data points from all the
groups put together, is much larger than the differences
between each point and its group mean.
F will be large and there is strong evidence that there is some
difference is the true means between the groups.
Next time: More on ANOVA, ANOVA tables, and examples.