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Testing means, part II
The paired t-test
Outline of lecture
• 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?
Ho: The coin is fair (pheads = 0.5).
Ha: The coin is not fair (pheads ≠ 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
8 e-04
6 e-04
Probability
4 e-04
2 e-04
0 e+00
498000
499000
500000
x
501000
502000
Test statistic
Binomial test
• P = 2*Pr[X≥501,823]
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
Normal approximation to the
binomial distribution
The binomial distribution, when number of trials n is large
and probability of success p is not close to 0 or 1, is
approximated by a normal distribution having mean np and
standard deviation 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= n p 1− p = 1,000,000∗ 0.5∗ 1− 0.5
• s = 500
Normal distributions
• Any normal distribution can be converted to
a standard normal distribution, by
Z=
Y− m
s
Z-score
Z=
Y− m
s
501,823− 500,000
Z=
= 3.646
500
From standard normal table:
P = 0.0001
Conclusion
• P = 0.0001, 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
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.
Result
Heads
Tails
Observed
Expected
501823
500000
498167
500000
Test statistic: 13.3
Critical value: 3.84
Coin toss example
Binomial test
Normal
approximation
Most accurate
Hard to calculate
Assumes:
Random sample
Approximate
Easier to calculate
Assumes:
Random sample
Large n
p far from 0, 1
Chi-squared
goodness of fit
test
Approximate
Easier to calculate
Assumes:
Random sample
No expected <1
Not more than 20%
less than 5
Coin toss example
Binomial test
Normal
approximation
Chi-squared
goodness of fit
test
in this case, n very large (1,000,000)
all P < 0.05, reject null hypothesis
Normal distributions
• Any normal distribution can be converted to
a standard normal distribution, by
Z=
Y− m
s
Z-score
t distribution
• We carry out a similar transformation on the
sample mean
mean under Ho
Y− m
t=
s/ n
estimated
standard error
How do we use this?
• t has a Student's t distribution
• Find confidence limits for the mean
• Carry out one-sample t-test
t has a Student’s t distribution*
t has a Student’s t distribution*
Uncertainty
makes the null
distribution
FATTER
* Under the null hypothesis
Confidence interval for a mean
Y ± SE Y t
2 ,df
(2) = 2-tailed significance level
df = degrees of freedom, n-1
SEY = standard error of the mean
Confidence interval for a mean
Y ± SE Y t
2 ,df
95 % Confidence interval:
Use α(2) = 0.05
Confidence interval for a mean
Y ± SE Y t
2 ,df
c % Confidence interval:
Use α(2) = 1-c/100
One-sample t-test
Null hypothesis
The population mean
is equal to o
Sample
Test statistic
t=
Y−
o
compare
Null distribution
t with n-1 df
s/ n
How unusual is this test statistic?
P < 0.05
Reject Ho
P > 0.05
Fail to reject Ho
The following are equivalent:
•
•
•
•
Test statistic > critical value
P < alpha
Reject the null hypothesis
Statistically significant
Quick reference summary:
One-sample t-test
• What is it for? Compares the mean of a numerical variable
to a hypothesized value, μo
• What does it assume? Individuals are randomly sampled
from a population that is normally distributed
• Test statistic: t
• Distribution under Ho: t-distribution with n-1 degrees of
freedom
• Formulae:Y = sample mean, s = sample standard deviation
t=
Y−
o
s/ n
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
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Paired vs. 2 sample comparisons
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Paired designs
• Data from the two groups are paired
• There is a one-to-one correspondence
between the individuals in the two groups
34
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
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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
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Paired comparisons - setup
• We have many pairs
• In each pair, there is one member that has
one treatment and another who has another
treatment
• “Treatment” can mean “group”
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Paired comparisons
• To compare two groups, we use the mean of
the difference between the two members of
each pair
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Example: National No Smoking
Day
• Data compares injuries at work on National
No Smoking Day (in Britain) to the same
day the week before
• Each data point is a year
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data
Year
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
Injur ies before No
Smoking Day
516
610
581
586
554
632
479
583
445
522
Injur ies on No
Smoking Day
540
620
599
639
607
603
519
560
515
556
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Calculate differences
Injur ies before No
Smoking Day
Injur ies on No
Smoking Day
Differ ence
516
540
(d)
24
610
620
10
581
586
599
639
18
53
554
632
479
607
603
519
53
-29
40
583
445
560
515
-23
70
522
556
34
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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.
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Hypotheses
Ho: Work related injuries do not change during
No Smoking Days (μ=0)
Ha: Work related injuries change during
No Smoking Days (μ≠0)
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Calculate differences
Injur ies before No
Smoking Day
Injur ies on No
Smoking Day
Differ ence
516
540
(d)
24
610
620
10
581
586
599
639
18
53
554
632
479
607
603
519
53
-29
40
583
445
560
515
-23
70
522
556
34
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Calculate t using d’s
d =25
2
d
s =1043 .78
n =10
25 -0
t
=
=2.45
1043 .78 /10
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Caution!
• 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
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Critical value of t
t 0.05 2
,9
= 2.26
Test statistic: t = 2.45
So we can reject the null hypothesis: Stopping smoking
increases job-related accidents in the short term.
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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.
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Quick reference summary:
Paired t-test
• What is it for? To test whether the mean difference in a
population equals a null hypothesized value, μdo
• What does it assume? Pairs are randomly sampled from a
population. The differences are normally distributed
• Test statistic: t
• Distribution under Ho: t-distribution with n-1 degrees of
freedom, where n is the number of pairs
• Formula:
d − do
t=
SE d