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STAT 101
Dr. Kari Lock Morgan
Normal Distribution
Chapter 5
• Normal distribution
• Central limit theorem
• Normal distribution for confidence intervals
• Normal distribution for p-values
• Standard normal
Statistics: Unlocking the Power of Data
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Re-grade Requests
 4e potential grading mistake: 0.025 is correct
 Requests for a re-grade must be submitted in
writing by class on Wednesday, March 5th
 Partial credit will NOT be adjusted
 Valid re-grade requests:
 You got points off but believe your answer is correct
 Points were added incorrectly
 Warning: scores may go up or down
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Bootstrap and Randomization Distributions
Correlation: Malevolent
uniforms
Measures from Scrambled Collection 1
Slope :Restaurant
tips
Measures from Scrambled RestaurantTips
-60
-40
Dot Plot
-20
0
20
slope (thousandths)
Mean :Body
Temperatures
Measures from Sample of BodyTemp50
98.2
98.3
98.4
40
-0.4
-0.2
0.0
r
0.2
0.4
What do you
Diff means: Finger taps
notice?
0.6
Dot Plot
Measures from Scrambled CaffeineTaps
98.5
98.6
Nullxbar
98.7
98.8
0.5
phat
0.6
98.9
Dot Plot
Dot Plot
99.0
-4
Proportion : Owners/dogs
0.4
60
-0.6
Measures from Sample of Collection 1
0.3
Dot Plot
-3
-2
-1
0
Diff
1
2
3
Mean : Atlanta commutes
Measures from Sample of CommuteAtlanta
0.7
0.8
Statistics: Unlocking the Power of Data
26
27
28
29
xbar
30
4
Dot Plot
31
32
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Normal Distribution
• The symmetric, bell-shaped curve we have
1000
0
500
Frequency
1500
seen for almost all of our bootstrap and
randomization distributions is called a
normal distribution
-3
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-2
-1
0
1
2
3
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Central Limit Theorem!
For a sufficiently large sample
size, the distribution of sample
statistics for a mean or a
proportion is normal
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Distribution of 𝒑
n 1
n  10
n  30
n  50
n  100
p  0.5
0.0
0.5
1.0
0.0
0.5
1.0 0.0
0.0
0.5
1.0
0.0
0.5
1.0
0.0
0.5
1.0
0.0
0.5
1.0
0.5
1.0 0.0
0.5
1.0
0.0
0.5
1.0
0.5
1.0 0.0
0.5
1.0
0.0
0.5
1.0
p  0.7
0.0
p  0.1
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0.0
0.5
1.0 0.0
0.5
1.0
0.0
0.5
1.0
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CLT for a Mean
Population
8
3.0
1.5
0 1 2
10
x
n = 30
2.0
3.0
2
3
4
5
1.5
2.0
2.5
3.0
25
1
0 2 4
Statistics: Unlocking the Power of Data
1.0
0 10
Frequency
0
n = 50
3 4 5 6
8
6
4
4
0
2
Frequency
0
Distribution of
Sample Means
0.0
n = 10
Frequency
Distribution of
Sample Data
6 8
12
1.4
1.8
2.2
2.6
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Central Limit Theorem
• The central limit theorem holds for ANY
original distribution, although “sufficiently large
sample size” varies
• The more skewed the original distribution is
(the farther from normal), the larger the sample
size has to be for the CLT to work
• For small samples, it is more important that the
data itself is approximately normal
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Central Limit Theorem
• For distributions of a quantitative variable that
are not very skewed and without large outliers,
n ≥ 30 is usually sufficient to use the CLT
• For distributions of a categorical variable,
counts of at least 10 within each category is
usually sufficient to use the CLT
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Accuracy
• The accuracy of intervals and p-values generated
using simulation methods (bootstrapping and
randomization) depends on the number of simulations
(more simulations = more accurate)
• The accuracy of intervals and p-values generated
using formulas and the normal distribution depends on
the sample size (larger sample size = more accurate)
• If the distribution of the statistic is truly normal and
you have generated many simulated randomizations,
the p-values should be very close
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Normal Distribution
• The normal distribution is fully
characterized by it’s mean and standard
deviation
N  mean,standard deviation 
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Bootstrap Distributions
If a bootstrap distribution is
approximately normally distributed, we
can write it as
a)
b)
c)
d)
N(parameter, sd)
N(statistic, sd)
N(parameter, se)
N(statistic, se)
sd = standard deviation of variable
se = standard error = standard deviation of statistic
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Hearing Loss
• In a random sample of 1771 Americans aged 12
to 19, 19.5% had some hearing loss (this is a
dramatic increase from a decade ago!)
• What proportion of Americans aged 12 to 19
have some hearing loss? Give a 95% CI.
Rabin, R. “Childhood: Hearing Loss Grows Among Teenagers,” www.nytimes.com, 8/23/10.
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Hearing Loss
(0.177, 0.214)
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Hearing Loss
N(0.195, 0.0095)
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Confidence Intervals
If the bootstrap distribution is normal:
To find a P% confidence interval , we just
need to find the middle P% of the
distribution
N(statistic, SE)
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Area under a Curve
• The area under the curve of a normal
distribution is equal to the proportion of the
distribution falling within that range
• Knowing just the mean and standard
deviation of a normal distribution allows
you to calculate areas in the tails and
percentiles
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Hearing Loss
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(0.176, 0.214)
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Standardized Data
 Often, we standardize the data to have mean 0
and standard deviation 1
 This is done with z-scores
From x to z :
x  mean
z
sd
From z to x:
x = mean + z ´ sd
 Places everything on a common scale
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Standard Normal
• The standard normal distribution is
the normal distribution with mean 0 and
standard deviation
1 of Statistic Assuming Null
Distribution
N  0,1
-3
-2
-1
0
1
2
3
Statistic
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Standardized Data
 Confidence Interval (bootstrap distribution):
mean = sample statistic, sd = SE
From z to x: (CI)
x = mean + z ´ sd
x  statistic  z  SE
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P% Confidence Interval
1. Find z-scores (–z*
and z*) that capture
the middle P% of the
standard normal
2. Return to
original scale with
statistic  z* SE
P%
-z*
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z*
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Confidence Interval using N(0,1)
If a statistic is normally distributed, we find a
confidence interval for the parameter using
statistic  z* SE
where the area between –z* and +z* in the
standard normal distribution is the desired
level of confidence.
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Confidence Intervals
Find z* for a 99% confidence interval.
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z* = 2.575
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z*
 Why use the standard normal?
 Common confidence levels:
 95%: z*
= 1.96 (but 2 is close enough)
 90%: z*
= 1.645
 99%: z* =
2.576
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Sin Taxes
In March 2011, a random sample of 1000 US
adults were asked
“Do you favor or oppose ‘sin taxes’ on soda and
junk food?”
320 adults responded in favor of sin taxes.
Give a 99% CI for the proportion of all US adults
that favor these sin taxes.
From a bootstrap distribution,
we find SE = 0.015
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Sin Taxes
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Sin Taxes
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Randomization Distributions
If a randomization distribution is
approximately normally distributed, we
can write it as
a) N(null value, se)
b) N(statistic, se)
c) N(parameter, se)
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p-values
If the randomization distribution is
normal:
To calculate a p-value, we just need to
find the area in the appropriate tail(s)
beyond the observed statistic of the
distribution
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First Born Children
• Are first born children actually smarter?
• Explanatory variable: first born or not
• Response variable: combined SAT score
• Based on a sample of college students, we
find 𝑥𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 − 𝑥𝑛𝑜𝑡 𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 = 30.26
• From a randomization distribution, we find
SE = 37
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First Born Children
𝑥𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 − 𝑥𝑛𝑜𝑡 𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 = 30.26
SE = 37
What normal distribution should we use
to find the p-value?
a)
b)
c)
d)
N(30.26, 37)
N(37, 30.26)
N(0, 37)
N(0, 30.26)
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Hypothesis Testing
Distribution of Statistic Assuming Null
Observed
Statistic
p-value
-3
-2
-1
0
1
2
3
Statistic
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First Born Children
N(0, 37)
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p-value = 0.207
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Standardized Data
 Hypothesis test (randomization distribution):
mean = null value, sd = SE
From x to z (test) :
x  mean
z
sd
x - null
z=
SE
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p-value using N(0,1)
If a statistic is normally distributed under H0,
the p-value is the probability a standard normal
is beyond
𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 − 𝑛𝑢𝑙𝑙 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟
𝑧=
𝑆𝐸
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First Born Children
𝑥𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 − 𝑥𝑛𝑜𝑡 𝑓𝑖𝑟𝑠𝑡 𝑏𝑜𝑟𝑛 = 30.26, SE = 37
1) Find the standardized test statistic
2) Compute the p-value
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First Born Children
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z-statistic
If z = –3, using  = 0.05 we would
(a) Reject the null
(b) Not reject the null
(c) Impossible to tell
(d) I have no idea
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z-statistic
•
Calculating the number of standard
errors a statistic is from the null value
allows us to assess extremity on a
common scale
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Confidence Interval Formula
IF SAMPLE SIZES ARE LARGE…
From N(0,1)
sample statistic  z  SE
*
From original
data
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From
bootstrap
distribution
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Formula for p-values
IF SAMPLE SIZES ARE LARGE…
From original
data
From H0
sample statistic  null value
z
SE
From
randomization
distribution
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Compare z to
N(0,1) for p-value
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Standard Error
• Wouldn’t it be nice if we could compute
the standard error without doing
thousands of simulations?
• We can!!!
• Or at least we’ll be able to next class…
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t-distribution
• For quantitative data, we use a tdistribution instead of the normal
distribution
•The t distribution is very similar to the
standard normal, but with slightly fatter
tails (to reflect the uncertainty in the
sample standard deviations)
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Degrees of Freedom
• The t-distribution is characterized by its
degrees of freedom (df)
• Degrees of freedom are based on sample size
• Single mean: df = n – 1
• Difference in means: df = min(n1, n2) – 1
• Correlation: df = n – 2
• The higher the degrees of freedom, the closer
the t-distribution is to the standard normal
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t-distribution
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Aside: William Sealy Gosset
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The Pygmalion Effect
Teachers were told that certain children
(chosen randomly) were expected to be
intellectual “growth spurters,” based on the
Harvard Test of Inflected Acquisition (a test
that didn’t actually exist). These children were
selected randomly.
The response variable is change in IQ over the
course of one year.
Source: Rosenthal, R. and Jacobsen, L. (1968). “Pygmalion in the Classroom:
Teacher Expectation and Pupils’ Intellectual Development.” Holt, Rinehart
and Winston, Inc.
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The Pygmalion Effect
Control Students
“Growth Spurters”
n
255
65
X
8.42
12.22
s
12.0
13.3
Can this provide evidence that merely
expecting a child to do well actually causes the
child to do better?
If so, how much better?
SE = 1.8
*s1 and s2 were not given, so I set them to give the correct p-value
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Pygmalion Effect
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Pygmalion Effect
From the paper:
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“The difference in gains could be ascribed
to chance about 2 in 100 times”
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Pygmalion Effect
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To Do
 Do Project 1 (due 3/7)
 Read Chapter 5
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