Download average weight from samples of 100

The Normal Distribution,
Central Limit Theorem, and
Introduction to statistical
inference
The Normal Distribution
‘Bell Shaped’
 Symmetrical
 Mean, Median and Mode
are Equal
m=mean

p(X)
s
X
m
s = standard deviation
The random variable has an infinite
theoretical range:
+  to  
Mean
= Median
= Mode
Examples:








height
weight
age
bone density
IQ (mean=100; SD=15)
SAT scores
blood pressure
ANYTHING YOU AVERAGE OVER A
LARGE ENOUGH # (Central Limit
Theorem)
Small standard deviation
Larger standard deviation
Even larger standard deviation
The Normal Distribution:
as mathematical function (pdf)
f ( x) 
1
s 2
Note constants:
=3.14159
e=2.71828
1 xm 2
 (
)
2
s
e
This is a bell shaped
curve with different
centers and spreads
depending on m and s
68-95-99.7 Rule
in Math terms…
m s

m s s

m  2s

m s s
2
m  3s

m s s
3
1
2
1
2
1
2
1 xm 2
 (
)
 e 2 s dx  .68
1 xm 2
 (
)
 e 2 s dx  .95
1 xm 2
 (
)
2
s
e
dx  .997
The 68-95-99.7 rule only gets
you so far…

For example, what’s the probability of
getting a math SAT score below 575 if SAT
scores are normally distributed with a mean
of 500 and a std dev of 50??
575
1
 P( X  575)  
e
  (50) 2
Solve this ... ?!
1 x 500 2
 (
)
2 50
dx
The Standard Normal Curve:
“Universal Currency”
All normal distributions can be converted into the
standard normal curve by subtracting the mean and
dividing by the standard deviation:
Z
X m
s
For example, 575 in math SAT units translates to 1.5
standard deviations above the mean.
The Standard Normal Curve:
“Universal Currency”
Z~Normal(m=0, s=1)
f (Z ) 
1
2
1
  Z 2
e 2
The Standard Normal Distribution (Z)
Somebody calculated all the integrals for the standard
normal and put them in a table! So we never have to
integrate!
Even better, computers now do all the integration.
Comparing X and Z units
500
0
575
1.5
X
Z
(m = 500, s = 50)
(m = 0, s = 1)
Example
So, What’s the probability of getting a math SAT score of
575 or less, m=500 and s=50?
575  500
Z
 1.5
50
575
1 x 500 2
)
50
 (
1
 P( X  575)  
e 2
  (50) 2
1.5
dx 
 

1
 Z2
1
 e 2 dz
2
No need to do the integration!
Just look up Z= 1.5 in standard normal chart  no
problem! = .9332
Looking up probabilities in the
standard normal table
What is the area to the
left of Z=1.50 in a
standard normal curve?
Z=1.50
Z=1.50
Area is 93.32%
Exercise (in groups of 2-3)
a.
b.
If birth weights in a population are normally
distributed with a mean of 109 oz and a
standard deviation of 13 oz,
What is the chance of obtaining a birth weight
of 141 oz or heavier when sampling birth
records at random?
What is the chance of obtaining a birth weight
of 120 or lighter?
Answer
a.
What is the chance of obtaining a birth
weight of 141 oz or heavier when
sampling birth records at random?
141  109
Z
 2.46
13
From the chart  Z of 2.46 corresponds to a right tail (greater than)
area of: P(Z≥2.46) = 1-(.9931)= .0069 or .69 %
Answer
b. What is the chance of obtaining a birth
weight of 120 or lighter?
120  109
Z
 .85
13
From the chart  Z of .85 corresponds to a left tail area of:
P(Z≤.85) = .8023= 80.23%
Are my data “normal”?


Not all continuous random variables are
normally distributed!!
It is important to evaluate how well the
data are approximated by a normal
distribution
Are my data normally
distributed?
1. Look at the histogram! Does it appear bell
shaped?
2. Compute descriptive summary measures—are
mean, median, and mode similar?
3. Do 2/3 of observations lie within 1 std dev of
the mean? Do 95% of observations lie within
2 std dev of the mean?
4. Look at a normal probability plot—is it
approximately linear?
5. Run tests of normality (such as KolmogorovSmirnov). But, be cautious, highly influenced
by sample size!
Example: Class coffee drinking
(n=21)
Mean=3.6 ounces/day
Std Dev=5.1 ounces/day
Range: 0 to 16
Example: Class coffee drinking
(n=21)
Mean1 Std Dev=
3.6  5.1 = -1.5 to 8.7
-1.5
8.7
(covers 90% of
observations)
Example: Class coffee drinking
(n=21)
Mean2 Std Dev=
3.6  10.2 = -6.6 to
13.8
(covers 90% of
observations)
13.8
Normal Probability Plot
Clearly not a straight line!
KolmogorovSmirnov test
agrees, not normal!
Example: Class wake-up times
(n=20)
Mean=7:20
Std Dev=0:56
Range: 5:00 to 9:00
Example: Class wake-up times
Mean1 Std Dev=
(n=20)
7:20  :56 = 6:24 to 8:16
(covers 80% of
observations)
6:24
8:16
Example: Class wake-up times
(n=20)
Mean2 Std Dev=
7:20  1:52 = 5:28 to 9:12
(covers 95% of
observations)
Normal Probability Plot
Pretty close to a straight
line!
KolmogorovSmirnov test
agrees, looks
normal!
Review Problem 1
Which of the following about the normal
distribution is NOT true?
a.
b.
c.
d.
Theoretically, the mean, median, and mode are the same.
About 2/3 of the observations fall within 1 standard deviation
from the mean.
It is a discrete probability distribution.
Its parameters are the mean, m , and standard deviation, s.
Review Problem 1
Which of the following about the normal
distribution is NOT true?
a.
b.
c.
d.
Theoretically, the mean, median, and mode are the same.
About 2/3 of the observations fall within 1 standard deviation
from the mean.
It is a discrete probability distribution.
Its parameters are the mean, m , and standard deviation, s.
Review Problem 2
For some positive value of Z, the probability
that a standard normal variable is between 0
and Z is 0.3770. The value of Z is:
a.
b.
c.
d.
0.18.
0.81.
1.16
1.47.
Review Problem 2
For some positive value of Z, the probability
that a standard normal variable is between 0
and Z is 0.3770. The value of Z is:
a.
b.
c.
d.
0.18.
0.81.
1.16
1.47.
Review Problem 3
The probability that a standard normal
variable Z is positive is ________.
a.
b.
c.
d.
50%
100%
0%
95%
Review Problem 3
The probability that a standard normal
variable Z is positive is ________.
a.
b.
c.
d.
50%
100%
0%
95%
Review Problem 4
Suppose Z has a standard normal distribution with a
mean of 0 and a standard deviation of 1. The
probability that Z values are larger than __________
is 0.6985.
a.
b.
c.
d.
e.
1.0
0
-0.6
+0.6
-2.0
Review Problem 4
Suppose Z has a standard normal distribution with a
mean of 0 and a standard deviation of 1. The
probability that Z values are larger than __________
is 0.6985.
a.
b.
c.
d.
e.
1.0
0
-0.6
+0.6
-2.0
Statistical Inference:
Hypothesis Testing and
Confidence Intervals
What is a statistic?


A statistic is any value that can be
calculated from the sample data.
Sample statistics are calculated to give
us an idea about the larger population.
Examples of statistics:

mean


difference in means


The difference in the average gas price in San
Francisco ($3.37) compared with Minneapolis
($2.65) is 72 cents.
proportion


The average cost of a gallon of gas in the US is
$2.87.
67% of high school students in the U.S. exercise
regularly
difference in proportions

The difference in the proportion of men who
approve of George W. (44%) and women who do
(38%) is 6%
What is a statistic?

Sample statistics are estimates of
population parameters.
Sample statistics estimate
population parameters:
Sample statistic: mean IQ
of 5 subjects
Truth (not
observable)
Mean IQ of
some population
of 100,000
people =100
110  105  96  124  115
 110
5
Sample
(observation)
Make guesses
about the whole
population
Sampling Distributions
Most experiments are one-shot deals. So, how do we know if
an observed effect from a single experiment is real or is just an
artifact of sampling variability (chance variation)?
**Requires a priori knowledge about how sampling variability
works…
Question: Why have I made you learn about probability
distributions and about how to calculate and
manipulate expected value and variance?
Answer: Because they form the basis of describing the
distribution of a sample statistic.
What is sampling variation?




Statistics vary from sample to sample due to
random chance.
Example:
A population of 100,000 people has an
average IQ of 100 (If you actually could
measure them all!)
If you sample 5 random people from this
population, what will you get?
Sampling Variation
120  160  180  95  95
90  85  95  92  88  130

90
5
100  105  86  104  95
110  105 596  124  115  98
5
5 (not
Truth
observable)
Mean
IQ=100
 110
Sampling Variation and
Sample Size




Do you expect more or less sampling
variability in samples of 10 people?
Of 50 people?
Of 1000 people?
Of 100,000 people?
Standard error


Standard error is the standard deviation
of a sample statistic.
It’s a measure of sampling variability.
What is statistical inference?

The field of statistics provides guidance
on how to make conclusions in the face
of this chance variation.
Examples of Sample Statistics:
Single population mean (known s)
Single population mean (unknown s)
Single population proportion
Difference in means (ttest)
Difference in proportions (Z-test)
Odds ratio/risk ratio
Correlation coefficient
Regression coefficient
…
The Central Limit Theorem:
If all possible random samples, each of size n, are
taken from any population with a mean m and a
standard deviation s, the sampling distribution of
the sample means (averages) will:
1. have mean:
mx  m
2. have standard deviation:
s
sx 
n
3. be approximately normally distributed regardless of the shape
of the parent population (normality improves with larger n)
Symbol Check
mx
sx
The mean of the sample means.
The standard deviation of the sample means. Also
called “the standard error of the mean.”
Computer simulation of the sampling
distribution of the sample mean:
1. Pick any probability distribution and specify a mean and
standard deviation.
2. Tell the computer to randomly generate 1000 observations
from that probability distributions
E.g., the computer is more likely to spit out values with high
probabilities
3. Plot the “observed” values in a histogram.
4. Next, tell the computer to randomly generate 1000 averagesof-2 (randomly pick 2 and take their average) from that
probability distribution. Plot “observed” averages in histograms.
5. Repeat for averages-of-10, and averages-of-100.
Uniform on [0,1]: average of 1
(original distribution)
Uniform: 1000 averages of 2
Uniform: 1000 averages of 5
Uniform: 1000 averages of 100
~Exp(1): average of 1
(original distribution)
~Exp(1): 1000 averages of 2
~Exp(1): 1000 averages of 5
~Exp(1): 1000 averages of 100
~Bin(40, .05): average of 1
(original distribution)
~Bin(40, .05): 1000
averages of 2
~Bin(40, .05): 1000
averages of 5
~Bin(40, .05): 1000
averages of 100
The Central Limit Theorem:
If all possible random samples, each of size n, are
taken from any population with a mean m and a
standard deviation s, the sampling distribution of
the sample means (averages) will:
1. have mean:
mx  m
2. have standard deviation:
s
sx 
n
Also called
standard error
of the mean!
3. be approximately normally distributed regardless of the shape
of the parent population (normality improves with larger n)
Review Question 5
I roll a die 10 times and record the average
of the 10 rolls. Then, I repeat this
experiment 100 times and record 100
averages. What is the distribution of the
averages? (recall: mean of die toss=3.5 and
std dev of a die toss=1.71)
a.
b.
c.
d.
Normal, mean=3.5, std dev= 1.71
Binomial, mean=5, std dev= 1.1
Uniform, mean=3.5, std dev= 1.1
Normal, mean=3.5, std dev= 1.71/ 10
Review Question 5
I roll a die 10 times and record the average
of the 10 rolls. Then, I repeat this
experiment 100 times and record 100
averages. What is the distribution of the
averages?
a.
b.
c.
d.
Normal, mean=3.5, std dev= 1.71
Binomial, mean=5, std dev= 1.1
Uniform, mean=3.5, std dev= 1.1
Normal, mean=3.5, std dev= 1.71/ 10
Review Question 6
I roll a die 100 times and record the
average of the 100 rolls. Then, I repeat this
experiment 10 times and record 10
averages. What is the distribution of the
averages?
a.
b.
c.
d.
Normal,
Normal,
Normal,
Normal,
mean=3.5, std dev= .54
mean=5, std dev= 1.71
mean=3.5, std dev= .171
mean=3.5, std dev= 1.71
Review Question 6
I roll a die 100 times and record the
average of the 100 rolls. Then, I repeat this
experiment 10 times and record 10
averages. What is the distribution of the
averages?
a.
b.
c.
d.
Normal, mean=3.5, std dev= .54
Normal, mean=5, std dev= 1.71
Normal, mean=3.5, std dev= .171
Normal, mean=3.5, std dev= 1.71
Review Question 7

I measure systolic blood pressure in a sample of 20 people from
my population of interest. The standard deviation of blood pressure
in this population is 10 and the mean is 150. If I could repeat my
experiment 100 times on new samples of 20, what would the
distribution of sample means be?
a.
Normal, mean=150, std dev=10
b.
Normal, mean=150, std dev= .5
c.
Normal, mean=150, std dev= 2.2
d.
Cannot determine since we do not know if blood
pressures are normally distributed in this population.
Review Question 7

I measure systolic blood pressure in a sample of 20 people from
my population of interest. The standard deviation of blood pressure
in this population is 10 and the mean is 150. If I could repeat my
experiment 100 times on new samples of 20, what would the
distribution of sample means be?
a.
Normal, mean=150, std dev=10
b.
Normal, mean=150, std dev= .5
c.
Normal, mean=150, std dev= 2.2
d.
Cannot determine since we do not know if blood
pressures are normally distributed in this population.
Example 1: Weights of doctors




Experimental question: Are practicing doctors
setting a good example for their patients in
their weights?
Experiment: Take a sample of practicing
doctors and measure their weights
Sample statistic: mean weight for the sample
IF weight is normally distributed in doctors
with a mean of 150 lbs and standard
deviation of 15, how much would you expect
the sample average to vary if you could
repeat the experiment over and over?
Relative frequency of 1000 observations of weight
mean= 150 lbs; standard deviation = 15 lbs
Standard
deviation reflects
the natural
variability of
weights in the
population
-1 SD
+1 SD
+2 SD
-2 SD
+3 SD
-3 SD
doctors’ weights
standard error of the mean  15
-1 SD
2
 10.6lbs
+1 SD
-2 SD
+2 SD
-3 SD
+3 SD
average
1000
weight
doctors’
fromweights
samples of 2
standard error of the mean  15
10
 4.74lbs
-1 SD
-2 SD
-3 SD
+1 SD
+2 SD
+3 SD
average weight from samples of 10
standard error of the mean  15
-1 SD
-2 SD
-3 SD
+1 SD
+2 SD
+3 SD
average weight from samples of 100
100
 1.5lbs
Using Sampling Variability


In reality, we only get to take one
sample!!
But, since we have an idea about how
sampling variability works, we can make
inferences about the truth based on one
sample.
Experimental results

Let’s say we take one sample of 100
doctors and calculate their average
weight….
Expected Sampling Variability for n=100
if the true weight is 150 (and SD=15)
What are we
going to think if
our 100-doctor
sample has an
average weight of
160?
average weight from samples of 100
Expected Sampling Variability for n=100
if the true weight is 150 (and SD=15)
If we did this
experiment 1000
times, we
wouldn’t expect to
get 1 result of 160
if the true mean
weight was 150!
average weight from samples of 100
“P-value” associated with this experiment
“P-value” (the
probability of our
sample average being
160 lbs or more IF the
true average weight is
150)
< .0001
Gives us evidence that
150 isn’t a good guess
average weight from samples of 100
Calculating the p-value
Formally,
160  150
Z
 6.6
1.5
p  .0001
P-value< .0001 gives us evidence against our null hypothesis.
P-value<.0001 means:
The probability of seeing what you saw or something
more extreme if the null hypothesis is true (due to
chance)<.0001
P(empirical data | null hypothesis) <.0001
The P-value
P-value is the probability that we would have seen our
data (or something more unexpected) just by chance if
the null hypothesis (null value) is true.
Small p-values mean the null value is unlikely given
our data.
The P-value


By convention, p-values of <.05 are often
accepted as “statistically significant” in the
literature; but this is an arbitrary cut-off.
A cut-off of p<.05 means that in about 5 of
100 experiments, a result would appear
significant just by chance (“Type I error”).
Hypothesis Testing
The Steps:
1. Define your hypotheses (null, alternative)
2. Specify your null distribution
3. Do an experiment
4. Calculate the p-value of what you observed
5. Reject or fail to reject (~accept) the null
hypothesis
Hypothesis Testing
1. Define your hypotheses (null, alternative)

The null hypothesis is the “straw man” that we are trying to shoot down.

Null here: “mean weight of doctors = 150 lbs”

Alternative here: “mean weight > 150 lbs” (one-sided)
2. Specify your sampling distribution (under the null)

If we repeated this experiment many, many times, the sample average
weights would be normally distributed around 150 lbs with a standard
15
 1.5
error of 1.5
100
3. Do a single experiment (observed sample mean = 160 lbs)
4. Calculate the p-value of what you observed (p<.0001)
5. Reject or fail to reject the null hypothesis (reject)
Expected Sampling Variability for n=2
What are we
going to think if
our 2-doctor
sample has an
average weight of
160?
average weight from samples of 2
Expected Sampling Variability for n=2
P-value = 17%
i.e. about 17 out of 100
“average of 2”
experiments will yield
values 160 or higher
even if the true mean
weight is only 150
average weight from samples of 2
Expected Sampling Variability for n=10
P-value = 2%
i.e. about 2 out of
100 “average of
10” experiments
will yield values
160 or higher
even if the true
mean weight is
only 150
Two sided pvalue=4%
average weight from samples of 100
Statistical Power
We found the same sample mean (160
lbs) in our 100-doctor sample, 10doctor sample, and 2-doctor sample.
But we only rejected the null based on
the 100-doctor and 10-doctor samples.
Larger samples give us more statistical
power…
Hypothesis testing: Error and
Power

Type-I Error


Rejecting the null when the effect isn’t real, e.g.
false positive.
Type-II Error


(also known as “α”):
(also known as “β “):
Failing to reject the null when the effect is real,
e.g. false negative.
POWER (the flip side of type-II error: 1- β):

The chance you’ll find an effect if it’s real.
Type I and Type II Error in a box
Your Statistical
Decision
Reject H0
True state of null hypothesis
H0 True
H0 False
Type I error (α)
Correct
Correct
Type II Error (β)
Do not reject H0
Statistical power = 1- type II error
Reminds me of…
Pascal’s Wager
The TRUTH
Your Decision
God Exists
God Doesn’t Exist
BIG MISTAKE
Correct
Correct—
Big Pay Off
MINOR MISTAKE
Reject God
Accept God
Review Question 8
Spine bone density is normally distributed in young
women, with a mean of 1.0 g/cm2 and a mean of 0.1
g/cm2. In my sample of 100 young women runners, the
average spine bone density is .93 g/cm2. Are runners
statistically different from the general population? What
is the p-value?
a.
b.
c.
d.
No, Z=-.1, p>.05
Yes, Z=-7, p<.0001
No, Z=-.7, p>.05
Yes, Z=-2, p<.05
Review Question 8
Spine bone density is normally distributed in young
women, with a mean of 1.0 g/cm2 and a mean of 0.1
g/cm2. In my sample of 100 young women runners, the
average spine bone density is .93 g/cm2. Are runners
statistically different from the general population? What
is the p-value?
a.
b.
c.
d.
No, Z=-.1, p>.05
Yes, Z=-7, p<.0001
0.93  1.0  0.07  0.07
No, Z=-.7, p>.05
Z


 7.0
.1
.1
.01
Yes, Z=-2, p<.05
10
100
Review Question 9
The p-value is:
a.
b.
c.
d.
The probability
The probability
hypothesis.
The probability
association.
The probability
one exists.
that my data are wrong.
of my data under the null
that I erroneously find an
that I find an association when
Review Question 9
The p-value is:
a.
b.
c.
d.
The probability that my data are wrong.
The probability of my data under the null
hypothesis.
The probability that I erroneously find an
association.
The probability that I find an association when
one exists.