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Sampling distributions
Cécile Ané
Stat 371
Spring 2006
Outline
1
Introduction
2
Sampling distribution of a proportion
3
Sampling distribution of the mean
4
Normal approximation to the binomial
5
The continuity correction
Sampling distributions
What does it mean to take a sample of size
n
Y
1
Y
n
form a random sample if they are independent and
have a common distribution.
From a sample, we can calculate a sample statistic suchas the sample mean
Y
Y
is random too! It can differ from sample to sample. The
textbook refers to a
meta-experiment
The distribution of
Y
is called a sampling distribution.
Sampling distribution of a proportion
Example: cross of two heterozygotes
Aa
×
Aa
. Probability
distribution of the offspring’s genotype:
Offspring genotype
AA
Aa
aa
An offspring is dominant if it has genotype
AA
or
Aa
Experiment: Get
n
2 offsprings, count the number
Y
of
dominant offspring, and calculate the sample proportion
p
Y
We would like
ˆ p
to be close to the “true” value
p
p
is random
Distribution of
ˆ p
(from the binomial distribution):
Y
p
IP
Sampling distribution of a proportion
Larger sample size:
Y
of dominant offspring out of
n
p
Y
20 the sample proportion.
We still want
ˆ p
to be close to the “true” value
p
p
is still random
What is the probability that
ˆ p
is within 0
05 of
p
? Translate
into a binomial question IP
p
IP
Y
IP
Y
IP
Y
IP
Y
IP
Y
Sample size of 20 better than sample size of 2 !!
Sampling distribution of the mean
Example: weight of seeds of some variety of beans.Sample size
n
Student #
Observations
sample mean
y
¯ y
¯ y
¯ y
Y
is random. How do we know its distribution?
We will see 3 key facts.
Key fact # 1
If
Y
1
Y
n
is a random sample, and if the
Y
i
’s have mean
μ
and standard deviation
σ
, then
Y
has mean
μ
¯ Y
μ
and variance var
Y
σ
2
n
, i.e. standard deviation
σ
¯ Y
σ
√
n
Seed weight example:
Assume beans have mean
μ
500 mg
and
σ
120 mg. In a sample of size
n
4, the sample mean
Y
has mean
μ
¯ Y
500 mg and standard deviation
σ
¯ Y
60 mg.
Key fact # 2
If
Y
1
Y
n
is a random sample, and if the
Y
i
’s are all from
N
μ, σ
, then
Y
∼ N
μ,
σ
n
Actually,
Y
1
Y
n
n
Y
∼ N
too.
Seed weight example: 100 students do the same experiment. 350
n=
n=
sample mean
The normal approximation to the binomial
Example:
X
of children with side effects after a vaccine, out
of
n
200 children. Probability of side effect:
p
- So
X
∼ B
What is
IP
X
Direct calculation:
IP
X
IP
X
IP
X
200
C
0
0
200
200
C
15
15
185
Heavy!Or we can use a trick: the binomial might be close to anormal distribution. Pretend
X
is normally distributed!
0
2
4
6
8
10
0.5 0.4 0.3 0.2 0.1 0.
n= 10 , p= 0.
Probability
0
2
4
6
8
10
0.25 0.20 0.15 0.10 0.05 0.
n= 50 , p= 0.
0
5
10
15
20
0.12 0.10 0.08 0.06 0.04 0.02 0.
n= 200 , p= 0.
0
5
10
15
20
0.25 0.20 0.15 0.10 0.05 0.
n= 20 , p= 0.
Probability
0
5
10
15
20
0.15 0.10 0.
n= 20 , p= 0.
Some Possible Values
0
5
10
15
0.25 0.20 0.15 0.10 0.05 0.
n= 20 , p= 0.
The normal approximation to the binomial
X
Y
1
Y
200
where
Y
1
if child #1 has side effects,
otherwise.
Y
200
if child #200 has side effects,
otherwise.
Apply key result #3: if
n
(# of children) is large enough,
then
Y
1
Y
n
has a normal distribution.
Use the normal distribution with
X
’s mean and variance:
μ
np
σ
np
p
If
X
∼ B
n
p
and if
n
is large enough:
if
np
and
n
p
(rule of thumb), then
X
’s distribution is approximately N
np
np
p
The normal approximation to the binomial
Back to our question:
IP
X
np
10 and
n
p
190 are both
5, so
X
≈ N
IP
X
IP
X
IP
Z
True value: > sum( dbinom(0:15,
size=200,
prob=0.05))
[1] 0.
The continuity correction
0
5
10
15
20
of children with side effect
0
5
10
15
20
The continuity correction
X
binomial
B
, and
Y
normal
N
No continuity correction:
IP
X
IP
Y
IP
Y
IP
Z
The continuity correction gives a better approximation.
IP
X
IP
Y
IP
Y
IP
Z
(true value was 0.9556)
The continuity correction
X
binomial
B
, and
Y
normal
N
What is the probability that between 8 and 15 children get sideeffects?
IP
X
IP
X
IP
Y
IP
Z
IP
Z
IP
Z
True value: > sum(
dbinom(8:15, size=200, prob=0.05) )
[1]
