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Transcript
```Parameter versus statistic


Population: the entire group
of individuals in which we are
interested but usually can’t
assess directly.
A parameter is a number
summarizing the population.
Parameters are usually
unknown.

Sample: the part of the
population we actually examine
and for which we do have data.

A statistic is a number
summarizing a sample. We
often use a statistic to estimate
an unknown population
parameter.
Population
Sample
Sampling distributions
Different random samples taken from the same population will give
different statistics.  A statistic computed from a random sample is a
random variable. But there is a predictable pattern in the long run.
The sampling distribution of a statistic is the
probability distribution of that statistic for samples
of a given size n taken from a given population.
Sampling distribution of x̅ (the sample mean)

The mean of the sampling distribution of x̅ is μ.
There is no tendency for a sample average to fall systematically
above or below μ, even if the population distribution is skewed.
 𝒙 is an unbiased estimate of the population mean μ.

The standard deviation of the sampling distribution of 𝒙 is
𝝈
.
𝒏
The standard deviation of the sampling distribution of 𝑥 measures
how much x̅ deviates from μ from sample to sample.
Note: Averages are less variable than individual observations.
For Normally distributed populations
When a variable in a population is Normally distributed, the sampling
distribution of the sample mean x̅ is also Normally distributed.
Sample means
population ~ N(m,s)
↓
sampling distribution of
𝒙 ~ N(m,s/√n)
population
The blood cholesterols of 14-year-old boys is ~ N(µ = 170, σ = 30) mg/dl.
The middle 99.7% of cholesterol levels in boys is 80 to 260 mg/dl.
80
110
140
170
200
230
260
Cholesterol level (mg/dl)
We consider random samples of 25 boys. The sampling distribution of average
cholesterol levels is ~ N(µ = 170, σ = 30/sqrt(25)) mg/dl.
The middle 99.7% of average cholesterol levels in boys is 152 to 188 mg/dl. ???
80
98
116
134
152
170
188
Average cholesterol level (mg/dl)
206
224
242
260
Deer mice (Peromyscus maniculatus) have a body length (excluding the tail)
known to vary Normally, with a mean body length µ = 86 mm, and standard
deviation σ = 8 mm.
For random samples of 20 deer mice, the distribution of the sample mean body
length is ???
A) Normal, mean 86, standard deviation 8 mm.
B) Normal, mean 86, standard deviation 20 mm.
C) Normal, mean 86, standard deviation 1.789 mm.
D) Normal, mean 86, standard deviation 3.9 mm.
Standardizing a Normal sampling distribution (z)
When the sampling distribution is Normal, we can standardize the
sample mean x̅ to obtain a z-score. This z-score can then be used to
find areas under the sampling distribution from Table B.
x
N(µ, σ/√n)
x m
z
s n
z
N(0,1)

Here, we work with the sampling distribution,
and s/√n is its standard deviation (indicative of spread).
Remember that s is the standard deviation of the original population.
Hypokalemia is diagnosed when blood potassium levels are low, below
3.5mEq/dl. Let’s assume that we know a patient whose measured
potassium levels vary daily according to N(m = 3.8, s = 0.2).
If only one measurement is made, what's the probability that this patient
will be misdiagnosed hypokalemic?
z
(x  m)
s

3.5  3.8
 1.5
0.2
P(z < 1.5) = 0.0668 ≈ 7%
If instead measurements are taken on four separate days, what is the
probability of such a misdiagnosis?
z
( x  m ) 3. 5  3 . 8

 3
s n
0.2 4
P(z < 3.0) = 0.0013 ≈ 0.1%
The central limit theorem
Central limit theorem: When randomly sampling from any population
with mean m and standard deviation s, when n is large enough, the
sampling distribution of x̅ is approximately Normal: N(m,s/√n).

The larger the sample size n, the better the approximation of Normality.

This is very useful in inference: Many statistical tests assume Normality for
the sampling distribution. The central limit theorem tells us that, if the
sample size is large enough, we can safely make this assumption even if
the raw data appear non-Normal.
How large a sample size?
It depends on the population distribution. More observations are
required if the population distribution is far from Normal.

A sample size of 25 or more is generally enough to obtain a Normal
sampling distribution from a skewed population, even with mild outliers in
the sample.

A sample size of 40 or more will typically be good enough to overcome an
extremely skewed population and mild (but not extreme) outliers in the
sample.
In many cases, n = 25 isn’t a huge sample. Thus,
even for strange population distributions we can
assume a Normal sampling distribution of the
sample mean, and work with it to solve problems.
Population with strongly
skewed distribution
Sampling distribution of
for n = 2 observations
x

Sampling distribution of x
for n = 10 observations
Sampling distribution of
for n = 25 observations
x
Even though the population (a) is strongly skewed, the sampling distribution of x
when n = 25 (d) is approximately
Normal, as expected from

 the central limit theorem.
How do we know if the population is Normal or not?

Sometimes we are told that a variable has an approximately Normal
distribution (e.g. large studies on human height or bone density).

Most of the time, we just don’t know. All we have is sample data.
 We can summarize the data with a histogram and describe its shape.
 If the sample is random, the shape of the histogram should be similar to the
shape of the population distribution.
 The central limit theorem can help guess whether the sampling distribution
should look roughly Normal or not.
of subjects
Number
Frequency
12
(a) Angle of big toe deformations
in 38 patients:
10
8
• Symmetrical, one small outlier
• Population likely close to Normal
• Sampling distribution ~ Normal
6
4
2
0
10
15
20
25
30
35
40
45
HAV angle
(b) Histogram of number of fruit
per day for 74 adolescent girls
• Skewed, no outlier
• Population likely skewed
• Sampling distribution ~ Normal
given large sample size
50
More
12
Sample of 28 acorns:
10
Describe the distribution of the sample.
What can you assume about the
population distribution?
Frequency
Atlantic acorn sizes (in cm3)
8
6
4
2
0
1.5
3
4.5
6
7.5
Acorn sizes
What would be the shape of the sampling distribution of sample mean?

For samples of size 5?

For samples of size 15?

For samples of size 50?
9
10.5 M
Chapter 12 reminder: Sampling distribution of a count
A population contains a proportion p of successes.
If the population size is much larger than the sample size, the count X of
successes in an SRS of size n has approximately the binomial dist’n
B(n, p) with mean m and standard deviation s:
m  np
s  npq  np(1  p)
If n is large, and p is not too close to 0 or 1, this binomial distribution can
be approximated by the Normal distribution:

N m  np, s  np(1  p)

Sampling distribution of a proportion p̂
When randomly sampling from a population with proportion p of
successes, the sampling distribution of the sample proportion p̂ [“p hat”]
has mean and standard deviation:
m pˆ  p
s pˆ 
p(1  p)
n

p̂ is an unbiased estimator the population proportion p.

Larger samples usually give closer estimates of the population proportion p.
Normal approximation
The sampling distribution of p̂ is never exactly Normal. But as the
sample size increases, the sampling distribution of p̂ becomes
approximately Normal.
The Normal approximation is most accurate for any fixed n when p is
close to 0.5, and least accurate when p is near 0 or near 1.
When n is large, and p is not too close to 0 or 1, the sampling
distribution of p̂ is approximately:

N m  p, s 
p(1  p) n

The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is about 8%.
We wish to take a random sample of size 125 from this
population. What is the probability that 10% or more in the
sample are color blind?
A sample size of 125 is enough to use of the Normal approximation (np = 10 and
n(1 – p) = 115).

Normal approximation for p̂ sampling distribution:

N p  0.08,


p(1  p) / n  0.024
z = (p̂ - p)/σ = (.10 – .08)/0.024 = .824
 P(z ≥ .82) = 0.2061 from Table B

Or P(p̂ ≥ .10) = 1 – pnorm(q=.82,mean=0,sd=1) = 1 - .7939 = .2061
The law of large numbers
Law of large numbers: As the number of randomly drawn observations
(n) in a sample increases,
the mean of the sample (x̅) gets
closer and closer to the population
mean m (quantitative variable).
the sample proportion ( p̂) gets
closer and closer to the population
proportion p (categorical variable).
Note: When sampling randomly from a given population:

The law of large numbers describes what would happen if we
took samples of increasing size n.

A sampling distribution describes what would happen if we took
all possible random samples of a fixed size n.
Both are conceptual ideas with many important practical applications.
We rely on their known mathematical properties, but we don’t actually
build them from data.
```
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