Transcript Sampling Distribution

Counting in probability
Permutations

The number of orderings of different events
Combinations

The number ways that outcomes can be
grouped
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Permutations
The number of orderings of different
events.

Three cards: AKQ
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Four digits: 7412
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AKQ, AQK, KQA, KAQ, QAK QKA = 6
24
Five symbols: # $ % & @
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120
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Three different cards: AKQ
AKQ, AQK, KQA, KAQ, QAK QKA = 6
Complete permutations
P  N!
N
N
N! ( N )(N  1)(N  2)(2)(1)
P  3* 2 *1  6
3
3
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Partial Permutations
Four Different Digits (of 10)
In the first spot: 10 different digits possible
 In the second spot: 9 possible left
 In the third spot: 8 possible left
 In the last spot: 7 possible left

10*9*8*7 = 5040
Pr
N
N!

( N  r )!
P410 
10! 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 *1

6!
6 * 5 * 4 * 3 * 2 *1
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Combinations
The number ways that outcomes can be
grouped
Calculate the number of combinations
of N objects taken r (r <= N) at a time.
Example: 13 cards (AKQ…2) taken 5 at
a time = 1287
N!
C 
r!*(N  r )!
N
r
13! 13 *12 *11 *10 * 9

5!*8!
5 * 4 * 3 * 2 *1
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Review the problem
We want to be able to say something
about the population from a single
sample that we have drawn.
This is the problem of statistical
inference
What can we infer from our sample?
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Our logic for proceeding
If the population has certain
characteristics,
then our sample will probably include
certain outcomes and probably not
include other outcomes
If our sample has outcomes that are
unlikely to come from that population, it
probably did not come from that
population
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Our logic (continued)
If our sample has outcomes that are
unlikely to come from that population, it
probably did not come from that
population
Our conclusion must be either that the
population is as hypothesized or it is not
as hypothesized.
We reject or fail to reject the
hypothesized population characteristic
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Once More
We reason from population
characteristics to probability distribution
of all possible samples -- the sampling
distribution.
We calculated this for the population of
7 black and 3 red marbles
We noted that the sampling distribution
began to look like a normal distribution
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Sampling distributions
There are a few sampling distributions
that occur very frequently
binomial
 normal
 student’s t statistic
 chi-square statistic
 F statistic

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Binomial
This is the distribution we have already
calculated for the red (p=.3) and black
(p=.7) marbles.
It is also the distribution for coin tosses
where p=.5 for heads and p=.5 for tails
It is the distribution for all binary
outcomes.
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Normal
This is the distribution for many
empirical distribution.
This is the distribution which many other
distributions approach when working
with large numbers: binomial, t, chisquare, etc.
This is the distribution for the sum of a
set of random variables.
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Normal distribution
characteristics
The formula
1
e
 2
1  x  x 


2   
2
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Normal distribution
characteristics
The shape
.4
68%
norm
.3
.2
.1
0
-5
-4
-3
-2
-1
0
x
1
2
3
4
5
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Normal distribution
characteristics
The shape
.4
95%
norm
.3
.2
.1
0
-3
-2
-1
0
x
1
2
3
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Normal distribution
characteristicsMean
The shape
.4
One unit is
one std. dev.
norm
.3
.2
.1
0
-3
-2
-1
0
x
1
2
3
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Why the normal distribution?
Central Limit Theorem
Sampling distribution of the mean
approaches normal distribution
Mean of sampling distribution
approaches mean of population
sd of sampling distribution approaches
population sd / sqrt(n) as n becomes
large
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Thus
pop mean = 5, pop sd = 1
If n=4, sampling dist mean = 5, sd = .5
 If n=9, sampling dist mean = 5, sd = .333
 If n=25, samp dist mean = 5, sd = .20
 If n=900, samp dist mean = 5, sd = .033

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.4
Pop
norm
.3
.2
.1
0
2
3
4
5
x
6
7
8
.4
norm
.3
N=4
.2
.1
0
2
3
4
5
xx
6
7
8
.4
norm
.3
N=9
.2
.1
0
2
3
4
5
xxx
6
7
8
.4
N=900
norm
.3
.2
.1
0
2
3
4
5
xxxx
6
7
8
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