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Transcript
```Sampling Distributions
Parameter
• A number that describes the population
• Symbols we will use for parameters
include
m - mean
s – standard deviation
p – proportion
Statistic
• A number that that can be computed from
sample data without making use of any
unknown parameter
• Symbols we will use for statistics include
x – mean
s – standard deviation
p – proportion
Identify the boldface values as parameter or statistic.
A carload lot of ball bearings has mean
diameter 2.5003 cm.
This is within the
Parameter
specifications for acceptance of the lot by
the purchaser. By chance, an inspector
chooses 100 bearings from the lot that
have mean diameter 2.5009 cm.
Because
Statistic
this is outside the specified limits, the lot
is mistakenly rejected.
Why do we take samples
• A census is not always accurate.
• Census are difficult or impossible
to do.
• Census are very expensive to do.
A distribution is all the
values that a variable can be.
The sampling distribution
of a statistic is the
distribution of values taken
by the statistic in all
possible samples of the
same size from the same
population.
Consider the population – the
length of fish (in inches) in my
pond - consisting of the values
2, 7, 10, 11, 14
What is the mean
mand
8.8
x =standard
deviation
of this
sx = 4.0694
population?
Let’s take samples of size 2
(n = 2) from this population:
How many samples of size 2 are
possible?
C = 10
5
2
Find all 10 of
these samples and
record the sample
means.
Permutations
7
2
10
14
11
Pairs
Pairs
x
2, 7 2, 10 2, 11 2, 14 7, 10 7, 11 7, 14 10, 11 10, 14 11, 14
4.5 6
6.5 8
8.5 9
mx = 8.8
sx = 2.4919
10. 10. 12
5
5
What is the mean
and standard
deviation of the
sample means?
12.
5
Repeat this procedure with sample
size n = 3
How many samples of size 3 are
possible?
C = 10
5
mx = 8.8
sx =
3
What
mean
Find
allisofthe
these
and standard
samples
and
deviation
of the
record
the
sample
1.66132 sample means?
means.
Permutations
2
2 2
2 2 2
11
10 10
7 7 7
14
11 14
10 11 14
7 7 7
10
10 10 11
11
11 14 14
14
Pairs
2,7 2,7,
2,7 2,7,
2,7 7,1
7,1 7,1
7,1 10,
10, 2,1
2,1 2,1
2,1 2,1
2,1 7,1
7,1
Pairs
2,7,
x
x
,14 0,1
0,1
10,10 11,11 14
11
6.3 6.6 7.6 9.3
1,1
1,1
44
10.
6
mx = 8.8
sx = 1.66132
11, 0,1
0,1 1,1
1,1
11,
14 11
14
44
11. 7.6 9
6
0, 0,1
0,1
0,
14 44
14
8.6 10.
3
What is the mean
and standard
deviation of the
sample means?
What do you notice?
• The mean of the sampling distribution
EQUALS the mean of the population.
mx = m
• As the sample size increases, the standard
deviation of the sampling distribution
decreases.
as n
sx
A statistic used to estimate a
parameter is unbiased if the
mean of its sampling
distribution is equal to the
true value of the parameter
being estimated.
General Properties
Rule 1:
mx = m
s
Rule 2: sx =
n
This rule is approximately correct as long
as no more than 10% of the population is
included in the sample
General Properties
Rule 3:
When the population distribution is
normal, the sampling distribution of x
is also normal for any sample size n.
General Properties
Rule 4: Central Limit Theorem
When n is sufficiently large, the
sampling distribution of x is well
approximated by a normal curve, even
when the population distribution is not
itself normal.
CLT can safely be applied if n exceeds 30.
If n is large or the population
distribution is normal, then
x  mx
x mx
z

s
sx
n
has approximately a standard
normal distribution.
Example 1
The army reports that the distribution of head
circumference among soldiers is
approximately normal with mean 22.8 inches
and standard deviation of 1.1 inches.
a) What is the probability that a randomly
selected soldier’s head will have a
circumference that is greater than 23.5 inches?
P(X > 23.5) = .2623
Normcdf(23.5,100,22.8,1.1)
b) What is the probability that a random
sample of five soldiers will have an
average head circumference that is greater
than 23.5 inches?
Do you expect the probability to
be moreWhat
or lessnormal
than the
curve are
to you
partnow
(a)? working
Explainwith?
P(X > 23.5) = .0774
Normcdf(23.5,100,22.8, (1.1/sqrt5))
A soft-drink bottler claims that, on average, cans
contain 12 oz of soda. Let x denote the actual
volume of soda in a randomly selected can.
Suppose that x is normally distributed with
s = .16 oz. Sixteen cans are to selected with a
mean of 12.1 oz. What is the probability that the
average of 16 cans will exceed 12.1 oz?
P(x >12.1) = .0062
Normcdf(12.1, 100, 12, (0.16/(sqrt 16))
Suppose a team of biologists has been
studying the Pinedale children’s fishing
pond. Let x represent the length of a
single trout taken at random from the
pond. This group of biologists has
determined that the length has a normal
distribution with mean of 10.2 inches and
standard deviation of 1.4 inches. What is
the probability that a single trout taken at
random from the pond is between 8 and
12 inches long?
P(8 < X < 12) = .8427
Normcdf(8, 12, 10.2, 1.4)
What is the probability that the mean
length of five trout taken at random is
between 8 and 12 inches long?
Do xyou
expect
the probability to
P(8<
<12)
= .9978
be more or less than the answer
to part (a)? Explain
What sample mean would be at the 95th
percentile? (Assume n = 5)
x = 11.23 inches
```