Transcript Chapter 18 Sampling distributions of means

Chapter 9
found online and modified slightly!
Sampling Distributions
Parameter
• A number that describes the population
• Symbols we will use for parameters
include
m - mean
s – standard deviation
p – proportion (p)
a – y-intercept of LSRL
b – slope of LSRL
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
a – y-intercept of LSRL
b – slope of LSRL
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 of 5
fish in my pond – the length of
fish (in inches):
2, 7, 10, 11, 14
What is the mean
mand
8.8
x =standard
deviation
of this
sx = 4.0694
population?
2, 7, 10, 11, 14
Let’s take samples of size 2
(n = 2) from this population:
How many samples of size 2 are
possible?
C = 10
5
mx = 8.8
sx =
2
2,7
2,10
2,11
2,14
7,10
7,11
7,14
10,11
10,14
11,14
Find
all 10
of
What
is the
mean
these
samples
of the
sampleand
record
the sample
means?
Can’t be calculated, because our population
is not at least 10 times
means.
as large as our population size.
Same as the population mean!
2, 7, 10, 11, 14
Repeat this procedure with sample size n = 3.
How many samples of size 3 are possible?
5C3
= 10
mx = 8.8 Same as our population mean.
sx =
What is the mean
Find all of these
and standard
Can’t be calculated, because samples
our populationand
is not at least 10
deviation of the
times as large as our population
size.the sample
record
sample means?
means.
2,7,10
2,7,11
2,7,14
2,10,11
2,10,14
2,11,14
7,10,11
7,10,14
7,11,14
10,11,14
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
We will see this occur in our “fishing activity”
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 population size is at least 10 times larger
than the sample size.
General Properties
Rule 3:
When the population distribution is
normal, the sampling distribution of x
is also normal for any sample size n.
Activity – drawing samples
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
How large is “sufficiently large”
itself normal.
anyway?
CLT can safely be applied if n exceeds 30.
EX) 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
normalcdf (23.5, e99, 22.8, 1.1)
lower b, upper b, mean,
s.d.
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 more or
less
than
the
answer
What normal curve are
normalcdf (23.5,
e99, (a)?
22.8, 1.1/√(5))
to part
Explain
you now working with?
P(X > 23.5) = .0774
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
normalcdf (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(8,or12,
less
than
the answer
Normalcdf
10.2,
1.4/√(5))
to part (a)? Explain
What sample mean would be at the 95th
percentile? (Assume n = 5)
x = 11.23 inches
invnorm (.95, 10.2, 1.4/√(5))
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 randomly
selected and a mean of 12.1 oz is calculated.
What is the probability that the mean of 16
cans will exceed 12.1 oz?
P(x >12.1) = .0062
normalcdf (12.1, e99, 12, .16/√(16))
A hot dog manufacturer asserts that one of its
brands of hot dogs has a average fat content of 18
grams per hot dog with standard deviation of 1
gram. Consumers of this brand would probably not
be disturbed if the mean was less than 18 grams, but
would be unhappy if it exceeded 18 grams. An
independent testing organization is asked to analyze
a random sample of 36 hot dogs. Suppose the
resulting sample mean is 18.4 grams. What is the
probability that the sample mean is greater than
18.4 grams?
P(x >18.4) = .0082
normalcdf (18.4, e99, 18, 1/√(36))
Does this result indicate that the
manufacturer’s claim is incorrect?
No. The manufacturer’s claim was that its
hotdogs had an average fat content of 18
grams.
We found that the probability (fat content was
greater than 18.4 grams) = .0082 which is less
than 1%. This means that approximately
99% of their hotdogs have fat content less
than 18.4 grams.