Transcript Ch 7 Powerpoint - Campbell County Schools

Ch 7 - Sampling
Distribution
Tuesday, January 3, 3012
The Airport Problem
• This activity will review binomial
calculations and introduce the idea of
sampling distribution of sample
proportions using simulation
• You will need your TI-83/84 calculators
At Guadalajara Airport in Mexico, passengers must claim their luggage and then
proceed to Customs. In the Customs area, each passenger will press a button that
activates a modified stoplight. This light has only red and green bulbs. If the green
lights shows, the passenger is free to go. If the light turns red, then Customs agents
will inspect the passenger’s luggage. Customs officials claim that the light has a
probability of 0.30 of showing red on any press of the button.
You have 20 minutes to start the activity that
I am passing out. It is due at the beginning
of class tomorrow.
You do not have to draw the graphs, if you
are describing them. Remember your
“SOCS!” You can look at them on your
calculators.
Vocabulary
• Population (review)
• Sample (review)
• Parameter
• Statistic
• *Hint: the p’s and the s’s go together!
Notation
• notation is very important, and you will
loose points on the exam if you use the
wrong notation (as well as if you use
the wrong word)

• Population mean:
_
• Sample mean: x
• Population Proportion: p
^
• Sample Proportion: p
7.1 - Continued
•
•
•
•
Each of you will randomly select 5 cards from a shuffled deck and note
the median value. Then replace your cards.
You will do this 2 times each
After you are finished, record the value of the sample median on the
dotplot on the board. Use a lower case “m” instead of a dot since we are
talking about a sample and not a population.
Then we will describe what we see: Shape, Center (mean), Spread
(standard deviation), and Outliers!
Wednesday, January 4, 2012
Sampling Distribution
The sampling distribution of a statistic is the distribution of values taken
by the statistic in all possible samples of the same size from a population.
For example, in our card activity if we were able to take every possible
sample of size 5 and record the outcomes, that would be the sampling
distribution of the median of cards 2 through 10 when selecting 5 cards.
That would be 36 nCr 5 = 376,992 different samples!
It’s very time consuming to take all possible samples, instead simulations
are done to imitate the process we just did. (FATHOM)
Using Fathom to simulate choosing
500 SRSs of size 5 from the deck of
cards 2 though 10 and finding the
sample medians produced the
following dotplot.
Distribution, Distribution, Distribution...
• There are 3 different types of
distributions:
• 1. Population Distribution
• 2. Distributions of Sample Data
• 3. Sampling Distributions
• *Yes, there is a difference between
sample distribution and sampling
distribution
What’s the difference between the 3 distributions?
•
•
A
distribution
of
sample
data
is
an
Sampling
distribution
describes
how
a
A population distribution is one graph
of
individual
graph
depicting
each
outcome
statistic
varies
in
many,
many
samples
everything possible as a whole. of the
from the sample
you
from
the at the
population.
You are
nodrew
longer
looking
population.
individual
elements
in thewe would have a
For
our card
activity,
sample/population.
bar graph with 9 bars (2 - 10) with a
For our card activity, we would have a
frequency
of
4
each.
separate
baractivity,
graph for
each sample
youthe
For
our card
it would
have been
drew showing
all the
cards you
drew and
simulation
we ran
on fathom
showing
the
their frequencies.
different
values of the medians of our
samples.
Describing Sampling
Distributions
Let’s describe the dotplot of the
simulation of 500 SRSs of size 5 from
our deck of cards:
Shape: roughly symmetric with
a single peak at 6
Center: the mean of the
sample medians is
about 6
Spread:most of the values fall between 4
and 8 with a few at 2 and 10
Outliers:
there doesn’t appear to
be any outliers
Why didn’t we find the standard
deviation since we estimated
the mean for the center?
More questions about our
card simulation:
• Was that a sampling distribution?
• If someone claims to set up the same
activity and they select a sample of size
5 and get a median of 4, is that
convincing evidence that they set their
deck up wrong?
Biased and Unbiased
Estimators
• This is different than the sampling
process being biased. When using an
estimator (i.e. a measure of center or
spread) we are assuming the sampling
process is not biased.
• The actual statistic we are finding can
be biased or unbiased as well.
Biased or Unbiased?
So which sample statistics are biased and which are unbiased?
To find out, let’s collect some quantitative data:
1. On the piece of paper given to you write how many hours of
sleep you got last night.
2. Each of you will randomly select a sample of 4 cards.
3. You will need to record the following information: the person’s
initials, the hours of sleep they got, the sample mean, and the
sample range.
4. Replace the cards, shake the bag, and draw another sample
of size 4 and record the same information.
5. Pass the bag to the next person then record your sample
mean and sample range on the corresponding dotplots on the
board.
6. Once everyone plots their data, we will analyze it and
compare it to the population mean and population range.
7.2 - Sample
Proportions
• Remember your notation:
•population parameter is p
^
•sample proportion is p
Thursday, January 5, 2012
Penny Activity
• Let’s look at the population dotplot.
Describe what you see.
Penny for you
thoughts
1. Take a sample of size 5 from the
population
^
2. Calculate p = the proportion of pennies
minted in the 2000s
3. Replace the pennies and repeat a
second and third time.
4. Record your
^ values on the dotplot on the
board with p instead of dots.
Penny for you
thoughts
1. Take a sample of size 10 from the
population
^
2. Calculate p = the proportion of pennies
minted in the 2000s
3. Replace the pennies and repeat a
second and third time.
4. Record your values on the dotplot on the
^
board with p instead of dots.
Penny for you
thoughts
1. Take a sample of size 20 from the
population
^
2. Calculate p = the proportion of pennies
minted in the 2000s
3. Replace the pennies and repeat a
second and third time.
4. Record your values on the dotplot on the
^
board with p instead of dots.
Compare the 4
graphs
Shapes?
Centers?
Spreads
?
More Notations
• We have a population, we take a
sample, and find some proportion.
• If we want to investigate those sample
proportions we can find the mean and
standard deviation of the sampling
distribution of the sample proportions.
•mean of sample proportions:
•standard deviation of sample
proportions:
^
p
 ^p
^
Sampling Distribution of p
• SHAPE: sometimes it can be
approximated by the Normal curve. It
depends on the sample size n and the
population proportion p.
•
^
^

CENTER:p
= p because p is an
unbiased estimator of p.
 ^p
• SPREAD:

gets
smaller
as
n
gets
^
p
larger. The value of
depends on both
n and
p.
p. 436 shows a good, small proof of why these are true
Sampling Distribution of a
Sample Proportion
•
Choose an SRS of size n from a population of size N with proportion p
^
of successes.
Let p be the sample proportion of successes. Then:
➡
^
the mean of the sampling distribution of
 p^ is
➡
p (1  p )
^
the standard deviation of the sampling distribution ofp^ p is
p
=p
1
n
➡ AS LONG AS THE 10% CONDITION IS SATISFIED:
➡ np
if  10
n
N
10
n(1  p )  10
and
, the Normal conditions are satisfied and
^
the sampling distribution
of p is approximately Normal.
Formulas are on your
formula sheet
district wants to know what proportion of
middle school students in her district are
planning on attending a four-year college
or university. Suppose that 80% of all
middle school students in her district are
planning on attending a four-year college
or university. What is the probability that
an SRS of size 125 will give a result
within 7 percentage points of the true
value?
• We will use the 4-step method to solve
this problem.
State
• We want to find the probability that the
percentage of middle school students
that plan to attend a 4-year college or
university falls between 73% and 75%
•
^
 p 0.78)
or in symbols: P(0.73
Plan
•
• Since the school district is large, we’ll
 ^p = 0.80.
assume the 10% condition is satisfied
and there are more than 1250 students.
.8 (.2 )
^ 
(10*125
=
1250).
So,
=
0.036
125
p
We can consider the distribution of p to
be approximately Normal since the
following are true:
np = 125(.8) = 100 > 10
n(1  p=) 125(.2) = 25 > 10
• P(0.73 ^p 
Do
0.87) = normalcdf(0.73, 0.87,
0.80, 0.036) = 0.948
• If you want full credit on the exam, you
must have clearly said everything in the
“Plan” step and these calculations will
receive full credit.
Sketching a Normal curve will help.
You can also use Table A to find the answer. Remember to standardize (z-score) first!
Conclude
• About 95% of all SRSs of size 125 will
give a sample proportion within 7
percentage points of the true proportion
of middle school students who want to
attend a four-year college or university.
7.3 - Sample Means
Friday, January 6, 2012
Back to our pennies!
• This activity is very similar to our first
activity. In the first one we compared the
population proportion of pennies minted
in the 2000s to the sample proportion of
pennies minted in the 2000s.
• This time we will look at the sample
distribution of the sample means of the
year the pennies were minted
• 1st: take an SRS of 5 pennies from the
population and record their years.
• 2nd: replace your sample then repeat
two more times.
• 3rd: find the mean year for each of your
3 samples. These are your sample _
means, x
• 4th: record your sample_ means on the
appropriate dotplot. Use x’s instead of
dots.
After everyone has done this, repeat this process again
with SRSs of size 10 and size 25.
Compare the 4
graphs
Shapes?
Centers?
Spreads
?
More Notation!
_
x
• Mean of the sampling distribution:
_
x
• Standard deviation of the sampling distribution of the sample means:
•
All of the notations in this chapter are very important and very similar.
You will loose credit for using the wrong notations on the exam. So if
you can’t remember it’s always best to write out what you are finding
rather than try to use a notation.
Suppose that x-bar is the mean of an
SRS of size n drawn from a large
population with mean and standard
deviation , it does not matter what
shape the population has.
The mean of the sampling
_
distribution of x is
 _x  
_
The standard deviation of the sampling
distribution of x is
_
x 

n
as long as the 10% condition is satisfied!
These formulas are on
your formula sheet
distribution of x, these means to state if it
is Normal and find the mean and
_
standard deviation.
• Hint: if the population itself is
approximately Normal, then so is the_
sampling distribution of x.
• Hint: Please read carefully! Make sure
you know if you are using the population
standard deviation or the sample means
standard deviation before you
standardize or use your normalcdf on
your calculator.
A grinding machine in an auto parts plant prepares axels with a
target diameter mu = 40.125 mm. The machine has some
variability, so the standard deviation of the diameters is sigma =
0.002 mm. The machine operator inspects a random_sample of 4
axles each hour for quality control purposes and records the
sample mean diameter x.
a) Assuming the process is working properly,
what are the mean and standard deviation of
the sampling distribution of x?
-
_
b) Can you find the probability that x is within .05
mm if you are choosing an SRS of 100 axels?
Explain
c) In order for you to pass this inspection the
standard deviation of the sampling
distribution of x needs to be 0.0005 mm.
How many axels would you have to sample?
•
The composite scores of individual students on the
ACT in 2009 followed a Normal distribution with
mean
and standardthat
deviation
5.1.
a) What
is 21.1
the probability
a single
student randomly chosen form all those
taking the test scores 23 or higher?
Show your work.
• b) Now take an SRS of 50 students who
took the test. What is the probability that
the mean score x of these students is 23
or higher?
What if the population shape is not Normal?
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Monday, January 9, 2012
Central Limit
Theorem
• Draw an SRS of size n from any
population with mean

standard deviation
•
and finite

_
CLT - when n is large, the sampling
distribution of the sample means x isNOTE: this is
of the sample
approximately Normal.
means, not
just any
sample!!!
How large is large?
• In order for the Normal conditions to
apply for the sample means, and the
population is not Normal the CLT will
n  30
apply
in most cases if
mean 1.6 flaws per square yard and
standard deviation 1.2 flaws per square
yard. The population distribution cannot
be Normal, because a count takes only
whole-number values. An inspector
studies 200 square yards of material,
_
records
the number of flaws found in
each square yard, and calculates x, the
mean number of flaws per square yard
inspected. Find the probability that the
mean number of flaws exceeds 2 per
square yard.
State
• What’s the probability that the mean
number of flaws per square yard of
carpet is more than 2?
Plan
• The mean of the sampling distribution of
the sample means
 x is1.6
_
• 10% condition is met since there’s more
 0.085
than 2000 square yards of carpet,so
x
_
• Since the sample size is large, 200 > 30,
we can safely use the Normal
distribution
as an approximation for the
_
sampling distribution of x
Do
• P(x > 2) = normalcdf(2, 100, 1.6, 0.085)
=0
Conclude
• There is virtually no chance that the
average number of flaws per yard in the
sample will be greater than 2.