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Transcript
Sampling
Distributions of
Proportions
•
•
The dotplot was a partial graph
Remember
thedistribution
skittles of
example.
of the sampling
all
sample proportions of sample size
We calculated the proportion of
40. If we found all the possible
orange
skittles &– this
marked
sample proportions
would it
be on
the dot
plots on the
board.
approximately
normal!
What shape did the n=40 dot
plot have?
Sampling Distribution
• Choose an SRS of size n from a
large population with population
proportion p having some
characteristic of interest. Let phat be the proportion of the
sample having that characteristic..
• We need to come up with some
formulas for the mean and
standard deviation.
Suppose we have a population of six people:
Melissa, Jake, Charles, Kelly, Mike, &
Brian
What is the proportion of females? 1/3
What is the parameter of interest in
this population?
Proportion of females
Draw samples of two from this population.
How many different samples are possible?
6C2
=15
Find the 15 different samples that are
possible & find the sample proportion of
the number of females in each sample.
Jake & Brian
Melissa & Jake
.5
Charles & Kelly
Melissa & Charles
.5
Melissa & Kelly
1
Charles & Mike
How does the mean of the
Melissa & Mike
.5
Charles & Brian
sampling distribution
(mp-hat)
Melissa & Brian
.5
Kelly
& Mike
compare to the
population
Jake & Charles
0
Kelly
& Brian
parameter
(p)?
Jake & Kelly
m.5p-hat = p
Mike & Brian
Jake & Mike
0
0
.5
0
0
.5
.5
0
Find the mean & standard deviation of all p-hats.
μpˆ
1

3
&
σ pˆ  0.29814
Formulas:
The mean of the
sampling
distribution.
X
pˆ 
n
m pˆ  p
The standard
deviation of the
sampling
distribution.
p1  p 
 pˆ 
n
Does the standard deviation of the
sampling distribution equal the equation?
NO -
σ pˆ 
 
1 2
3 3 1
2
3
 0.29814
WHY?
So – in order to calculate the
standard deviation of the
sampling distribution, we
MUST be sure that our sample
size is less than 10% of the
population!
We are sampling more
than 10% of our
population!
Assumptions (Rules of Thumb)
• Use this formula for standard
deviation when the population is
sufficiently large, at least 10 times as
large as the sample.
• Sample size must be large enough to
insure a normal approximation can be
used. We can use the normal
approximation when
np > 10 & n (1 – p) > 10
Why does the second assumption insure
an approximate normal distribution?
Remember back to binomial distributions
Suppose n = 10 & p = 0.1
(probability of a success),
a histogram of this
np > 10 & n(1-p) > 10
distribution is strongly
insures that the sample
skewed
right!
Now
use
n
100 &enough
p = 0.1
size is=large
to
(Now np >have
10!) While
the
a normal
histogramapproximation!
is still strongly
skewed right – look what
happens to the tail!
Based on past experience, a
bank believes that 7% of the
μpˆ  .07
people who receive loans will
not make payments on
.07 time.
.93
σ


.
01804
ˆ
p
The bank recently approved
200 loans.
Yes
–
200
np = 200(.07) = 14
and
standard deviation
n(1 - p) = 200(.93) = 186
What are the mean
of the proportion of clients in this group
who may not make payments on time?
Ncdf(.10,
Are assumptions
met? 1E99, .07, .01804) =
.0482
What is the probability that over 10% of
these clients will not make payments on
time?
Example #1
A polling organization asks an SRS of 1500 first year college
students whether they applied for admission to any other college.
In fact, 35% of all first-year students applied to colleges besides
the one they are attending. What is the probability that the
random sample of 1500 students will give a result within 2
percentage points of the true value?
STATE
PLAN
DO
CONCLUDE
Example #1
STATE:
We want to know the probability that a random
sample yields a result within 2 percentage points
of the true proportion.
We want to determine
P (.33  pˆ  .37)
Example #1
PLAN:
We have drawn an SRS of size 1500 from the population
of interest.
The mean of the sampling distribution of p-hat is 0.35:
mpˆ  0.35
Example #1
PLAN:
We can assume that the population of first-year college
students is over 15,000, and are safe to use the standard
deviation formula:
p(1  p )
(0.35)(0.65)
 pˆ 

 0.0123
n
1500
In order to use a normal approximation for the sampling
distribution, the expected number of successes and failures
must be sufficiently large:
np  10 and n(1  p)  10
1500(.35)  10 and 1500(.65)  10
Therefore, pˆ  N (0.35,0.0123)
Example #1
DO: Perform a normal distribution calculation
to find the desired probability:
P (.33  pˆ  .37)  .8961
Example #1
CONCLUDE: About 90% of all SRS’s of size
1500 will give a result within 2 percentage points
of true proportion.
Example #2
Suppose one student tossed a coin 200
times and found only 42% heads. Do you
believe that this is likely to happen?
 = 100 & n(1-p)
200(.5) = 100
.
5
(.
5
)
np = 200(.5)
=
  .0118
ncdf   ,.42,.5,
Since both
a normal curve!
 > 10, I can use
200
 m &  using the formulas.

Find
No – since there is
approximately a 1% chance of
this happening, I do not
believe the student did this.
Example #3
Assume that 30% of the students at
HH wear contacts. In a sample of 100
students, what is the probability that
more than 35% of them wear
contacts?
mp-hat = .3
& p-hat = .045826
np = 100(.3) = 30 & n(1-p) =100(.7) = 70
Check assumptions!
Ncdf(.35, 1E99, .3, .045826) = .1376
Example #4 (Your turn)
• About 11% of American adults are black.
Therefore, the proportion of blacks in an SRS
of 1500 adults should be close to .11.
If a national sample contains only 9.2%
black, should we suspect that the sampling
procedure is somehow under-representing
blacks?