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Sampling
Distributions of
Proportions
Sampling Distribution
• Is the distribution of possible values
of a statistic from all possible
samples of the same size from the
same population
• In the case of the pennies,
the
We it’s
will use:
p for the
population
distribution of all possible
sample
proportion
proportions (p)
and
p-hat for the sample
proportion
Suppose we have a population of six
people: Alice, Ben, Charles, Denise,
Edward, & Frank
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.
Ben & Frank
Alice & Ben
.5
Charles & Denise
Alice & Charles
.5
Alice & Denise
1
Charles & Edward
Alice & Edward
.5
Charles & Frank
the mean of the
Alice & Frank How does
.5
Denise & Edward
(mp-hat)
Ben & Charlessampling
0 distribution
Denise & Frank
Ben & Denise compare
.5 to the population
Edward & Frank
parameter
(p)?
m
=
p
Ben & Edward
0p-hat
0
.5
0
0
.5
.5
0
Find the mean & standard deviation of all p-hats.
μpˆ
1

3
&
σ pˆ  0.29814
Formulas:
μpˆ  p
σ pˆ 
These are found on
the formula chart!
p 1  p 
n
Does the standard deviation of the
sampling distribution equal the
equation?
NO -
σ pˆ 
 
1 2
1
3So3– in order
calculator
to 0
.29814 the
2 standard
3 deviation of the
sampling distribution, we
WHY?
MUST be sure that our sample
Correction
factor
– multiply
size is
less than
10%
of the by
We are sampling more than 10% of our population!
If we
N n
population!
use the correction factor, we will see that weNare
 1 correct.
σ pˆ 
 
1 2
3 3  6  2  0.29814
2
6 1
Assumptions (Rules of Thumb)
• Sample size must be less than 10% of
the population (independence)
• Sample size must be large enough to
ensure a normal approximation can be
used.
np > 10 & n (1 – p) > 10
Why does the second assumption ensure
an approximate normal distribution?
Remember back to binomial distributions
Suppose n = 10 & p = 0.1
(probability of a success), a
histogram of this distribution
> 10 &skewed
n(1-p) >right!
10
isnp
strongly
insures that the sample size
Now
use
n
=
100
&
p
=
0.1
(Now
is large enough to have a
np normal
> 10!) While
the histogram is
approximation!
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?
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 & s using the formulas.

Find
No – since there is
approximately a 1% chance of
this happening, I do not
believe the student did this.
Assume that 30% of the students
at BHS wear contacts. In a
sample of 100 students, what is
the probability that more than 35%
of them wear contacts?
mp-hat = .3
& sp-hat = .045826
Check assumptions!
np = 100(.3) = 30 & n(1-p) =100(.7) = 70
Ncdf(.35, 1E99, .3, .045826) = .1376