#### Transcript Proportions

```Sampling
Distributions of
Proportions
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
A distribution is all the
values that a variable can be.
•
•
The dotplot is a partial graph of the
Toss
a
penny
20
times
and
record
sampling distribution of all sample
theproportions
sample size 20. If I
found all
possible sample
Calculate
thetheproportion
proportions
this plot
wouldon
bethe
& mark
it on the– dot
approximately normal!
board.
What shape do you think the
dot plot will have?
Sampling Distribution
• Is the distribution of possible
values
x
ˆ
p
 possible
of a statistic from all
n
samples of the same size from the
Where x is the number in the
same population
sample & n is the sample size
• 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
We are interested in the proportion of
females. This is called
The parameter of interest
What is the proportion of females? 1/3
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
Suppose we have a population of six
people: Alice, Ben, Charles, Denise,
Edward, & Frank
Draw samples of three from this
population.
How many
different
samples
are
What do you notice about the
possible?
20
means
standard
6C&
3=
deviations?
Find the mean & standard deviation of all p-hats.
m pˆ
1

3
&
s pˆ  0.2108
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
3 So3– in
order

 to 0calculate
.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)
(Population > 10n)
• Sample size must be large enough to
insure a normal approximation can be
used.
np > 10 & n (1 – p) > 10
Why does the second assumption insure
an approximate normal distribution?
What would happen if
the fixed
numberdistributions
was
Remember back
to binomial
100?
Suppose n = 10 & p = 0.1
(probability of a success), a
histogram of this distribution
is strongly skewed right!
Suppose a binomial distribution has
n = 100 and p = 0.1.
What is the mean and standard
these when
bars are
deviation ofNotice
thisthat
distribution?
However,
n isextremely
large enough,
m=
small and
extend
out
to 100
– so
this
the
tail
will
into
an
10distribution
& s = 3 is skewed right
approximate normal curve
Graph a histogram of this binomial
distribution. What shape to do you
expect this to be?
Since p = .1, we would expect this
distribution to be skewed right
Why do we need to also check
n(1 – p)?
Consider what the histogram looks
like when n = 10 and p = .9.
We must also check that the upper
tail will spread out into an
approximate normal curve.
Assumptions (Rules of Thumb)
• Sample size must be less than 10% of
the population (independence)
(Population > 10n)
• Sample size must be large enough to
insure a normal approximation can be
used.
np > 10 & n (1 – p) > 10
Chip Activity:
•Select three samples of size 5, 10, and 15
and record the number of blue chips.
•Place your proportions on the appropriate
dotplots.
What do you notice about these
distributions?
Some proportion distributions
where p = 0.2
Let p be the proportion of successes in a random
sample of size n from a population whose
proportion of S’s (successes) is p.
n = 10
n = 20
n = 50
n = 100
0.2
0.2
0.2
0.2
Based on past experience, a
bank believes that 7% of the
μpˆ  .07
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? 10^99, .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? Find the probability that a
coin would
than 42%


.5(.5=) 200(.5)
np = 200(.5)
= 100 & n(1-p)
= 100


ncdf


,.
42
,.
5
,

.
0118
of Since
the
time.
both
curve!
 > 10, I can use a normal

200 

Find m & s using the formulas.
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 SHS 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, 10^99, .3, .045826) = .1376
Example
If the true proportion of defectives produced
by a certain manufacturing process is 0.08
and a sample of 400 is chosen, what is the
probability that the proportion of defectives
in the sample is greater than 0.10?
Since np  400(0.08)  32 > 10 and
n(1-p) = 400(0.92) = 368 > 10,
it’s reasonable to use the normal
approximation.
Example
(continued)
mp  p  0.08
p(1  p)
0.08(1  0.08)
sp 

 0.013565
n
400
p  mp 0.10  0.08
z

 1.47
sp
0.013565
P(p > 0.1)  P(z > 1.47)
 1  0.9292  0.0708
Example
Suppose 3% of the people contacted by phone are
receptive to a certain sales pitch and buy your
product. If your sales staff contacts 2000 people,
what is the probability that more than 100 of the
people contacted will purchase your product?
Clearly p = 0.03 and p = 100/2000 = 0.05 so



0.05  0.03 
P(p > 0.05)  P  z >

(0.03)(0.97) 


2000


0.05  0.03 

 P z >
  P(z > 5.24)  0
0.0038145 

Example - continued
If your sales staff contacts 2000 people, what
is the probability that less than 50 of the
people contacted will purchase your
product?
Now p = 0.03 and p = 50/2000 = 0.025 so



0.025  0.03 
P(p  0.025)  P  z 

(0.03)(0.97) 


2000


0.025  0.03 

 P z 
  P(z  1.31)  0.0951
0.0038145 

```