Download Chapter 9 Section 9.2 – Sample Proportions

CHAPTER 9
Section 9.2 – Sample Proportions
THE SAMPLING DISTRIBUTION OF 𝑝

As you learned in section 9.1, in order to estimate
a population parameter, p, you use a sample
statistic, 𝑝.

Recall that 𝑝 is the sample proportion of successes:
count of "successes" in sample 𝑋
𝑝=
=
size of sample
𝑛


Since the values of X and 𝑝 will vary with
repeated samples, both X and 𝑝 are random
variables.
Provided that the population is much larger than
the sample (by at least 10 times), the count X will
follow a binomial distribution.
SAMPLING DISTRIBUTION OF A SAMPLE
PROPORTION

Choose an SRS of size n from a large population with
population proportion p having some characteristic of
interest. Let 𝑝 be the proportion of the sample having that
characteristic. Then:

The mean of the sampling distribution is exactly p.
𝜇𝑝 = 𝑝

The standard deviation of the sampling distribution is:
𝜎𝑝 =

𝑝(1 − 𝑝)
𝑛
Rule of thumb #1:

In order to use the standard deviation of 𝑝, the
population must be at least 10 times as large as the
sample.
USING THE NORMAL APPROXIMATION FOR 𝑝


Recall from section 9.1 that the shape of the
sampling distribution of 𝑝 is approximately
normal and is closer to normal when the sample
size n is larger.
Rule of thumb #2:

We will use the normal approximation to the
sampling distribution of 𝑝 for values of n and p that
satisfy 𝑛𝑝 ≥ 10 and 𝑛(1 − 𝑝) ≥ 10
EXAMPLE 9.7 – APPLYING TO COLLEGE

A polling organization asks an SRS of 1,500 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 1,500 students will give
a result within 2 percentage points of this true value?
The normal approximation to
the sampling distribution of p.

First check rule of thumb #1 & #2:
The population is all first year college students in the U.S. and
since the sample size is 1500, the population would have to
contain at least 10(1500) =15,000 first year college students in
order to use the standard deviation formula. Since there are
over 1.7 million first year college students we can proceed.
 We can use the normal approximation since np = 1500(.35) 525
and n(1-p) = 1500(.65) = 975 are both greater than 10.

EXAMPLE 9.7 – APPLYING TO COLLEGE

We want to find the probability that 𝑝 falls between .33 and
.37.

𝑧=
𝑝−𝑝
𝜎𝑝
=
.33−.35

𝑧=
𝑝−𝑝
𝜎𝑝
=

𝑃 0.33 ≤ 𝑝 ≤ 0.37 = 𝑃 −1.63 ≤ 𝑧 ≤ 1.63
= .9484 − .0516
= .8968
(.35)(1−.35)
1500
.37−.35
(.35)(1−.35)
1500
= −1.63
= 1.63
This means that almost 90% of all samples will give a result
within 2 percentage points of the truth about the population.
EXAMPLE 9.8 – SURVEY UNDERCOVERAGE



One way of checking the effect of undercoverage, nonresponse, and
other sources of error in a sample is to compare the sample with
known facts about the population.
About 11% of American adults are black. The proportion p of blacks
in a SRS of 1,500 adults should therefore be close to .11. It is
unlikely to be exactly .11 because of sampling variability. If a
national sample contains only 9.2 % blacks, should we suspect that
the sampling procedure is somehow underrepresenting blacks?
Test both rule of thumbs:
 The population of all black American adults is larger than
10(1500) = 15,000 so the standard deviation formula can be used
 A normal approximation can be used since np = 1500(.11) = 165
and n(1-p) = 1500(.89) = 1335.
EXAMPLE 9.8 – SURVEY UNDERCOVERAGE
𝑧
𝑃
=
𝑝−𝑝
𝜎𝑝
=
.092−.11
(.11)(1−.11)
1500
= −2.23
𝑝 ≤ 0.092 = 𝑃 𝑧 ≤ −2.23 = .0129
Only 1.29% of all samples would have so few blacks
which would give good reason to believe that the
sampling procedure underrepresented blacks.


Homework: p.511 #’s 19-21, 23, 25, 26, & 30
Here is a video lesson if you would like to see
more examples:

http://www.youtube.com/watch?v=xZx6wKUEShs