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Section 7.2
Sample Proportions
ˆ
The Sampling Distribution of p
How good is the statistic pˆ as an estimate of the parameter p? The
sampling distribution of pˆ answers this question.
Consider the approximate sampling distributions generated by a
simulation in which SRSs of Reese’s Pieces are drawn from a
population whose proportion of orange candies is 0.45.
What do you notice about the shape, center, and spread of each?
ˆ
The Sampling Distribution of p
What did you notice about the shape, center, and spread of each
sampling distribution?
Shape : In some cases, the sampling distribution of pˆ can be
approximated by a Normal curve. This seems to depend on both the
sample size n and the population proportion p.
Center : The mean of the distribution is m pˆ = p. This makes sense
because the sample proportion pˆ is an unbiased estimator of p.
Spread : For a specific value of p , the standard deviation s pˆ gets
smaller as n gets larger. The value of s pˆ depends on both n and p.
There is an important connection between the sample proportion p̂ and
the number of "successes" X in the sample.
pˆ =
count of successes in sample X
=
size of sample
n
ˆ
The Sampling Distribution of p
In Chapter 6, we learned that the mean and standard
deviation of a binomial random variable X are
mX = np
sX = np(1 - p)
Since pˆ = X /n = (1/n) × X, we are just multiplying the random variable X
by a constant (1/n) to get the random variable pˆ . Therefore,
1
m pˆ = (np) = p
n
pˆ is an unbiased estimator or p
The Sampling Distribution of p̂
The standard deviation of the sampling distribution is:
1
np(1 - p)
s pˆ =
np(1 - p) =
=
2
n
n
p(1 - p)
n
As sample size increases, the
spread decreases.
As long as the 10% condition is satisfied:
n < (1/10)N or 10n < N
As n increases, the sampling distribution becomes
approximately Normal. Before you perform Normal
calculations, check that the Normal condition is satisfied:
np ≥ 10 and n(1 – p) ≥ 10.
ˆ
The Sampling Distribution of p
An SRS of 1500 first-year college students were
asked whether they applied for admission to any
other college. In fact, 35% of all first-year
students applied to colleges beside the one they
are attending. What is the probability that the
poll will be within 2 percentage points of the
true p?
Step 1: State the distribution and the values of
interest.
𝜇𝑝 = 𝑝 = 0.35
𝑃 0.33 < 𝑝 < 0.37 = ?
Check conditions!!!!!
10n < N
np ≥ 10 and n(1 – p) ≥ 10.
10(1500) = 15,000 <
population of all first
year college students 
1500(.35) = 525 ≥ 10  and 1500(.65) = 975 ≥ 10 
Since we meet our conditions we can consider
the distribution of 𝑝 to be approximately Normal.
Step 2: Perform calculationsshow your work!!!!!
𝜇𝑝 = 𝑝 = 0.35
(0.35)(0.65)
= .0123
 pˆ 
1500
Normalcdf (lwr: 0.33, upr: 0.37, μ: 0.35, : 0.0123) = 0.896
𝑃 0.33 < 𝑝 < 0.37 = 0.896
Step 3: About 90% of all college student SRSs of size 1500
fall within 2% of the real p of students who applied to other
schools .
Planning for college
The superintendent of a large school district wants to know
what proportion of middle school students in her district
are planning to attend a four-year college or university.
Suppose that 80% of all middle school students in her
district are planning to attend 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?
Solution:
Step 1: State the distribution and the values of interest. We want to find the
probability that the percentage of middle school students who plan to attend a fouryear college or university falls between 73% and 87%. Converting to proportions, we
want to find, P(0.73 ≤ 𝑝 ≤ 0.87). The mean of the sampling distribution of 𝑝 is
𝜇𝑝 = 𝑝= 0.80.
Step 2: Check Conditions
Because the school district is large, we can assume that there are more than 10(125) =
1250 middle school students. We can consider the distribution of 𝑝 to be
approximately Normal because np = 125(0.80) = 100 > 10 and n(1 – p) = 125(0.20) =
25 > 10
0.80(0.20)
 0.036


Step 3: Answer the question.
p̂
125
The standardized scores for the boundary values are
z
0.73  0.80
 1.94
0.036
z
0.87  0.80
 1.94
0.036
P(−1.94 ≤ 𝑧≤ 1.94)= 0.9476
Using technology: The command normalcdf(lower: 0.73, upper: 0.87, μ: 0.80, :
0.036) gives and area of 0.9482.
Answer in Context!!! 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 are
planning to attend a four-year college or university.
The Gallup Poll asked a random sample of 1785 adults whether they
attended church during the past week. Let 𝑝 be the proportion of
people in the sample who attended church. A newspaper report
claims that 40% of all U.S. adults went to church last week. Suppose
this claim is true.
a. Describe the sampling distribution of 𝑝.
b. Of the poll respondents, 44% said they did attend church last
week. Find the probability of obtaining a sample of 1785 adults in
which 44% or more say they attended church last week if the
newspaper report’s claim is true. Does this poll give convincing
evidence against the claim? Explain.
c. What sample size would be required to reduce the standard
deviation of the sampling distribution to one-third the value you
found? Justify your answer.
Have you ever noticed that some goldfish crackers
have smiles and some don’t? We are going to
estimate the proportion of all goldfish crackers that
have faces.
Select a Dixie cup of cheddar Goldfish crackers.
Don’t eat any yet!