Section 7.2 – Sample Proportions
Sample Proportions:
 Related to binomial
 Deals mostly with CATEGORICAL variables
 Can use normal approximation
Sampling Distribution of a Sample Proportion, p̂
Choose an SRS of size n from a population of size N with proportion p of successes. Let p̂ be the sample
proportion of successes. Then:
 The mean of the sampling distribution of  p̂  p .

The standard deviation of the sampling distribution of p̂ is  pˆ 
p(1  p)
as long as the 10%
n
condition is satisfied: n  1 N .
10

As n increases, the sampling distribution of p̂ becomes approximately Normal. Before you perform
Normal calculations, check that the Large Counts condition is satisfied: np  10 and n(1  p)  10 .
p. 444
* p̂ is less variable in large samples because as n increases,  p̂ decreases.
Check Your Understanding, p. 445 About 75% of young adult Internet users (ages 18 to 29) watch online
videos. Suppose that a sample survey contacts an SRS of 1000 young adult Internet users and calculates the
proportion p̂ in this sample who watch online videos.
1. What is the mean of the sampling distribution of p̂ ?
𝜇𝑝̂ = 𝑝 = 0.75
2. Find the standard deviation of the sampling distribution of p̂ . Check that the 10% condition is met.
1
10% condition: 1000 ≤ 10 · Young adult Internet users; this condition is met, so
𝑝(1−𝑝)
𝜎𝑝̂ = √
𝑛
0.75(0.25)
=√
1000
≈ 0.0137
3. Is the sampling distribution of p̂ approximately Normal? Check that the Large Counts condition is met.
np = (1000)(0.75) = 750 ≥ 10
and
n (1 – p) = (1000)(0.25) = 250 ≥ 10
Yes, the sampling distribution of 𝑝̂ is approximately Normal.
4. If the sample size were 9000 rather than 1000, how would this change the sampling distribution of p̂ ?
The sampling distribution would still be approximately Normal with mean 0.75, but the standard
deviation would be smaller by a factor of 3: 𝜎𝑝̂ = √
𝑝(1−𝑝)
𝑛
0.75(0.25)
=√
9000
≈ 0.0046.
̂
Using Normal Approximation for 𝒑
Example, p. 445
Going to College
A polling organization asks an SRS of 1500 first-year college students how far away their home is. Suppose
that 35% of all first-year students attend college within 50 miles of home. Find the probability that the random
sample of 1500 students will give a result within 2 percentage points of this true value. Show your work.
We want to find the probability that 𝑝̂ falls between 0.33 and 0.37 (within 2 percentage points, or 0.02, of
0.35). So P (0.33 ≤ 𝑝̂ ≤ 0.37). We have an SRS of size n = 1500 drawn from a population in which the
proportion p = 0.35 attend college within 50 miles of home.
𝜇𝑝̂ = 𝑝 = 0.35
1
Check 10% condition: 1500 ≤ 10 · All first-year college freshmen, safe to assume the condition is met.
𝑝(1−𝑝)
𝜎𝑝̂ = √
𝑛
0.35(0.65)
=√
1500
≈ 0.0123
Check Large Numbers condition: (1500)(0.35) = 525 ≥ 10 and (1500)(0.65) = 975 ≥ 10, so the
sampling distribution of 𝑝̂ is approximately Normal.
So…
P (0.33 ≤ 𝑝̂ ≤ 0.37)
0.33−0.35
0.37−0.35
= 𝑃 ( 0.0123 ≤ 𝑧 ≤ 0.0123 )
= 𝑃(−1.63 ≤ 𝑧 ≤ 1.63)
= 0.9484 – 0.0516 = 0.8968
About 90% of all SRSs of size 1500 will give
results within 2 percentage points of the truth
about the population.
HW p. 448 # 33, 34, 35, 43, 44, 45
Due: Monday