Section 9.2 – Sample Proportions
Sample Proportions:
 Related to binomial
 Deals mostly with CATEGORICAL variables
 Can use normal approximation
 p̂  p
 pˆ 
Depend on p, population
proportion and n, sample size
p(1  p)
n
Both are on the AP formula chart
Won’t work if sample is large part of
the population (Ex. 50 out of 100)
* p̂ is less variable in large samples
How can we tell when it will work?
 1st Rule of Thumb: use only when the population is at least 10 times as large as the sample
(population ≥ 10∙n)

2nd Rule of Thumb: (When to use normal approximation) When
np ≥ 10
and
n (1 – p) ≥ 10
Exercise 9.19, p. 511
Do you drink the cereal milk?
p  0.7
pˆ  0.67
n  1012
0.7(1  0.7)
 0.0144
1012
(b) We can use the formula for standard deviation because it follows our first rule of thumb:
Population (U.S. adults) ≥ 10∙(1012) = 10,120. There are more than 10,120 U.S. adults.
(a)  pˆ  0.7
 pˆ 
(c) Normal Approximation? (1012)(0.7) = 708.4 ≥ 10 ☺,
Yes, we can use normal approximation.
(1012)(1 – 0.7) = 303.6 ≥ 10
☺
 pˆ  0.7 0.67  0.7 
(d) P( pˆ  0.67 ) = P 

 = P(z ≤ -2.08) = 0.0188
0.0144 
 0.0144
Standardizing because we’re using normal approximation
Proportion of 1012 that
drink the cereal milk
(e)
1
2
 pˆ   pˆ 
new
 Use Normal Distribution Table
1 0.7(0.3)
1 0.7(0.3)
0.7(0.3)



2 1012
4 1012
4048
1
1

2
4
New n
So a sample size of 4048 would reduce the standard deviation to one-half the original standard deviation.
(f) Probably higher because more teenagers probably drink the cereal milk. Some adults may find it to be
childish or rude.
Example 9.7, p. 507
SRS of 1500,
35%  p = 0.35
-
Within 2% points 
35 + 2 = 37
and
35 – 2 = 33
Looking for: P(0.33 ≤ p̂ ≤ 0.37)
Rule of Thumb 1: 10(1500) = 15,000 There are more first-year college students than this, so we can use the
standard deviation formula.
 pˆ  0.35
 pˆ 
0.35(1  0.35)
 0.0123
1500
Rule of Thumb 2: 1500(0.35) = 525 ≥ 10
and
1500(1 – 0.35) = 975 ≥ 10
Since both are true, we can use Normal Approximation for this problem.
 0.33  0.35 pˆ  0.35 0.37  0.35 


P(0.33 ≤ p̂ ≤ 0.37) = P 
 = P(-1.63 ≤ z ≤ 1.63) = 0.9484 – 0.0516 = 0.8968
0.0123
0.0123 
 0.0123
About 89.7% of all samples give a result within 2 percentage points of the true population proportion.