Sample Proportions:

Related to binomial

Deals mostly with CATEGORICAL variables

Can use normal approximation
  p
  p (1
 p )
Depend on p, population proportion and n, sample size n
Both are on the AP formula chart
*
ˆ
is less variable in large samples
Won’t work if sample is large part of the population (Ex. 50 out of 100)
How can we tell when it will work?

1 st
Rule of Thumb: use only when the population is at least 10 times as large as the sample
(population ≥ 10∙ n )

2 nd
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?
ˆ 
0.67
p

0.7
n

1012
(a)
 
0.7
 
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.
(c) Normal Approximation? (1012)(0.7) = 708.4 ≥ 10 ☺
, (1012)(1 – 0.7) = 303.6 ≥ 10
☺
Yes, we can use normal approximation.

(d) P(
ˆ 
0.67
) = P
0.0144
0.0144
Proportion of 1012 that drink the cereal milk
= P( z ≤ -2.08) = 0.0188

Use Normal Distribution Table
Standardizing because we’re using normal approximation
(e)
 new

1
2
 
1 0.7(0.3)
2 1012

1

4
0.7(0.3)

0.7(0.3)
1 012 404 8
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 ≤ ˆ
≤ 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.
 
0.35
 
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.
P(0.33 ≤ ˆ
≤ 0.37) = P
0.0123


0.35

0.37 0.35
0.0123
0.0123
= P(-1.63 ≤ z ≤ 1.63) = 0.9484 – 0.0516 = 0.8968
About 89.7% of all samples give a result within 2 percentage points of the true population proportion.