AP Statistics
Ch. 18 Central Limit Theorem
CENTRAL LIMIT THEOREM
Two assumptions:
1.
The sample values must be independent
2.
The sample size, n, must be large enough

The mean of a random sample has a sampling distribution whose shape can be approximated by a Normal
model

The larger the sample, the better the approximation will be

This is regardless of the shape of the distribution of the population being sampled from or the shape of
the distribution of the sample
DISTRIBUTION OF SAMPLE PROPORTIONS
 Population has fixed proportion (p)

To find the population proportion a sample is taken and a sample proportion is calculated  p̂ 
If samples are repeatedly taken with the same sample size

The mean of the sampling distribution would be the population proportion

The standard deviation would be  p̂ 
p̂  p
pq
n
Conditions to check for the assumptions:
1. Randomization: The sample was obtained through random sample techniques or we can at least assume
that the sample was representative.
2. Success/Failure: The expected number of successes/failure is both greater than or equal to 10
np  10
nq  10
3. 10% Condition: Each sample is less than 10 percent of the population
All conditions have been met to use the Normal model for the distribution or sample proportions.
• If samples were repeatedly taken with the same sample size then from the CLT, the distribution would be
approximately Normal

pq 
pˆ ~ N  p,

n 

Example Skittles:
• According to the manufacturer of the candy Skittles, 20% of the candy produced is the color red. What is
the probability that given a large bag of skittles with 58 candies that we get at least 17 red?
n = 58
Conditions:
1. SRS
p = .20
q = .80
pˆ 
17
58
1. Assumed representative
(58)(.20)  10
(58)(.80)  10
2.
np & nq  10
2.
3.
10% condition
3. There are more than 580 skittles produced, so we are sampling less than 10%
pop  10n
All conditions met to use the Normal model

Model:   p  .20
N (.20,.0525)

Answer question:
pˆ 
17
 .293
58
P( pˆ  .293)  .0383
Practice: p. 433 #12, 16

pq
(.20)(.80)

 .0525
n
58
AP STATISTICS
Ch. 19: CONFIDENCE INTERVALS: For single sample proportions

However, most of the time we don’t know… the population proportion (p)

We take samples and calculate _ p̂ _____ in order to try and find __p______

Since we don’t know p, we can’t find … the standard deviation

__ p̂ _____ is the estimate for _p______

So… we can estimate the std. deviation with _standard error_____________________________:
o
SE =
ˆˆ
pq
n
CONFIDENCE INTERVALS… The basics:
 Based on … p̂ and the sampling distribution of p̂

Start with _ p̂ _____ and give ourselves _a margin of error on either side__________

Size of the interval (of the MOE) is based on _sample size_______ and _level of confidence______
o
Larger sample size = smaller interval (more accurate)
o
Larger confidence level = larger (wider) interval
(need more room for error)

Basic Setup:
Estimate  MOE

Specifically, for 1-proportion sample:

pˆ  ( Z * )( SE pˆ )  pˆ  ( Z * ) 


ˆˆ 
pq
 (a, b)
n 
Z* = critical value
_______ % of data is between  Z (confidence level)


3 main confidence levels:
Level of Confidence
90%
95%
99%
𝑍 ∗ value
1.645
1.960
2.576
Other confidence levels… how do we find Z*?
EX: 92%
Invnorm(.04, 0 ,1) = 1.751

SENTENCE INTERPRETATION:
We are _______% confident that the true % of ______________ is between ___a____ and ____b____ %
*must write this sentence every time
CONDITIONS: (for 1-Prop Z-Interval)
1. Simple Random Sample (SRS)
2.
npˆ & nqˆ  10
*must use p̂ because we don’t know p
3. 10 % condition
pop  10n

If all conditions are met, then we can say that p̂ has the model: N  pˆ ,


And we can use the Normal Model for the 1-Prop Z-interval
ˆˆ 
pq

n 
Example: We want to know the real improvement rate for a new medication. We conduct an experiment and find
that out of 53 subjects, 27% of them report improvement with the new medications. Create a 95% confidence
interval (and interpret).
n = 53
pˆ  .27
qˆ  .73
C=95%
Conditions:
1. Simple Random Sample (SRS)
2.
npˆ & nqˆ  10
3. 10 % condition
pop  10n
1. Assumed representative
2.
(53)(.27)  10
(53)(.73)  10
3. There are more than 530 people taking medication so we are
sampling less than 10% of the population.
Conditions met N(.27, .061)1-Prop. Z-Int
 (.27)(.73) 
.27  (1.960) 

53


.27  .11956
(.15044,.38956)
We are 95% confident that the true proportion of people seeing improvement with the new medication is
between 15.044% and 38.956%.
Example: We take a simple random sample of 95 Bucks county residents and find that only 20 of them approve of
a new property tax to pay for repairs to local roads. Estimate with 99% confidence the true percent of people
who approve of the tax.
n = 95
pˆ 
20
95
qˆ 
75
95
C=99%
Conditions:
1. Simple Random Sample (SRS)
2.
npˆ & nqˆ  10
3. 10 % condition
pop  10n
1. Stated SRS
 20 
(95)    10
 95 
2.
 75 
(95)    10
 95 
3. There are more than 900 residents in Bucks County so we are
sampling less than 10% of the population.
Conditions met N(.211, .0419)1-Prop. Z-Int
 (.211)(.789) 
.211  (2.576) 

95


(.1031,.3189)
We are 99% confident that the true proportion of people who approve of a new property tax to pay for
repairs to local roads is between 10.31% and 31.89%.
What does ____% confidence mean?
Sentence:
In repeated samples of size ___n____, the confidence interval created will catch the true
__parameter (p)___ __CI____% of the time.
For the previous example:
In repeated samples of size 95, the confidence interval created will catch the true proportion of
those who approve a new property tax to pay for repairs to local roads 99% of the time.
MORE ABOUT MARGIN OF ERROR:
Things that affect MOE (margin of error):
MOE
*sample size

pˆ  ( Z * ) 

ˆˆ
pq
n

  ( a, b)

* confidence level
If we want a particular MOE, we can set a level of confidence and a sample size in order to attain that MOE.
Example: Let's go back to our example about the improvement rate with a new medication. We found 27%
improvement. How many subjects would we need in a new experiment to make a 98% conf. interval while still
keeping a 5% MOE?
MOE  ( Z * )
ˆˆ
pq
n
 (.27)(.73) 
.05  (2.326) 

n


.1971
n
n  426.392
n  427 people
.0215 
*always round up!
NOTE: If you are not given a value of p̂ , you can use __0.50_________
-
It doesn’t favor either outcome (success or failure)
it will give you the largest sample size
Example: What sample size must be used to estimate the outcome of a political election with a margin of error of
3% and 99% confidence?
 (.50)(.50) 
.03  (2.576) 

n


.25
n
n  1843.271
n  1844 people
.0116 
Example: What sample size must be used to estimate the true percent of left-handed people in the nation with
90% confidence and a margin of error of 8%? Assume that it has been shown in previous research that the
percent of left-handed people was 38%.
 (.38)(.62) 
.08  (1.645) 

n


.2356
n
n  99.616
n  100 people
.0486 