Download Central limit Theorem

Central Limit Theorem
Two assumptions
1. The sampled 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.
Central limit Theorem
Sample Distribution Models for
Means and Proportions
Distribution of sample proportions
• Population has a fixed proportion ( p )
• To find population proportion a sample is taken
$)
and a sample proportion is calculated ( p
If samples are repeatedly taken with the same
sample size
• The mean of the sample distribution would be
the population proportion, μ p̂ = p
• The standard deviation would be σ p̂ =
pq
n
Conditions to check for the assumptions
1. Success/Failure: The expected number of
successes and failure is both greater than 10
np ≥ 10 and nq ≥ 10
2. 10% Condition: Each sample is less than 10%
of the population
3. Randomization: The sample was obtained
through random sample techniques or we can
at least assume that the sample is
representative.
All conditions have been met to use the Normal
model for the distribution of sample proportions.
Example: Skittles
• 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 ⎠
⎝
Conditions:
1. 10% condition: 58 skittles is less than 10% of
all skittles produced.
2. Success/Failure:
np = 58 × 0.20 = 11.6 ≥ 10
nq = 58 × 0.80 = 46.4 ≥ 10
There are at least 10 successes and failures
3. Randomization: Though not from a random
sample we can assume the bag is
representative of the population.
All conditions have been met to use the Normal
model for the distribution of sample proportions.
• 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?
• Mean: μ = p = 0.20
• Standard Deviation:
pq
0.20 × 0.80
=
= 0.0525
σ=
n
58
• So the model for p̂ becomes N(0.20,0.0525)
• Sample proportion: pˆ =
17
= 0.293
58
• Then to find the probability that we get a sample
proportion of 0.293 or higher:
P ( pˆ ≥ 0.293) = 0.0383
p. 433 #12,16
12. Smoking
p = 0.264 n = 50
1. Random: stated as a random sample
2. 10% condition: 50 adults is less than 10% of all adults
3. Success/Failure: np = 50(.264) = 13.2 ≥ 10
nq = 50(.736) = 36.8 ≥ 10
There are at least 10 successes and failures.
All conditions have been met to use the Normal model for
the distribution of sample proportions.
⎛
(0.264)(0.736) ⎞
pˆ ~ N ⎜ 0.264,
⎟⎟ = N ( 0.264,0.0623 )
⎜
50
⎝
⎠
16. Contacts
p = 0.30 n = 100
1. Random: stated as a random sample
2. 10% condition: 100 college students are less than 10%
of all students at a university
3. Success/Failure: np = 100(0.3) = 30 ≥ 10
nq = 100(0.7) = 70 ≥ 10
There are at least 10 successes and failures.
All conditions have been met to use the Normal model for
the distribution of sample proportions.
⎛
(0.3)(0.7) ⎞
pˆ ~ N ⎜ 0.3,
⎟⎟ = N ( 0.3,0.0458 )
⎜
100
⎝
⎠
P ( pˆ > 1/ 3) = 0.2334