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AMS7: WEEK 4. CLASS 3
Sampling distributions and estimators.
Central Limit Theorem
Normal Approximation to the Binomial Distribution
Friday April 24th, 2015
Sampling distributions and estimators
• REMEMBER: We want to learn about the population from
the sample.
∑
• BUT: Lets consider the sample mean =
• n is the sample size.
• If we take two different samples from the same
population, we will obtain most probably two different
values
REMEMBER
• The sample mean is a statistic
• If we take many samples of size n, we have different
values of the sample mean Variability
Sampling
• We can talk about the probability distribution of the
Sample Mean
• IN GENERAL: We can talk about the probability
distribution of a statistic
Some definitions
• Sampling distributions of the mean: Is the probability
distribution of the sample means, for all samples of size n
• Sampling variability: This is the variability of a statistics
when its value changes from sample to sample.
Example
• If I take a random sample from this class to estimate the
mean heights of the students (), the value of the sample
mean for one sample of size n, will vary from the value
obtained from another sample of size n.
• REMEMBER:
is a Parameter
from the Population
is a Statistic
from the Sample
IMPORTANT RESULT
• WHEN THE SAMPLE SIZE INCREASES, THE SAMPLE
DISTRIBUTION OF THIS SAMPLE MEANS TENDS TO
BECOME A NORMAL DISTRIBUTION
• Other Statistics:
1. Sample Mean
2. Sample Median
3. Sample Variance
4. Sample Standard Deviation
5. Sample Proportions
Sampling distributions of proportions
• Is the probability distribution of sample proportions, with
all samples having the same sample size n
• Example: Use a random sample from the population, for
example, 1,000 people, to estimate the proportion who
will vote for candidate A in the following elections (using
Polls)
• IMPORTANT RESULT: Under certain conditions, the
distribution of sample proportions approximates a
normal distribution
The Central Limit Theorem
1) Take a sample from a population
• : Population Mean
• 2: Population Variance
• Random Variable X: Person’s height
• X may or may not be normally distributed
• X has a Mean and a Variance 2
2) For a sample of size n consider the sample mean
∑
=
Central Limit Theorem
• As the sample size increases the distribution of
approaches a normal distribution with mean and a
Variance
మ
(standard deviation is
)
• NOTATION:
1) Mean of the sample mean
= 2) Standard deviation of the sample mean
=
Example of application of the Central Limit
Theorem
• Assume that men’s weights are normally distributed with a
given mean = 172 lb. and a standard deviation given by =29 lb.
1) If one man is selected at random, find the probability that
his weight is less than 167 lb. Random variable X is
mean’s weight
a) Draw a picture:
167
172
Weight
Example (Cont.)
b) Calculate z Score
=
=
=
-0.17
c) Area= P(Weight < 167)=P(z<-0.17)=0.4325
2) If 36 men are randomly selected, find the probability that
they have a mean weight less than 167 lb.
Now we have a sample of 36 men
Example (Cont.)
a) Use Central Limit Theorem:
has a normal distribution with mean ௑ത = and a standard deviation ௑ത =
ఙ
௡
=
ଶଽ
ଷ଺
=
ଶଽ
଺
= 4.83
b) Find the z Score for :
‫=ݖ‬
௫ିఓ೉
ഥ
ఙ೉
ഥ
=
ଵ଺଻ିଵ଻ଶ
=
ସ.଼ଷ
3) Find the area below the z Score:
Area=P(Mean weight < 167)=P(z<-1.04)=0.1492
-1.04
Normal approximation to the binomial
• When working with a binomial distribution if np ≥ 5 and
nq ≥ 5 the binomial random variable has a probability
distribution that can be approximated by a normal
distribution, with mean and standard deviation:
ߤ = n.p
= . . We must verify that it is reasonable to approximate the
binomial distribution by the normal distribution
Normal approximation to the binomial
(cont.)
EXAMPLE: Suppose that X is a random variable with a binomial
distribution (Example: Number of left-handed in a Statistics
class)
1) With n=200, p=0.5
• q=1-p=0.5
• np= 200⨯0.5=100 (np ≥ 5)
• qn= 200⨯0.5=100 (nq ≥ 5)
2) Then we can calculate:
ߤ = n.p= 200⨯0.5=100
࣌ = ࢔. ࢖. ࢗ= ૛૙૙ × ૙. ૞ × ૙. ૞= 7.07
Normal approximation to the binomial
(cont.)
c) Find the probability that X is less than 120 (P(X<120).
We can use the normal approximation (it is easier!) but we
need to have a continuity correction because the binomial
distribution is discrete and the normal distribution is
continuous.
Number 120 will be represented by the interval (120-0.5,
120+0.5)= (119.5,120.5) in the normal curve.
Normal approximation to the binomial
(cont.)
100
(119.5 120.5)
120
Normal approximation to the binomial
(cont.)
• We want the z Score:
=
=
.
=
.
2.76
• P(z<2.76)=P(X<120)
• P(x≤119.5)= P(z≤ 2.76)=0.9971 (Using the continuity
correction)