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Day 51 Agenda:
DG22 --- 15 minutes
Pick up THQ #5
Advanced Placement
Statistics
Section 9.3: Sample Means
EQ: What does the Central Limit
Theorem tell us about sampling
distributions?
Form groups of 2 to complete the
• Use a cup to scoop out a sample of pennies
from the sampling frame.
• Randomly select 25 for your group then
return the extra pennies to container. KEEP
YOUR CUP.
• Follow the directions on your worksheet.
Creating dot plots to show shape of our sampling distributions: Go to the
board and place a dot at the age for each of your pennies. Use the correct
color marker to plot your five means for penny samples of size 1, size 5,
and size 10 and your one mean for sample size 25. After everyone has done
this, sketch the shape of each histogram below.
 CONCLUSION:
Our original population distribution was not described as
Normal nor was it bell-shaped. In fact it was
Skewed Right
__________________________.
Sample size
However, as we increased the __________________,
the
distribution got closer and closer to a
Bell-shaped
_________________
curve and could be approximated
using a Normal
___________.
Approximation
This property is called the _________.
Central Limit
Theorem
Sample Means --- averages of observations
 Sample Means are less variable than
single observations.
 Sample Means have a more normal
distribution than single observations.
A visual comparison of the distribution of sample means
as the sample size increases.
RECALL: Sampling Proportions
What about Sampling Means?
x
μ
Conclusion about Sampling Distributions:


n
True no matter what the shape of
the population distribution.
Central Limit Theorem ---SRS of size n
taken from population with mean µ and
standard deviation σ:
Sample Size LARGE ENOUGH
Law of Large Numbers --- draw observations
at random from any population with finite
mean µ:
Central Limit Theorem  Law of Large Numbers
Sample Size Large Enough to
use Normal Approximation
Sample Mean approaches
True Population Mean as
observations increase.
SPARK NOTES FOR THIS SECTION:
SPARK NOTES FOR THIS SECTION:
SPARK NOTES FOR THIS SECTION:
In Class Assignment: We will
do Worksheet: Sample Means
together. Finish it for HW.
Recall: State, Plan, Do
Worksheet: Sample Means
Follow the template we used in class for sampling
distributions for proportions.
1.
A survey found that the American family generates an
average of 17.2 pounds of glass garbage each year. Assume the
distribution is normal with a standard deviation of 2.5 pounds.
STATE:
a. What is the probability that a randomly selected American
family will generate more than 18 pounds of garbage?
PLAN:
parameter of interest
μ = the true mean pounds of glass garbage produced by
American families annually
1.
A survey found that the American family generates an
average of 17.2 pounds of glass garbage each year. Assume the
distribution is normal with a standard deviation of 2.5 pounds.
a. What is the probability that a randomly selected family will
generate more than 18 pounds of garbage?
PLAN:
randomness
The problem states a family will be selected randomly.
independence
Standard Error not needed. Only selecting a single
family.
1.
A survey found that the American family generates an
average of 17.2 pounds of glass garbage each year. Assume the
distribution is normal with a standard deviation of 2.5 pounds.
a. What is the probability that a randomly selected family will
generate more than 18 pounds of garbage?
PLAN:
large counts
Problem states that distribution of garbage consumption
for American families is Normally distributed.
1.
A survey found that the American family generates an
average of 17.2 pounds of glass garbage each year. Assume the
distribution is normal with a standard deviation of 2.5 pounds.
a. What is the probability that a randomly selected family will
generate more than 18 pounds of garbage?
Do:
μ = 17.2
σ = 2.5
n=1
18  17.2
P( X  18)  P( z 
)  P( z  0.32)  0.3745  37.45%
2.5
The probability that a randomly selected American family will
generate more than 18 pounds of garbage each year is 37.45%.
μ = 17.2
σ = 2.5
n=1
18  17.2
P( x  18)  P( z 
)  P( z  0.32)  0.3745  37.42%
2.5
The probability that a randomly selected family will generate
more than 18 pounds of garbage each year is 37.42%
b. What is the probability that the mean sample of 55
families selected randomly will be between 17 and 18
pounds?
PLAN:
randomness
The problem states the families will be selected randomly.
independence
all families > 10(55)
all families > 550
Condition met for Independence
PLAN:
n = 55
large counts
55 > 30
CLT states that sample
size is large enough to
use Normal Approx.
Do:
n  55
  17.2
2.5
SE 
 0.3371
55
17  17.2
18  17.2
P(17  x  18)  P(
z
)
0.3371
0.3371
P(0.593  z  2.37)  0.715  71.5%
The probability that the mean weight of yearly garbage of 55
randomly selected American families will be between 17 and 18
pounds is 71.5%.
c. If the distribution of glass garbage produced by
the population were not normal, describe the
distributions for a) and b).
a) The sample size is only n = 1, therefore this distribution
will not be normal.
b) The sample size is n = 55. The CLT says that sample sizes
of at least 30 will create a distribution that is
approximately normal, even if the population distribution
was not normal.
2.
The average yearly cost per household of owning a
dog is $186.80. Assume the standard deviation of the
distribution is $32. Suppose we randomly select 50
households that own a dog. What is the probability that the
sample mean for these 50 households is less than $175?
What is the probability that the sample mean for these
50 households is less than $175?
PLAN:
parameter of interest
μ = the true mean annual cost per household for owning a dog
randomness
The problem states the families will be selected randomly.
2.
The average yearly cost per household of owning a
dog is $186.80. Assume the standard deviation of the
distribution is $32. Suppose we randomly select 50
households that own a dog. What is the probability that the
sample mean for these 50 households is less than $175?
PLAN:
independence
All households > 10(50)
All households > 500
Condition met for Independence
large counts
n = 50
50 > 30 CLT says normal
distribution appropriate.
2.
The average yearly cost per household of owning a
dog is $186.80. Assume the standard deviation of the
distribution is $32. Suppose we randomly select 50
households that own a dog. What is the probability that the
sample mean for these 50 households is less than $175?
Do:
n = 50
  186.8
32
SE 
 4.53
50
175  186.8
P( x  175)  P( z 
)  P( z  2.61)  0.005  0.5%
4.53
The probability that a random sample of 50 households will have
an average yearly cost for owning a dog less than $175 is 0.5%.
3. The average teacher’s salary in New Jersey (ranked
first among states) is $52,174. Assume the distribution
is normal with a standard deviation of $700.
a. What is the probability that a randomly selected
teacher makes less than $50,000 a year?
PLAN:
parameter of interest
μ = the true mean salary of teachers in New Jersey
randomness
The problem states a teacher will be selected randomly.
3. The average teacher’s salary in New Jersey (ranked
first among states) is $52,174. Assume the distribution
is normal with a standard deviation of $700.
PLAN:
independence
Standard Error not needed. Only selecting a single
teacher.
large counts
Problem states that distribution of teacher’s salaries in NJ
is Normally distributed.
3. The average teacher’s salary in New Jersey (ranked
first among states) is $52,174. Assume the distribution
is normal with a standard deviation of $700.
Do:
μ = 52,175
σ = 700
n=1
50,000  52,174
P( X  50,000)  P( z 
)
700
P( z  3.106)  0.0009  0.09%
The probability that a randomly
selected teacher in NJ makes
less than $50,000 is 0.09%.
b. If we randomly sample 100 teachers’ salaries, what is
the probability that the sample mean is less than $50,000?
PLAN:
parameter of interest
μ = the true mean salary of teachers in New Jersey
randomness
The problem states a random sample of 100 teachers’
salaries will be obtained
independence
All teachers in NJ > 10(100)
All teachers in NJ > 1000
Condition met for Independence
b. If we randomly sample 100 teachers’ salaries, what
is the probability that the sample mean is less than
$50,000?
large counts
PLAN:
n = 100
100 > 30 CLT says normal
distribution appropriate.
Do we need to reference the CLT?
Do:
n = 100
  52,174
700
SE 
 70
100
b. If we randomly sample 100 teachers’ salaries, what
is the probability that the sample mean is less than
$50,000?
Do:
n = 100
  52,174
700
SE 
 70
100
50,000  52,174
P( x  50,000)  P( z 
)  P( z  31)  0.000  0%
70
The probability that a random sample of 100
teachers in NJ have a mean salary less than $50,000
is 0%.
Assignment:
p. 595 - 596
p. 601 - 602
#31 – 34
#35 – 40