Lecture14

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Transcript Lecture14

Review
• Normal Distributions
– Draw a picture.
– Convert to standard normal (if necessary)
– Use the binomial tables to look up the value.
– In the case of a reverse look up we may have to
now solve for x.
Problems
Problems 5.36, 5.40, 5.48
Estimation
This is our introduction to the field of
inferential statistics.
We already know why we want to study
samples instead of entire populations, (e.g.
limited resources, destructive sampling etc.).
By studying the sample and its statistics,
can we make inferences about the
population and its parameters.
Estimator and Point Estimate
An estimator is a “sample statistic” (such as
the sample mean, or sample standard
deviation) used to approximate a population
parameter.
A Point Estimate is a single value or point
used to approximate a population
parameter.
“The sample
mean x is the
best point
estimate of the
population mean
m.”
Age of STFX Students
• There are approximately 4000 students at STFX. Our goal is to
determine the mean age of all STFX students, (the parameter
to be estimated is mean age, m ).
• Can we use a statistic of the sample to estimate a parameter of
a population?
• We take a sample of 50 STFX students and calculate the mean
age of the sample, we find that the sample mean is 21.4 years.
• Therefore the best point estimate of the population mean m of
the ages of all STFX students is the sample mean = 21.4
• We see that 21.4 is the best point estimate of the age of all
STFX students, but we have no indication of just how good that
estimate really is.
The Concept of Sampling
Distributions
Parameter – numerical measure of a
population. It is usually unknown
Sample Statistic - numerical descriptive
measure of a sample. It is usually known.
Taking all the possible sample statistics, we
get a sample distribution.
Sampling Distribution of Sample
Means
The sampling distribution of Sample
Means is the distribution of the sample
means obtained when we repeatedly draw
samples of the same size from the same
population. For sufficiently large samples,
this distribution is almost always Bell
Shaped, that is, it is almost always Normal.
Central Limit Theorem
Take ANY random variable X and compute m and s
for this variable. If samples of size n are randomly
selected from the population, then:
1) For large n, the distribution of the sample means, x
will be approximately a normal distribution,
2) The mean of the sample means will be the
population mean m and
3) The standard deviation of the sample means will
be
s
n
Notation
Let samples of size n be selected from a
population with mean m and standard deviation
s,
The mean of the sample means is denoted as
mx
Therefore the CLT says that
mx  m
Notation
Let samples of size n be selected from a
population with mean m and standard deviation
s,
The standard deviation of the sample means is
denoted as
s
x
According to the CLT, we have that
for large populations.
sx s
n
Application of the CLT
Therefore, the distribution of the sample
mean x of a random sample drawn from
practically any population with mean m
and standard deviation s can be
approximated by a normal distribution
with mean m , and standard deviation
s
provided
the
population
is
large.
n
Finding the probability that a
sample mean is between
a and b
P ( a  x  b)  P ( a  m x  x  m x  b  m x )
P
(
a  mx

x  mx

b  mx
)
sx
sx
sx
am
x  mx b  m
 P(


)
s
s
sx
n
n
Example:
In human engineering and product design, it is often
important to consider the weights of peoples so that
airplanes or elevators are not overloaded, chairs don’t
break and other unpleasant happenings don’t occur.
Assume that the population of men and women has
normally distributed weights with a mean of 173
pounds and a standard deviation of 30 pounds.
– A) If 1 person is randomly selected, find the probability that
their weight is greater than 180 pounds.
– B) If 36 different people are randomly selected from this
population find the probability that their mean weight is
greater than 180 pounds.
Solutions: A) .4090
B) .0808
Example
To avoid false advertisement suits, a beverage
bottler must make reasonably certain that 500 ml
bottles actually contain 500 ml. To infer whether a
bottling machine is working satisfactorily, the
bottler randomly samples 10 bottles per hour and
measures the amount of beverage in each bottle.
The mean of the 10 measurements is used to
determine whether to readjust the amount of
beverage delivered to each bottle. Records show
that the amount of fill per bottle is normally
distributed with a standard deviation of 1.5 ml and
a mean fill of 501 ml. What is the probability that
the sample mean of 10 test bottles is less than 500
ml. Solution : P( x  500)  0.0174
Practice Problems
• #6.34 page 310
• #6.41 page 311
Overview
• Central Limit Theorem.
Homework
• Review Chapter 6.1-6.3
• Read Chapters 7.1-7.3
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