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LECTURE 11
Interval Estimation
This lecture covers material on interval estimation of a population mean and
proportion  large sample and small sample cases, confidence intervals, t
distribution, and determination of the sample size.
Microsoft Excel is used for some
applications.
Read:
Chapter 8, Sections 8.1 through 8.4.
1
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View Full Document This lecture builds further on the issue of statistical inference.
The manager knows
that point estimates of population parameters that are based on sample statistics
derived from a single sample cannot be expected to be equal to the population
parameter values. Therefore it is important to know
how precise are these estimates.
That is where interval estimation plays a role.
The objective is to say with some
degree of confidence that the estimate of a population parameter (e.g., mean,
proportion etc.,) lies within an interval.
Interval Estimation of Population Means
In statistical inference, we need to make estimates of population parameters.
So we
take a random sample and use its sample statistics to make these point estimates.
From our discussion in Lecture 9:
The sample mean
is a point estimate of the population mean
μ
.
may not be exactly equal to
μ
.
The interval estimate for the population mean is obtained by:
±
Margin of Error
Large Sample Case  Population Standard Deviation known:
In this case sample size is large or
n
≥
30.
Population Mean
=
μ
is not known.
Population standard deviation
=
σ
is known.
Sample mean
=
Sample standard deviation
=
s
.
The sampling distribution of sample means can be used to make probability
statements about the margin of error.
Calculating Probability Estimates of Margin of Error:
By the central limit theorem, the sampling distribution of sample means will be
normally distributed with a mean
μ
and standard deviation
σ
x
‾
=
σ
/
√
n.
Suppose we want to know that given a sample mean of , the probability that the
margin of error is 6.
2
Note:
This means that the maximum difference between and
μ
or (

μ
) =
± 6.
If
is >
μ
, then (

μ
)
≤
+ 6;
If
is <
μ
, then (

μ
)
≥
 6.
We want to find P(


μ

≤
6).
P(


μ

≤
6)
=
P(0 <
≤
6)
+
P(0 >
≥
6).
Converting to the standard normal table for the positive value of :
z
=
( 
μ
)/
σ
x
‾
=
6/
σ
x
‾
.
(Note:
Therefore ‘6’ =
z
*
σ
x
‾
).
We need to find P(0 <
z
≤
6/
σ
x
‾
).
As the normal curve is symmetrical, this is also the same as P(0>
z
≥
6/
σ
x
‾
)
Therefore
P(


μ

≤
6)
=
P(0 <
z
≤
6/
σ
x
‾
)
+
P(0 >
z
≥
6/
σ
x
‾
).
3
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View Full Document EXAMPLE
1
We take a sample of size 64 from a population having a known standard deviation of
16.
In this case sample size
n
= 64 is large or
n
≥
30.
Population Mean
=
μ
is unknown.
Population standard deviation
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This note was uploaded on 09/21/2011 for the course OM 210 taught by Professor Singer during the Summer '08 term at George Mason.
 Summer '08
 SINGER

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