Download Chapter 8, part A

Chapter 8
Interval Estimation
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I. Introduction
The previous chapter found a point estimate x of the
population parameter . But we know that for
every sample, we will never have a point estimate
that is exactly equal to the true population
parameter.
Thus this chapter shows us how interval estimates of
 and p can be developed to provide information
about the precision of that estimate.
II. Interval Estimation of 
In a large sample case, n>=30.
We will show how a sampling distribution of x can
be used to develop an interval estimate of .
A. An example
CJW is a mail-order sporting equipment firm that
conducts a monthly customer service survey. Their
scale is 0-100 where 100 is “Excellent” service.
They find that the population =20, x changes every
month and  is unknown.
Most recent survey: x =82, =20, n=100
B. Sampling Error
Question: How good is the sample estimate of the
population parameter?
Sampling Error = x  
We don’t precisely know the sampling error because
we don’t know  . But we can draw some
probability conclusions about the size of the
sampling error.
How?
The Central Limit Theorem allows us to conclude
that the sampling distribution of x can be
approximated by a normal probability distribution.
n=100, =20
20
x 
2
100

x
C. Probability Statements
From the standard normal probability table (z-table)
we know that 95% of all values are within z=1.96* x
of the mean.
Precision statement: “There is
95% of all sample x-bar
values fall here.
a 95% probability that x
will provide a sampling
error of 3.92 or less.”
-3.92

+3.92
x
D. General Lingo
• : probability that the sampling error is larger
than the sampling error in the precision statement.
• /2: probability in each tail of the distribution
• (1-): probability that a sample mean will provide
a sampling error less than or equal to the sampling
error in the precision statement.
• Z/2 is the value of the standard normal random
variable corresponding to an area of /2 in the
upper tail.
A picture should help...
In our example
(1-) =.95,
=.05, /2=.025.
(1- ) of all xbar values.
/2
/2

x
Precision statement: There is a (1-) probability that the value of a sample mean will
provide a sampling error of z/2*  x or less.
E. Interval Estimate: large sample,
 known.
x  z /2

n
Confidence
Level
90%
95%
99%

/2
z/2
.10
.05
.01
.05
.025
.005
1.645
1.96
2.576
z/2 is the z-value providing an area of /2 in the
upper tail of the standard normal probability
distribution.
F. Interval Estimate: large sample,
 unknown.
x  z /2
s
n
If the population standard deviation
is unknown, use the sample value.
Example: Suppose a 50-point exam is given to 36
statistics students. The mean is 39.5 and the standard
deviation is 7.77.
Construct a 95% confidence interval.
Solution
With 95% confidence, we find a z.025=1.96. Using
the rest of our information, we construct:
39.5 ± 1.96(7.77/6)
What does this interval tell
us?
or: 39.5 ± 2.54
36.96
39.5
42.04