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Chapter
6
Confidence Intervals
Elementary Statistics
Larson Farber
Inferential Statistics
Inferential Statistics-the branch of statistics that uses
sample statistics to make inferences about population
parameters.
Applications of Inferential Statistics
 Estimating Parameters
 Hypothesis Testing
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Point Estimate
DEFINITION:
A point estimate is a single
value estimate for a population
parameter. The best point
estimate of the population mean
 is the sample mean
x.
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Example: Point Estimate

A random sample of airfare prices (in dollars)
for a one-way ticket from Atlanta to Chicago. Find a point
estimate for the population mean, .
99 102 105 105 104 95 100 114 108 103 94 105 101
109 103 98 96 98 104 87 101 106 103 90 107 98
101 107 105 94 111 104 87 117 101
The sample mean is
x 3562
x

 101.77
n
35
The point estimate for the price of all one way
tickets from Atlanta to Chicago is $101.77.
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Interval Estimates
Point estimate
•
101.77
An interval estimate is an interval, or range of values
used to estimate a population parameter
(
•
101.77
)
The level of confidence, c is the probability that the
interval estimate contains the population parameter
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Distribution of Sample Means
When the sample size is at least 30, the sampling
distribution for x is normal
Sampling distribution x
For c = 0.95
0.95
0.025
-1.96
0.025
0 1.96
z
95% of all sample means will have standard
scores between z = -1.96 and z = 1. 96
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Maximum Error of Estimate
DEFINITION: Given a level of confidence, c, the maximum
error of estimate E is the greatest possible distance between
the point estimate and the value of the parameter it is

estimating.
Ez z
c
x
c
n
When n  30, the sample standard deviation, s can be used for .
 Find E, the maximum error of estimate for the one-way plane fare
from Atlanta to Chicago for a 95% level of confidence given s = 6.69
Using zc=1.96, s = 6.69 and n = 35,
E  zc

n
 1.96
6.69
35
 2.22
You are 95% confident that the maximum error of estimate is $2.227
Confidence Intervals for

Definition: A c-confidence interval for the population mean is
x  E 
  x E
 Find the 95% confidence interval for the one-way plane fare from
Atlanta to Chicago.
You found
x = 101.77 and E = 2.22
Right endpoint
Left endpoint
x  E  101.77  2.22  99.55
(
99.55
•
101.77
x  E  101.77  2.22  103.99
)
103.99
99.55    103.99
With 95% confidence, you can say the mean one-way fare from Atlanta
8
to Chicago is between $99.55 and $103.99
Sample Size
Given a c-confidence level and an maximum error of estimate,
E, the minimum sample size n, needed to estimate , the
2
population mean is
 z c 
n

 E 
 You want to estimate the mean one-way fare from Atlanta
to Chicago. How many fares must be included in your sample
if you want to be 95% confident that the sample mean is
within $2 of the population mean?
 z c   1.96  6.69 
n
 
  42.98
2

 E  
2
2
You should include at least 43 fares in your sample. Since9
you already have 35, you need 8 more.
The t-distribution
If the distribution of a random variable x is normal and
n < 30, then the sampling distribution of x is a tdistribution with n-1 degrees of freedom.
Sampling distribution x
n =13
d.f.=12
c=90%
.05
-1.782
.90
0 1.782
.05
t
The critical value for t is 1.782. 90% of the sample means10 with
n = 12 will lie between t = -1.782 and t = 1.782
Confidence Interval-Small Sample
Maximum error of estimate
E  tc
s
n
 In a random sample of 13 American adults, the mean waste
recycled per person per day was 4.3 pounds and the standard
deviation was 0.3 pound. Assume the variable is normally
distributed and construct a 90% confidence interval for .
1. The point estimate is x = 4.3 pounds
2. The maximum error of
estimate is
E  tc
s
n
 1.782
0.3
13
 0.148
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Confidence Interval-Small Sample
1. The point estimate is x=4.3 pounds
2. The maximum error of estimate is
E  tc
s
n
 1.782
0.3
13
 0.148
Right endpoint
Left endpoint
x  E  4.3  0.148  4.152
(
4.152
x  E  4.3  0.148  4.448
•
4.3
4.15 <  < 4.45
)
4.448
With 90% confidence, you can say the mean waste recycled
12
per person per day is between 4.15 and 4.45 pounds.
Confidence Intervals for Population Proportions
The point estimate for p, the population proportion of
successes is given by the proportion of successes in a sample
x
pˆ 
n
(Read as p-hat)
q̂ is the point estimate for the proportion of failures where
ˆ  1 p
ˆ
q
If np  5 and nq  5 , the sampling distribution for
p̂ is
normal.
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Confidence Intervals for Population Proportions
The maximum error of estimate, E for a c-confidence
interval is:
E  zc
pˆ qˆ
n
A c-confidence interval for the population proportion, p is
ˆ E  p  p
ˆ  E
p
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Confidence Interval for p
 In a study of 1907 fatal traffic accidents, 449 were alcohol
related. Construct a 99% confidence interval for the proportion
of fatal traffic accidents that are alcohol related.
1. The point estimate for p is
qˆ  1  0.235  0.765
x 449
pˆ  
 0.235
n 1907
2. 1907(.235) 5 and 1907(.765) 5, so the sampling
distribution is normal.
3.
E  zc
pˆ qˆ
(.235)(.765)
 2.575
 0.025
n
1907
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Confidence Interval for p
 In a study of 1907 fatal traffic accidents, 449 were alcohol
related. Construct a 99% confidence interval for the proportion of
fatal traffic accidents that are alcohol related.
Left endpoint
Right endpoint
pˆ  E  0.235  0.025  0.21
(
.21
ˆ  E  0.235  0.025  0.26
p
•
.235
)
.26
0.21 < p < 0.26
With 99% confidence, you can say the proportion of fatal
accidents that are alcohol related is between 21% and 26%.
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Minimum Sample Size
If you have a preliminary estimate for p and q the
minimum sample size given a c-confidence interval
and a maximum error of estimate needed to estimate
p is:
 zc 
ˆ qˆ 
n p

 E 
If you do not have a preliminary estimate, use 0.5 for
ˆ and qˆ
both p
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Example: Minimum Sample Size
 You wish to estimate the proportion of fatal accidents that
are alcohol related at a 99% level of confidence. Find the
minimum sample size needed to be be accurate to within 2%
of the population proportion.
With no preliminary estimate use 0.5 for pˆ and qˆ
 zc 
n  pˆ qˆ  
E
2
=
 2.575 
(0.5)(0.5)
  4414.14
 0.02 
You will need at least 4415 for your sample.
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Example: Minimum Sample Size
 You wish to estimate the proportion of fatal accidents that
are alcohol related at a 99% level of confidence. Find the
minimum sample size needed to be be accurate to within 2%
of the population proportion. Use a preliminary estimate of p =
0.235
 zc 
n  pˆ qˆ  
E
2
 2.575 
n  (0.235)(0.765)
  2980.05
 0.02 
With a preliminary sample you need at least n =2981for your sample.
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The Chi-Square Distribution
The point estimate for 2 is s2 and the point estimate for  is s.
If the sample size is n, use a chi-square 2 distribution with n-1 d.f. to
form a c-confidence interval.
.95
0
10
20
30
40
6.908
28.845
 Find R2 the right- tail critical value and L2 the left-tail critical value
for c = 95% and n = 17.
When the sample size is 17, there are 16 d.f.
Area to the right of R2 is (1- 0.95)/2 = 0.025 and area to the right of L2
is (1+ 0.95)/2 = 0.975
L2 =6.908
R2 =28.845
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 and 
Confidence Intervals for
2
(n  1) s 2
(n  1) s 2
2
A c-confidence interval for a



2
population variance is:
R
L2
To estimate the standard deviation take the square root of each endpoint.
You randomly select and record the prices of 17 CD players.
The sample standard deviation is $150. Construct a 95%
2
confidence interval for  and 
2
(17  1)150 2
(
17

1
)
150
2 
28.845
6.908
12480.50 <  < 52113.49
Find the square root of each endpoint
2
$117.72 <  < $228.28
You can say with 95% confidence that  is between 21
12480.50 and 52113.49 and  between $117.72 and $228.28.
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