### Chapter08

```Chapter 8
Estimation
with Single
Samples
Slide 8-1
Learning Objectives
• Know the difference between point and
interval estimation.
• Estimate a population mean from a sample
mean for large sample sizes.
• Estimate a population mean from a sample
mean for small sample sizes.
• Estimate a population proportion from a
sample proportion.
• Estimate the minimum sample size
necessary to achieve given statistical
goals.
Slide 8-2
Statistical Estimation
• Point estimate -- the single value of a
statistic calculated from a sample
• Interval Estimate -- a range of values
calculated from a sample statistic(s) and
standardized statistics, such as the Z.
– Selection of the standardized statistic is
determined by the sampling distribution.
– Selection of critical values of the
standardized statistic is determined by
the desired level of confidence.
Slide 8-3
Confidence Interval
to Estimate  when n is Large
• Point estimate
• Interval
Estimate
X 
X
X Z

n
or
X Z
n

n
 X Z

n
Slide 8-4
Distribution of Sample Means
for (1-)% Confidence


2
2



X
Z
Z
2
0
Z
2
Slide 8-5
Z Scores for Confidence Intervals
in Relation to 


2
.5 

.5 
2
2

2


X
Z
Z
2
0
Z
2
Slide 8-6
Distribution of Sample Means
for (1-)% Confidence


2
1 
1 
2
2
2


X
Z
Z
2
0
Z
2
Slide 8-7
Probability Interpretation
of the Level of Confidence
P r ob[ X 
Z
2

n
   X 
Z
2

] 1
n
Slide 8-8
Distribution of Sample Means
for 95% Confidence
.025
.025
95%
.4750
.4750

X
Z
-1.96
0
1.96
Slide 8-9
Example: 95% Confidence
Interval for 
X  4 .26 ,   1.1, and n  60 .
X Z
4 .26  1.96

n
1.1
60
   X Z

n
   4 .26  1.96
1.1
60
4 .26  0 .28    4 .26  0 .28
3.98    4 .54
Slide 8-10
95% Confidence Intervals for 
95%

X
X
X
X
X
X
X
Slide 8-11
95% Confidence Intervals for 
Is our interval,
95%
3.98  4.54,
in the red?

X
X
X
X
X
X
X
Slide 8-12
Demonstration Problem 8.1
X  10 .455 ,   7 .7 , and n  44 .
90% confidence  Z  1.645
X Z

n
7 .7
1 0 .4 5 5  1.6 4 5
44
   X Z

n
   1 0 .4 5 5  1.6 4 5
7 .7
44
1 0 .4 5 5  1.9 1    1 0 .4 5 5  1.9 1
8 .5 4 5    1 2 .3 6 5
P r ob[ 8.545    12 .365 ]  0.90
Slide 8-13
Demonstration Problem 8.2
X  34 .3 ,   8 , N = 800 and n  50.
98% confidence  Z  2 .33
X Z

N n
N 1
n
34.3  2.33
8
50
800  50
800  1
   X Z

n
   34.3  2.33
N n
N 1
8
50
800  50
800  1
34.3  2.55    34.3  2.55
31.75    36.85
Slide 8-14
Confidence Interval to Estimate 
when n is Large and  is Unknown
X  Z
2
S
n
or
X  Z
2
S
n
   X  Z
2
S
n
Slide 8-15
Z Values for Some of the More
Common Levels of Confidence
Confidence
Level
Z Value
90%
1.645
95%
1.96
98%
2.33
99%
2.575
Slide 8-16
Estimating the Mean of a Normal
Population: Small n and Unknown 
• The population has a normal
distribution.
• The value of the population standard
deviation is unknown.
• The sample size is small, n < 30.
• Z distribution is not appropriate for
these conditions
• t distribution is appropriate
Slide 8-17
The t Distribution
• A family of distributions -- a unique
distribution for each value of its
parameter, degrees of freedom (d.f.)
• Symmetric, Unimodal, Mean = 0, Flatter
than a Z
• t formula
X  
t 
S
n
Slide 8-18
Comparison of Selected t Distributions
to the Standard Normal
Standard Normal
t (d.f. = 25)
t (d.f. = 5)
t (d.f. = 1)
-3
-2
-1
0
1
2
3
Slide 8-19
Table of Critical Values of t
t0.100 t0.050 t0.025 t0.010 t0.005
df
1
2
3
4
5
3.078
1.886
1.638
1.533
1.476
6.314
2.920
2.353
2.132
2.015
12.706
4.303
3.182
2.776
2.571
31.821
6.965
4.541
3.747
3.365
63.656
9.925
5.841
4.604
4.032
1.714
25
1.319
1.318
1.316
1.708
2.069
2.064
2.060
2.500
2.492
2.485
2.807
2.797
2.787
29
30
1.311
1.310
1.699
1.697
2.045
2.042
2.462
2.457
2.756
2.750
40
60
120
1.303
1.296
1.289
1.282
1.684
1.671
1.658
1.645
2.021
2.000
1.980
1.960
2.423
2.390
2.358
2.327
2.704
2.660
2.617
2.576
23
24

1.711


t
Slide 8-20
Confidence Intervals for  of a Normal
Population: Small n and Unknown 
X t
S
n
or
X t
S
n
  X t
S
n
df  n  1
Slide 8-21
Solution for Demonstration Problem 8.3
X  2 .1 4 , S  1.2 9 , n  1 4 , d f  n  1  1 3

2
t

1  .9 9
. 0 0 5 ,1 3
2
 0 .0 0 5
 3.0 1 2
X t
2 .1 4  3.0 1 2
S
n
1.2 9
14
   X t
S
n
   2 .1 4  3.0 1 2
1.2 9
14
2 .1 4  1.0 4    2 .1 4  1.0 4
1.1 0    3.1 8
Slide 8-22
Solution for Demonstration Problem 8.3
X t
2 .1 4  3.0 1 2
S
n
1.2 9
14
   X t
S
n
   2 .1 4  3.0 1 2
1.2 9
14
2 .1 4  1.0 4    2 .1 4  1.0 4
1.1 0    3.1 8
P r ob [1.10    3.18 ]  0 .99
Slide 8-23
Confidence Interval to Estimate
the Population Proportion
p 
Z
2
p q
n
 P  p 
Z
2
p q
n
w h ere:
p = sam p le p ro p o rtio n
q = 1 - p
P = p o p u latio n p ro p o rtio n
n = sam p le siz e
Slide 8-24
Solution for Demonstration Problem 8.5
n  2 1 2 , X  3 4 , p 
X
n
q = 1 - p  1  0 .1 6  0 .8 4

34
212
 0 .1 6
9 0 % C o n fid en ce  Z  1.6 4 5
p  Z
0 .1 6  1.6 4 5
p q
n
( 0 .1 6 )( 0 .8 4 )
212
 P  p  Z
p q
n
 P  0 .1 6  1.6 4 5
( 0 .1 6 )( 0 .8 4 )
212
0 .1 6  0 .0 4  P  0 .1 6  0 .0 4
0 .1 2  P  0 .2 0
P r ob[ 0.12  P  0.20 ]  0.90
Slide 8-25
Determining Sample Size
when Estimating 
• Z formula
Z 
X 

n
• Error of Estimation
E 
(tolerable error)
• Estimated Sample Size
X 
n 
Z
2
• Estimated 

1
4
 Z  


 E 
2
2
2
E
2
2
range
Slide 8-26
Example: Sample Size when
Estimating 
E  1,   4
90% confidence  Z  1.645
Z
2
n 
2
2
2
E
(1.6 4 5) ( 4 )
1
2

2
2
 4 3.3 0 o r 4 4
Slide 8-27
Solution for Demonstration Problem 8.6
E  2 , ra n g e  2 5
9 5 % co n fid en ce  Z  1.9 6
estim a ted  :
 1
ra n g e     2 5   6 .2 5
 4
4
1
Z 
E
(1.9 6 ) ( 6.2 5)
2
2
2
n 
2
2

2
2
 3 7 .5 2 o r 3 8
Slide 8-28
Determining Sample Size
when Estimating P
•Z
formula
Z 
pP
P Q
n
• Error of Estimation
(tolerable
error)
• Estimated Sample
Size
E  pP
n
Z
2
PQ
E
2
Slide 8-29
Solution for Demonstration Problem 8.7
E  0 .03
98% C onfidence  Z  2 .33
estim ated P  0 .40
Q  1  P  0 .60
n

Z
2
PQ
2
E
( 2 .33)
2
 0 .40  0 .60 
.003
2
 1, 447 .7 or 1,448
Slide 8-30
Determining Sample Size when
Estimating P with No Prior Information
P
PQ
0.5
0.25
Z = 1.96
E = 0.05
400
350
300
0.4
0.24
0.3
0.21
250
n 200
150
0.2
0.16
0.1
0.09
100
50
0
0
n 
Z
2
E
0.1
0.2
0.3
0.4
0.5
P
0.6
0.7
0.8
0.9
1
1
4
2
Slide 8-31
Solution for Demonstration Problem 8.8
E  0 .05
90% C onfidence  Z  1.645
w ith no prior estim ate of P , use P  0 .50
Q  1  P  0 .50
n

Z
2
PQ
2
E
(1.6 4 5)
2
 0 .5 0  0 .5 0 
.0 5
2
 2 7 0 .6 o r 2 7 1