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General Business 704
Estimation:
Confidence Intervals
Based in part on Chapter 6
Objectives:
Estimation

Distinguish point & interval estimates

Explain interval estimates

Compute confidence interval estimates
Population mean & proportion
 Population total & difference


Determine necessary sample size
Thinking Challenge
Suppose you’re
interested in the
average amount of
money that students
in this class (the
population) have in
their possession.
How would you find
out?
Statistical
Methods
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
Hypothesis
Testing
Estimation
Process
Population


Mean, , is
unknown

 
Sample


 
Random Sample
Mean 
X = 50
I am 95%
confident that
 is between
40 & 60.
Population Parameter
Estimates
Estimate population
parameter...
Mean

with sample
statistic
x
Proportion
p
ps
Variance

2
Differences
2
12
s
x1 -x2
Estimation
Methods
Estimation
Point
Estimation
Interval
Estimation
Confidence
Interval
Bootstrapping
Estimation
Methods
Estimation
Point
Estimation
Interval
Estimation
Confidence
Interval
Bootstrapping
Point Estimation

Provides single value



Based on observations from 1 sample
Gives no information about how
close value is to the unknown
population parameter
Example: Sample meanX = 3 is
point estimate of unknown
population mean
Estimation
Methods
Estimation
Point
Estimation
Interval
Estimation
Confidence
Interval
Bootstrapping
Interval Estimation

Provides range of values


Gives information about closeness to
unknown population parameter


Based on observations from 1 sample
Stated in terms of probability
Example: Unknown population mean
lies between 40 & 60 with 95%
confidence
Key Elements of
Interval Estimation
A probability that the population parameter
falls somewhere within the interval.
Confidence
interval
Confidence
limit (lower)
Sample statistic
(point estimate)
Confidence
limit (upper)
Confidence Limits
for Population Mean
Parameter =
Statistic ± Error
© 1984-1994
T/Maker Co.
(1)
  X  Error
(2)
Error  X   or X  
X 
(3)
Z
(4)
Error  Z x
(5)
  X  Z x
x

Error
x
Many Samples Have
Same Interval
X =  ± Zx
x_
-2.58x
-1.65x
-1.96x

+1.65x
+2.58x
+1.96x
90% Samples
95% Samples
99% Samples
X
Level of Confidence


Probability that the unknown
population parameter falls within
interval
Denoted (1 - 


is probability that parameter is not
within interval
Typical values are 99%, 95%, 90%
Intervals &
Level of Confidence
Sampling
Distribution /2
of Mean
x_
1 -
/2
x = 
_
X
(1 - ) % of
intervals
contain .
Intervals
extend from
X - ZX to
X + ZX
 % do not.
Large number of intervals
Factors Affecting
Interval Width

Data dispersion


Sample size


Measured by 
Intervals extend from
X - ZX toX + ZX
X =  / n
Level of confidence
(1 - )

Affects Z
© 1984-1994 T/Maker Co.
Confidence
Interval Estimates
Confidence
Intervals
Mean
Known
Proportion
 Unknown
Variance
Finite
Population
Confidence
Interval Estimates
Confidence
Intervals
Mean
 Known
Proportion
 Unknown
Variance
Finite
Population
Confidence Interval
Mean ( Known)

Assumptions
Population standard deviation is known
 Population is normally distributed
 If not normal, can be approximated by
normal distribution (n  30)


Confidence interval estimate
X  Z / 2 

n
   X  Z / 2 

n
Note: 99% Z=2.58, 95% Z=1.96 , 90% Z=1.65
Estimation Example
Mean ( Known)
The mean of a random sample of n = 25
isX = 50. Set up a 95% confidence
interval estimate for  if  = 10.
X  Z / 2 

   X  Z / 2 

n
n
10
10
50  1.96 
   50  1.96 
25
25
46.08    53.92
Thinking Challenge
You’re a Q/C inspector for
Gallo. The  for 2-liter
bottles is .05 liters. A
random sample of 100
bottles showedX = 1.99
liters. What is the 90%
confidence interval
estimate of the true mean
amount in 2-liter bottles?
2
liter
© 1984-1994 T/Maker Co.
Confidence Interval
Solution for Gallo
X  Z / 2 

n
   X  Z / 2 

n
.05
.05
1.99  1.645 
   1.99  1.645 
100
100
1.982    1.998
Confidence
Interval Estimates
Confidence
Intervals
Mean
s Known
Proportion
Unknown
Variance
Finite
Population
Confidence Interval
Mean ( Unknown)

Assumptions
Population standard deviation is
unknown
 Population must be normally distributed



Use Student’s t distribution
Confidence interval estimate
S
S
X  t  / 2, n 1 
   X  t  / 2, n 1 
n
n
Student’s t Distribution
Standard
Bellnormal
shaped
Symmetric
t (df = 13)
‘Fatter’ tails
t (df = 5)
0
Z
t
Note: As d.f. approach 120, Z and t become very similar
Student’s t Table
Upper Tail Area
df
.25
.10
Assume:
n=3
df = n - 1 = 2
 = .10
/2 =.05
/2
.05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
.05
3 0.765 1.638 2.353
0
t values
2.920
t
Degrees of Freedom


Number of observations that are free
to vary after sample statistic has
been calculated
degrees of freedom
Example

Sum of 3 numbers is 6
X1 = 1 (or any number)
X2 = 2 (or any number)
X3 = 3 (cannot vary)
Sum = 6
= n -1
= 3 -1
=2
Estimation Example
Mean ( Unknown)
A random sample of n = 25 hasX = 50
& S = 8. Set up a 95% confidence
interval estimate for .
S
X  t  / 2, n 1 
   X  t  / 2, n 1 
n
8
50  2.0639 
   50  2.0639 
25
46.69    53.30
S
n
8
25
Thinking Challenge
You’re a time study analyst
in manufacturing. You’ve
recorded the following
task times (min.): 3.6, 4.2,
4.0, 3.5, 3.8, 3.1.
What is the 90%
confidence interval
estimate of the population
mean task time?
Confidence Interval
Solution for Time Study
X = 3.7
S = 3.8987
n = 6, df = n - 1 = 6 - 1 = 5
S / n = 3.8987 / 6 = 1.592
t.05,5 = 2.0150
3.7 - (2.015)(1.592) 3.7 + (2.015)(1.592)
0.492  6.908
Confidence
Interval Estimates
Confidence
Intervals
Mean
Known
Proportion
 Unknown
Variance
Finite
Population
Estimation for
Finite Populations

Assumptions

Sample is large relative to population
 n / N > .05

Use finite population correction factor

Confidence interval (mean,  unknown)
S
Nn
S
Nn
X  t  / 2, n 1 

   X  t  / 2, n 1 

n N 1
n N 1
Confidence
Interval Estimates
Confidence
Intervals
Mean
Known
Proportion
 Unknown
Variance
Finite
Population
Confidence Interval
Proportion

Assumptions
Two categorical outcomes
 Population follows binomial distribution
 Normal approximation can be used
 n·p  5 & n·(1 - p)  5


Confidence interval estimate
ps  (1  ps )
ps  (1  ps )
ps  Z 
 p  ps  Z 
n
n
Estimation Example
Proportion
A random sample of 400 graduates
showed 32 went to grad school. Set
up a 95% confidence interval estimate
for p.
ps  (1  ps )
ps  (1  ps )
ps  Z  / 2 
 p  ps  Z  / 2 
n
n
.08  (1 .08)
.08  (1 .08)
.08  1.96 
 p  .08  1.96 
400
400
.053  p  .107
Thinking Challenge
You’re a production
manager for a newspaper.
You want to find the %
defective. Of 200
newspapers, 35 had
defects. What is the
90% confidence interval
estimate of the population
proportion defective?
Confidence Interval
Solution for Defects

n·p  5
n·(1 - p)  5
ps  (1  ps )
ps  (1  ps )
ps  Z  / 2 
 p  ps  Z  / 2 
n
n
.175  (.825)
.175  (.825)
.175  1.645 
 p  .175  1.645 
200
200
.1308  p  .2192
Estimation Methods
Estimation
Point
Estimation
Interval
Estimation
Confidence
Interval
Bootstrapping
Bootstrapping Method



Used if population is not normal
Requires significant computer power
Steps
Take initial sample
 Sample repeatedly from initial sample
 Compute sample statistic
 Form resampling distribution
 Limits are values that cut off smallest &
largest /2 %

Finding Sample Sizes
For Estimating 
(1)
Z
X 
x

Error
x
(2)
Error  Z x  Z
(3)
Z 
n
Error 2
2
2
I don’t want to
sample too much
or too little!

n
Sample Size Example
What sample size is needed to be 90%
confident of being correct within  5?
A pilot study suggested that the
standard deviation is 45.
a f a f  219.2  220
af
2
1645
.
45
Z 
n

2
2
Error
5
2
2
2
Thinking Challenge
You work in Human
Resources at Merrill Lynch.
You plan to survey employees
to find their average medical
expenses. You want to be
95% confident that the
sample mean is within ± $50.
A pilot study showed that 
was about $400. What
sample size do you use?
Sample Size Solution
Medical Expenses
Z 2 2
n
2
Error
1.96f a
400f
a

a50f
2
2
2
 245.86  246
Finding Sample Sizes For
Estimating Proportions
Z p(1  p)
n
2
Error
2
I don’t want to
sample too much
or too little!
Remember
•Error is acceptable error
•Z is based on confidence level chosen
•p is the true proportion of “success”
•Never under-estimate p
•When in doubt, use p=.5
Sample Size Example
for Estimating p
What sample size is needed to be 90%
confident (Z=1.645) of being correct
within proportion of .04 when using
p=.5 (since no useful estimate of p is
available)?
Z p(1  p) 1.645 (.5)(.5)
n


422
.
82

423
2
2
Error
.04
2
2
Estimation of
Population Total

In auditing, population total is more
important than mean


Total = NX
Confidence interval (population total)
S
Nn
S
Nn
NX  t 

 Total  NX  t 

n N 1
n N 1

Degrees of freedom = n - 1
Estimation
of Differences

Used to estimate the magnitude of
errors

Steps
Determine sample size
 Compute average difference,D
 Compute standard deviation of
differences
 Set up confidence interval estimate

Estimation of
Differences Equations
Mean Difference:
Standard Deviation:
n
D
 Di
i 1
n
n
sD 
2
D
 i  nD
i 1
n1
Interval Estimate:
SD N  n
SD N  n
ND  Nt 

  D  ND  Nt 

n N 1
n N 1
Objectives:
Estimation

Distinguish point & interval estimates

Explain interval estimates

Compute confidence interval estimates
Population mean & proportion
 Population total & difference


Determine necessary sample size