Download Chapter 6: Confidence Intervals

Estimation and
Confidence Intervals
Types of Estimators
• Point Estimate
A single-valued estimate.
A single element chosen from a sampling distribution.
Conveys little information about the actual value of the
population parameter, about the accuracy of the estimate.
• Confidence Interval or Interval Estimate
An interval or range of values believed to include the
unknown population parameter.
Associated with the interval is a measure of the confidence
we have that the interval does indeed contain the parameter of
interest.
Confidence Interval or
Interval Estimate
A confidence interval or interval estimate is a range or interval of
numbers believed to include an unknown population parameter.
Associated with the interval is a measure of the confidence we have
that the interval does indeed contain the parameter of interest.
• A confidence interval or interval estimate has two
components:
A range or interval of values
An associated level of confidence
Confidence Interval for 
When  Is Known
•
If the population distribution is normal, the sampling
distribution of the mean is normal.
If the sample is sufficiently large, regardless of the shape of
the population distribution, the sampling distribution is
normal (Central Limit Theorem).
In either case:
Standard Normal Distribution: 95% Interval
0.4

 

P   196
.
 x    196
.
  0.95

n
n
0.3
f(z)

or

 

P x  196
.
   x  196
.
  0.95

n
n
0.2
0.1
0.0
-4
-3
-2
-1
0
z
1
2
3
4
Confidence Interval for  when
 is Known (Continued)
Before sampling, there is a 0.95probability that the interval
  1.96

n
will include the sample mean (and 5% that it will not).
Conversely, after sampling, approximately 95% of such intervals
x  1.96

n
will include the population mean (and 5% of them will not).
That is, x  1.96

n
is a 95% confidence interval for  .
A 95% Interval around the Population
Mean
Sampling Distribution of the Mean
Approximately 95% of sample means
can be expected to fall within the
interval   1.96  ,   1.96  .
0.4
95%
f(x)
0.3
0.2

0.1
2.5%
2.5%
0.0
  196
.


  196
.
n

x
n
x
x
2.5% fall below
the interval
n 
n
Conversely, about 2.5% can be

.
expected to be above   196
and
n
2.5% can be expected to be below

  1.96
.
n
x
x
x
2.5% fall above
the interval
x
x
x
x
95% fall within
the interval
So 5% can be expected to fall outside

 
the interval   196
.
.
,   196
.

n
n
95% Intervals around the Sample
Mean
Sampling Distribution of the Mean
0.4
95%
f(x)
0.3
0.2
0.1
2.5%
2.5%
0.0
  196
.


  196
.
n

x
n
x
x
x
x
x
x
x
x
x
x
x
x
x
*5% of such intervals around the sample
x
*
Approximately 95% of the intervals
 around the sample mean can be
x  1.96
n
expected to include the actual value of the
population mean, . (When the sample
mean falls within the 95% interval around
the population mean.)
*
mean can be expected not to include the
actual value of the population mean.
(When the sample mean falls outside the
95% interval around the population
mean.)
The 95% Confidence Interval for 
A 95% confidence interval for  when  is known and sampling is
done from a normal population, or a large sample is used:
x  1.96
The quantity 1.96
sampling error.

n

n
is often called the margin of error or the
For example, if: n = 25
 = 20
x = 122
A 95% confidence interval:

20
x  1.96
 122  1.96
n
25
 122  (1.96)(4 )
 122  7.84
 114.16,129.84
A (1-a )100% Confidence Interval for 
We define za as the z value that cuts off a right-tail area of a under the standard
2
2
normal curve. (1-a) or (1-a)100% is called the confidence level. a is called the
level of significance.


P z > za   a/2


2


P z  za   a/2


2





P  za z za   (1  a)
 2

2
S tand ard Norm al Distrib ution
0.4
(1  a )
f(z)
0.3
0.2
0.1
a
a
2
2
(1- a)100% Confidence Interval:
0.0
-5
-4
-3
-2
-1
z a
2
0
1
Z
za
2
2
3
4
5
x  za
2

n
Critical Values of z and Levels of
Confidence
0.99
0.98
0.95
0.90
0.80
2
0.005
0.010
0.025
0.050
0.100
Stand ard N o rm al Distrib utio n
za
0.4
(1  a )
2
2.576
2.326
1.960
1.645
1.282
0.3
f(z)
(1  a )
a
0.2
0.1
a
a
2
2
0.0
-5
-4
-3
-2
-1
z a
2
0
1
2
Z
za
2
3
4
5
The Level of Confidence and the
Width of the Confidence Interval
When sampling from the same population, using a fixed sample size, the
higher the confidence level, the wider the confidence interval.
St an d ar d N or m al Di stri b uti o n
0.4
0.4
0.3
0.3
f(z)
f(z)
St an d ar d N or m al Di s tri b uti o n
0.2
0.1
0.2
0.1
0.0
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5
-4
-3
-2
-1
Z
1
2
3
4
Z
80% Confidence Interval:
x  128
.
0

n
95% Confidence Interval:
x  196
.

n
5
The Sample Size and the Width of the
Confidence Interval
When sampling from the same population, using a fixed confidence
level, the larger the sample size, n, the narrower the confidence
interval.
S a m p lin g D is trib utio n o f th e M e an
S a m p lin g D is trib utio n o f th e M e an
0 .4
0 .9
0 .8
0 .7
0 .3
f(x)
f(x)
0 .6
0 .2
0 .5
0 .4
0 .3
0 .1
0 .2
0 .1
0 .0
0 .0
x
95% Confidence Interval: n = 20
x
95% Confidence Interval: n = 40
Point Estimates and Confidence Intervals for a
Mean – σ Known
EXAMPLE
The American Management Association wishes to
have information on the mean income of
middle managers in the retail industry. A
random sample of 256 managers reveals a
sample mean of $45,420. The standard
deviation of this population is $2,050. The
association would like answers to the following
questions:
x  sample mean
z  z - value for a particular confidence level
σ  the population standard deviation
n  the number of observatio ns in the sample
1.
What is the population mean?
In this case, we do not know. We do know the
sample mean is $45,420. Hence, our best
estimate of the unknown population value is
the corresponding sample statistic.
2.
What is a reasonable range of values for the
population mean? (Use 95% confidence level)
1. The width of the interval is
determined by the level of
confidence and the size of
the standard error of the
mean.
2. The standard error is affected
by two values:
- Standard deviation
- Number of observations
in the sample
The confidence limit are $45,169 and $45,671
The ±$251 is referred to as the margin of error
3.
What do these results mean?
If we select many samples of 256 managers,
and for each sample we compute the mean
and then construct a 95 percent confidence
interval, we could expect about 95 percent of
these confidence intervals to contain the
population mean.
Confidence Interval or Interval Estimate for 
When  Is Unknown - The t Distribution
If the population standard deviation, , is not known, replace
 with the sample standard deviation, s. If the population is
normal, the resulting statistic: t  X  
s
n
has a t distribution with (n - 1) degrees of freedom.
•
•
•
•
The t is a family of bell-shaped and symmetric
distributions, one for each number of degree of
freedom.
The expected value of t is 0.
The t is flatter and has fatter tails than does the
standard normal.
The t distribution approaches a standard normal
as the number of degrees of freedom increases
Standard normal
t, df = 20
t, df = 10


Confidence Intervals for  when  is
Unknown- The t Distribution
A (1-a)100% confidence interval for  when  is not known
(assuming a normally distributed population):
s
x t
n
a
2
where ta is the value of the t distribution with n-1 degrees of
2
a
freedom that cuts off a tail area of 2 to its right.
The t Distribution
t0.005
-----63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.704
2.660
2.617
2.576
t D is trib utio n: d f = 1 0
0 .4
0 .3
Area = 0.10
0 .2
Area = 0.10
}
t0.010
-----31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.624
2.602
2.583
2.567
2.552
2.539
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.457
2.423
2.390
2.358
2.326
}
t0.025
-----12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
2.042
2.021
2.000
1.980
1.960
f(t)
t0.050
----6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.684
1.671
1.658
1.645
0 .1
0 .0
-2.228
Area = 0.025
-1.372
0
t
1.372
2.228
}

t0.100
----3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
1.303
1.296
1.289
1.282
}
df
--1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
40
60
120
Area = 0.025
Whenever  is not known (and the population is
assumed normal), the correct distribution to use is
the t distribution with n-1 degrees of freedom.
Note, however, that for large degrees of freedom,
the t distribution is approximated well by the Z
distribution.
Example
A stock market analyst wants to estimate the average return on a
certain stock. A random sample of 15 days yields an average
(annualized) return of x  10.37% and a standard deviation of s = 3.5%.
Assuming a normal population of returns, give a 95% confidence
interval for the average return on this stock.
df
--1
.
.
.
13
14
15
.
.
.
t0.100
----3.078
.
.
.
1.350
1.345
1.341
.
.
.
t0.050
----6.314
.
.
.
1.771
1.761
1.753
.
.
.
t0.025
-----12.706
.
.
.
2.160
2.145
2.131
.
.
.
t0.010
-----31.821
.
.
.
2.650
2.624
2.602
.
.
.
t0.005
-----63.657
.
.
.
3.012
2.977
2.947
.
.
.
The critical value of t for df = (n -1) = (15 -1)
=14 and a right-tail area of 0.025 is:
t0.025  2.145
The corresponding confidence interval or
s
x

t
interval estimate is:
0 . 025
n
35
.
 10.37  2.145
15
 10.37  1.94
 8.43,12.31
Confidence Interval for the Mean – Example
using the t-distribution
EXAMPLE
A tire manufacturer wishes to investigate the tread life of its tires. A
sample of 10 tires driven 50,000 miles revealed a sample mean of
0.32 inch of tread remaining with a standard deviation of 0.09
inch.
Construct a 95 percent confidence interval for the population mean.
Would it be reasonable for the manufacturer to conclude that after
50,000 miles the population mean amount of tread remaining is
0.30 inches?
Large Sample Confidence Intervals for
the Population Mean
df
--1
.
.
.
120

t0.100
----3.078
.
.
.
1.289
1.282
t0.050
----6.314
.
.
.
1.658
1.645
t0.025
-----12.706
.
.
.
1.980
1.960
t0.010
-----31.821
.
.
.
2.358
2.326
t0.005
-----63.657
.
.
.
2.617
2.576
Whenever  is not known (and the population is
assumed normal), the correct distribution to use is
the t distribution with n-1 degrees of freedom.
Note, however, that for large degrees of freedom,
the t distribution is approximated well by the Z
distribution.
Large Sample : Sample size > 30
Large Sample Confidence Intervals for
the Population Mean
A large - sample (1 - a )100% confidence interval for :
s
x  za
n
2
Example : An economist wants to estimate the average amount in
checking accounts at banks in a given region. A random sample of
100 accounts gives x-bar = $357.60 and s = $140.00. Give a 95%
confidence interval for , the average amount in any checking account
at a bank in the given region.
x  z0.025
s
140.00
 357.60  1.96
 357.60  27.44   33016,385
.
.04
n
100
Large-Sample Confidence Intervals
for the Population Proportion, p
The estimator of the population proportion, p, is the sample proportion, pˆ . If the
sample size is large, pˆ has an approximately normal distributi on, with E( pˆ ) = p and
pq
V( pˆ ) = , where q = (1- p). When the population proportion is unknown, use the
n
estimated value, pˆ , to estimate the standard deviation of pˆ .
For estimating p, a sample is considered large enough when both n  p an n  q are greater
than 5.
Large-Sample Confidence Intervals
for the Population Proportion, p
A large - sample (1 - a )100% confidence interval for the population proportion, p :
pˆ  z
α
2
pˆ qˆ
n
where the sample proportion, p̂, is equal to the number of successes in the sample, x,
divided by the number of trials (the sample size), n, and q̂ = 1 - p̂.
Large-Sample Confidence Interval for the
Population Proportion, p
A marketing research firm wants to estimate the share that foreign companies
have in the American market for certain products. A random sample of 100
consumers is obtained, and it is found that 34 people in the sample are users
of foreign-made products; the rest are users of domestic products. Give a
95% confidence interval for the share of foreign products in this market.
p  za
2
pq
( 0.34 )( 0.66)

 0.34  1.96
n
100
 0.34  (1.96)( 0.04737 )
 0.34  0.0928
  0.2472 ,0.4328
Thus, the firm may be 95% confident that foreign manufacturers control
anywhere from 24.72% to 43.28% of the market.
Sample-Size Determination
Before determining the necessary sample size, three questions must
be answered:
• How close do you want your sample estimate to be to the unknown
•
•
parameter? (What is the desired bound, B?)
What do you want the desired confidence level (1-a) to be so that the
distance between your estimate and the parameter is less than or equal to B?
What is your estimate of the variance (or standard deviation) of the
population in question?

n
}
For example: A (1- a ) Confidence Interval for : x  z a
2
Bound, B
Sample Size and Standard Error
The sample size determines the bound of a statistic, since the standard
error of a statistic shrinks as the sample size increases:
Sample size = 2n
Standard error
of statistic
Sample size = n
Standard error
of statistic

Minimum Sample Size: Mean and
Proportion
Minimum required sample size in estimating the population
mean, :
za2 2
n 2 2
B
Bound of estimate:
B = za
2

n
Minimum required sample size in estimating the population
proportion, p
za2 pq
n 2 2
B
Sample-Size Determination:
Example
A marketing research firm wants to conduct a survey to estimate the average
amount spent on entertainment by each person visiting a popular resort. The
people who plan the survey would like to determine the average amount spent by
all people visiting the resort to within $120, with 95% confidence. From past
operation of the resort, an estimate of the population standard deviation is
σ = $400. What is the minimum required sample size?
za 
2
n
2
2
B
2
2
(1.96) ( 400)

120
2
 42.684  43
2
Sample-Size for Proportion:
Example
The manufacturers of a sports car want to estimate the proportion of people in a
given income bracket who are interested in the model. The company wants to
know the population proportion, p, to within 0.01 with 99% confidence. Current
company records indicate that the proportion p may be around 0.25. What is the
minimum required sample size for this survey?
n
za2 pq
2
B2
2.5762 (0.25)(0.75)

010
. 2
 124.42  125