Download EGR252S09 Chapter9 Lecture 1 JMB

Chapter 9:
One- and Two- Sample Estimation
Statistical Inference


Estimation
Tests of hypotheses
Interval estimation: (1 – α) 100% confidence
interval for the unknown parameter.


Example: if α = 0.01, we develop a 99%
confidence interval.
Example: if α = 0.05, we develop a 95%
confidence interval.
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 1
Single Sample: Estimating the Mean
Given:

σ is known and X is the mean of a random sample
of size n,
Then,

the (1 – α)100% confidence interval for μ is
 
 
X  z / 2 
    X  z / 2 

 n
 n
Z
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 2
Example
A traffic engineer is concerned about the delays at an
intersection near a local school. The intersection is
equipped with a fully actuated (“demand”) traffic light
and there have been complaints that traffic on the main
street is subject to unacceptable delays.
To develop a benchmark, the traffic engineer randomly
samples 25 stop times (in seconds) on a weekend day.
The average of these times is found to be 13.2 seconds,
and the variance is known to be 4 seconds2.
Based on this data, what is the 95% confidence interval
(C.I.) around the mean stop time during a weekend
day?
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 3
Example (cont.)
X = ______________
σ = _______________
α = ________________
α/2 = _____________
Z
Z
Z0.025 = _____________
Z0.975 = ____________
_______________ < μ < _________________
z0.025 = -1.96
13.2-(1.96)(2/sqrt(25)) = 12.416
EGR 252 Ch. 9 Lecture1
z0.975 = 1.96
13.2+(1.96)(2/sqrt(25)) = 13.984
JMB Spring 2009
Slide 4
Your turn …
 What is the 90% C.I.? What does it mean?
90%
-5
-4
-3
-2
Z(.05) = + 1.645
EGR 252 Ch. 9 Lecture1
-1
0
1
2
3
4
5
12.542 < μ < 13.858
JMB Spring 2009
Slide 5
What if σ2 is unknown?
 For example, what if the traffic engineer doesn’t
know the variance of this population?
1. If n is sufficiently large (n > 30), then the large
sample confidence interval is calculated by using
the sample standard deviation in place of sigma:
 s 
X  z / 2 

 n
2. If σ2 is unknown and n is not “large”, we must use
the t-statistic …
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 6
Single Sample: Estimating the Mean
(σ unknown, n not large)
Given:

σ is unknown and X is the mean of a random
sample of size n (where n is not large),
Then,

the (1 – α)100% confidence interval for μ is:
 s 
 s 
X  t / 2,n 1 
    X  t / 2,n 1 

 n
 n
-5
-4
EGR 252 Ch. 9 Lecture1
-3
-2
-1
0
1
2
JMB Spring 2009
3
4
5
Slide 7
Recall Our Example
A traffic engineer is concerned about the delays at an
intersection near a local school. The intersection is
equipped with a fully actuated (“demand”) traffic light
and there have been complaints that traffic on the main
street is subject to unacceptable delays.
To develop a benchmark, the traffic engineer randomly
samples 25 stop times (in seconds) on a weekend day.
The average of these times is found to be 13.2 seconds,
and the sample variance, s2, is found to be 4 seconds2.
Based on this data, what is the 95% confidence interval
(C.I.) around the mean stop time during a weekend day?
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 8
Small Sample Example (cont.)
X = ______________
s = _______________
α = ________________
α/2 = _____________
-5
-4
-3
-2
-1
0
1
2
3
4
5
t0.025,24 = _____________
_______________ < μ < ________________
t 0.025,24 = 2.064
13.2-(2.064)(2/sqrt(25)) = 13.374
EGR 252 Ch. 9 Lecture1
13.2+(2.064)(2/sqrt(25)) = 14.026
JMB Spring 2009
Slide 9
Your turn
A thermodynamics professor gave a physics
pretest to a random sample of 15 students
who enrolled in his course at a large state
university. The sample mean was found to be
59.81 and the sample standard deviation
was 4.94.
Find a 99% confidence interval for the mean
on this pretest.
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 10
Solution
X = ______________
s = _______________
α = ________________
α/2 = _____________
(draw the picture)
T___ , ____ = _____________
__________________ < μ < ___________________
X = 59.81
s = 4.94
α = .01
α/2 = .005 t (.005,14) = 2.977
Lower Bound 59.81 - (2.977)(4.94/sqrt(15)) = 56.01
Upper Bound 59.81 + (2.977)(4.94/sqrt(15)) = 63.61
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 11
Standard Error of a Point Estimate
Case 1: σ known

The standard deviation, or standard error of X is

n
Case 2: σ unknown, sampling from a normal
distribution

The standard deviation, or (usually) estimated
standard error of X is
s
n
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 12
9.6: Prediction Interval
For a normal distribution of unknown mean μ,
and standard deviation σ, a 100(1-α)%
prediction interval of a future observation, x0 is
1
1
X  z / 2 1 
 x0  X  z / 2 1 
n
n
if σ is known, and
1
1
X  t / 2,n 1s 1   x0  X  t / 2,n 1s 1 
n
n
if σ is unknown
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 13
9.7: Tolerance Limits
For a normal distribution of unknown mean
μ, and unknown standard deviation σ,
tolerance limits are given by
x + ks
where k is determined so that one can assert
with 100(1-γ)% confidence that the given
limits contain at least the proportion 1-α of
the measurements.
Table A.7 (page 761) gives values of
k for (1-α) = 0.9, 0.95, or 0.99 and
γ = 0.05 or 0.01 for selected values of n.
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 14
Example 9.8 (Page 284)
 Find the 99% tolerance limits that will contain 95% of
the metal pieces produced by the machine, given a
sample mean of 1.0056 and a sample standard
deviation of 0.0246.
 Table A.7 (page 761)


(1-α ) = 0.95
( 1-γ ) = 0.99

n = 9.
k = 4.550

x ± ks = 1.0056 ± (4.550) (0.0246)

 We can assert with 99% confidence that the
tolerance interval from 0.894 to 1.117 will contain
95% of the metal pieces produced by the machine
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 15
Summary
Confidence interval 
population mean μ
Prediction interval 
a new observation x0
Tolerance interval 
a (1-α) proportion of the
measurements can be
estimated with a 100(1-γ)%
confidence
EGR 252 Ch. 9 Lecture1
JMB Spring 2009
Slide 16