EGR252S10 Lecture 12 Chapter9 Part1 JMB publish
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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 2010
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
1-
/2
/2
Z
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
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 2010
Slide 3
Example (cont.)
X = ______________
σ = _______________
α = ________________
α/2 = _____________
Z
Z
Z0.025 = _____________
Z0.975 = ____________
Z0.025 = -1.96
13.2-(1.96)(2/sqrt(25)) = 12.416
Z0.975 = 1.96
13.2+(1.96)(2/sqrt(25)) = 13.984
Solution:
12.416 < μSTOP TIME < 13.984
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
Slide 4
Your turn …
What is the 90% C.I.? What does it mean?
90%
5%
-5
-4
-3
-2
-1
0
5%
1
2
3
4
5
Z(.05) = + 1.645 All other values remain the same.
The 90 % CI for μ = (12.542,13.858)
Note that the 95% CI is wider than the 90% CI.
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
Slide 5
What if σ 2 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 σ 2 is unknown and n is not “large”, we must use
the t-statistic.
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
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 2010
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 2010
Slide 8
Small Sample Example (cont.)
n = _______ df = _______ X = ______ s = _______
α = _______
α/2 = ____
-5
-4
-3
-2
-1
13.2 - (2.064)(2/sqrt(25)) = 13.374
0
t0.025,24 = _______
1
2
3
4
5
13.2 + (2.064)(2/sqrt(25)) = 14.026
_______________ < μ < ________________
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
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 2010
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 2010
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 2010
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,n1 s 1 x0 X t / 2,n1 s 1
n
n
if σ is unknown
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
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 2010
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 diameter of 1.0056 cm and a sample
standard deviation of 0.0246.
Table A.7 (page 761)
(1 - α ) = 0.95
(1 – Ƴ ) = 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 cm will contain
95% of the metal pieces produced by the machine.
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
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 100( 1-Ƴ )%
confidence
EGR 252 Ch. 9 Lecture1
JMB Spring 2010
Slide 16