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Chapter 9
One and Two Sample Estimation
Point Estimates
Unknown
Parameter
Symbol
Population mean
Population
standard deviation
Population
proportion
Point Estimator
m
s
Symbol
Sample mean
Sample standard
deviation
s
Sample proportion
p
Point estimates are computed from sample information
and used to estimate population parameters.
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 1
Interval Estimation
Interval estimation: The interval in which you
would expect to find the value of a
parameter.
By taking a random sample, we can compute
an interval with upper and lower limits called
a (1-a) 100% confidence interval for the
unknown parameter
 In most a cases the parameter for intervals
to be created will be m
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 2
Confidence Intervals
Point
Estimates
• Single value
• Mean
• Degree of uncertainty
• Range of certainty around the point estimate
Intervals
• Point Estimates (mean)
• Confidence level (1-a)
Based On • Standard deviation
Expressed
As
• The mean score of the students was 80.1 with a 95% CI of 77.4 – 82.5
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 3
Confidence Intervals
(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
MDH 2015
9th edition
Slide 4
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
s 
s 
X  za / 2 
  m  X  za / 2 

 n
 n
Values
obtained
from Z-table
1-a
a/2
a/2
Z
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 5
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
MDH 2015
9th edition
Slide 6
Example (cont.)
X = ______________
σ = _______________
α = ________________
α/2 = _____________
Z
Z
Z0.025 = _____________
Z0.975 = ____________
Enter values into equation
s 
s 
X  za / 2 
  m  X  za / 2 

 n
 n
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 7
Example (cont.)
X = ______13.2________
σ = _____2________
α = ______0.05________
α/2 = __0.025_______
Z
Z
Z0.025 = ___-1.96______
Z0.975 = ____1.96____
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
MDH 2015
9th edition
Slide 8
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
MDH 2015
9th edition
Slide 9
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  za / 2 

 n
2. If σ 2 is unknown and n is not “large”, we must use
the t-statistic.
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 10
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  ta / 2,n 1 
  m  X  ta / 2,n 1 

 n
 n
Values
obtained
from t-table
1-a
a/2
a/2
-5
-4
EGR 252 Ch. 9 Lecture1
-3
-2
-1
0
MDH 2015
1
2
9th edition
3
4
5
Slide 11
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
MDH 2015
9th edition
Slide 12
Small Sample Example (cont.)
n = _______ v = _______ X = ______ s = _______
α = _______
α/2 = ____
-5
-4
-3
-2
-1
0
t0.025,24 = _______
1
2
3
4
5
Find interval using equation below
 s 
 s 
X  ta / 2,n 1 
  m  X  ta / 2,n 1 

 n
 n
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 13
Small Sample Example (cont.)
n = __25___ v = _24___ X = __13.2_ s = __2__
α = __0.05___
-5
α/2 = 0.025_
-4
-3
-2
-1
13.2 - (2.064)(2/sqrt(25)) = 13.374
0
1
2
3
4
t0.025,24 = _2.064_
5
13.2 + (2.064)(2/sqrt(25)) = 14.026
__12.374_____ < μ < ___14.026____
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 14
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
MDH 2015
9th edition
Slide 15
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
MDH 2015
9th edition
Slide 16
Standard Error of a Point Estimate
Case 1: σ known

The standard deviation, or standard error of X is
s
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
MDH 2015
9th edition
Slide 17
Prediction Intervals
Used to predict the possible value of a future
observation
Example: In quality control, an experimenter
may need to use the observed data to predict
a new observation.
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 18
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  za / 2 s 1   x0  X  za / 2 s 1 
n
n
if σ is known, and
1
1
X  ta / 2,n1 s 1   x0  X  ta / 2,n1 s 1 
n
n
if σ is unknown
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 19
Tolerance Limits (Intervals)
What if you want to be 95% sure that the
interval contains 95% of the values? Or 90%
sure that the interval contains 99% of the
values?
These questions are answered by a
tolerance interval. To compute, or
understand, a tolerance interval you have to
specify two different percentages. One
expresses how sure you want to be, and
the other expresses what fraction of the
values the interval will contain.
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 20
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 745) 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
MDH 2015
9th edition
Slide 21
Tolerance Limits
How to determine 100(1-γ)% and 1-α.
For a sample size of 8, find the tolerance interval that
gives two-sided 95% bounds on 90% of the distribution
or population. X is 15.6 and s is 1.4
1-g (boundary or the limits)
1-a (proportion of the distribution)
From table on pg. 745, find the corresponding value:
n = 8, g = .05, a = 0.1 corresponding k…k = 3.136
x + ks = 15.6 + (3.136)(1.4)
Tolerance interval 19.99 – 11.21
We are 95% confident that 90% of the population falls within the limits
of 11.21 and 19.99
EGR 252 Ch. 9 Lecture1
MDH 2015
9th edition
Slide 22
Case Study 9.1c (Page 281)
 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 745)





(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
MDH 2015
9th edition
Slide 23
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
MDH 2015
9th edition
Slide 24