1. Chapter 9 One and Two Sample Estimation
Point Estimates
Unknown Parameter
Symbol
Point Estimator
Population mean m
Sample mean
Population standard deviation s
Sample standard deviation
Population proportion p
Sample proportion
Point estimates are computed from sample information and used to estimate population parameters.
EGR 252 Ch. 9 Lecture1
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2. 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
Updated for 8th edition by JMB
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3. Confidence Intervals Single value Mean Degree of uncertainty
Point Estimates
Single value
Mean
Intervals
Degree of uncertainty
Range of certainty around the point estimate
Based On
Point Estimates (mean)
Confidence level (1-a)
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
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4. 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.
Updated for 8th edition by JMB
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5. 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
show α/2 above and below and 1-α in the center.
Values obtained from Z-table
1 - a
a/2
a/2
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6. 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?
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7. Example (cont.) X = ______________ σ = _______________
α = ________________ α/2 = _____________
Z0.025 = _____________ Z0.975 = ____________
Enter values into equation
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8. Example (cont.) X = ______13.2________ σ = _____2________
α = ______0.05________ α/2 = __0.025_______
Z0.025 = ___-1.96______ Z0.975 = ____1.96____
Solution: < μSTOP TIME <
X = σ = 2
α = .05 α/2 = .025
Draw picture
z.025 = z.975 = 1.96
13.2-(1.96)(2/sqrt(25)) = (1.96)(2/sqrt(25)) =
Interpretation: there is a 95% chance that the “true mean” of traffic delay time is somewhere between and
Z0.025 = Z0.975 = 1.96
13.2-(1.96)(2/sqrt(25)) = (1.96)(2/sqrt(25)) =
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9. Your turn … What is the 90% C.I.? What does it mean?
Z =
< μ < 5% 5%
Z(.05) = 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.
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10. What if σ 2 is unknown?
For example, what if the traffic engineer doesn’t know the variance of this population?
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:
If σ 2 is unknown and n is not “large”, we must use the t-statistic.
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11. 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:
show α/2 above and below and 1-α in the center.
Values obtained from t-table
1 - a
a/2
a/2
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12. 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
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13. Small Sample Example (cont.)
n = _______ v = _______ X = ______ s = _______
α = _______ α/2 = ____ t0.025,24 = _______
Find interval using equation below
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14. Small Sample Example (cont.)
n = __25___ v = _24___ X = __13.2_ s = __2__
α = __0.05___ α/2 = 0.025_ t0.025,24 = _2.064_
__12.374_____ < μ < ___14.026____
X = s = 2
α = .05 α/2 = .025
Draw picture
t.025,24 = 2.064
13.2-(2.064)(2/sqrt(25)) = (2.064)(2/sqrt(25)) =
Interpretation: there is a 95% chance that the “true mean” of traffic delay time is somewhere between and
(2.064)(2/sqrt(25)) = (2.064)(2/sqrt(25)) =
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15. 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 and the sample standard deviation was 4.94.
Find a 99% confidence interval for the mean on this pretest.
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16. Solution X = ______________ s = _______________
α = ________________ α/2 = _____________
(draw the picture)
t___ , ____ = _____________
__________________ < μ < ___________________
X = s = 4.94
α = .01 α/2 = .005
Draw picture
t.005,14 = 2.977
59.81-(2.977)(4.94/sqrt(15)) = (2.977)(4.94/sqrt(15)) = 63.61
Interpretation:
X = s = α = α/2 = t (.005,14) = 2.977
Lower Bound (2.977)(4.94/sqrt(15)) = 56.01
Upper Bound (2.977)(4.94/sqrt(15)) = 63.61
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17. Standard Error of a Point Estimate
Case 1: σ known
The standard deviation, or standard error of X is
Case 2: σ unknown, sampling from a normal distribution
The standard deviation, or (usually) estimated standard error of X is
???? Z =
< μ <
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18. 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
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19. 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
if σ is known, and
if σ is unknown
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20. 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
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21. 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 and γ = 0.05 or 0.01 for selected values of n.
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22. 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
From table on pg. 745, find the corresponding value:
n = 8, g = .05, a = 0.1 corresponding k…k = 3.136
x + ks = (3.136)(1.4)
Tolerance interval – 11.21
We are 95% confident that 90% of the population falls within the limits of and 19.99
1-g (boundary or the limits)
1-a (proportion of the distribution)
EGR 252 Ch. 9 Lecture1
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23. 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 cm and a sample standard deviation of
Table A.7 (page 745)
(1 - α ) = 0.95
(1 – Ƴ ) = 0.99
n = 9
k = 4.550
x ± ks = ± (4.550) (0.0246)
We can assert with 99% confidence that the tolerance interval from to cm will contain 95% of the metal pieces produced by the machine.
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24. 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
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