EGR 252 Ch. 9 Lecture1 JMB Fall 2011 9th edition Slide 1 Chapter 9: One- and Two- Sample Estimation ...

EGR 252 Ch. 9 Lecture1 JMB Fall 2011 9th edition Slide 1

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.


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

n

zXn

zX

2/2/

Z

1 -


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?


Example (cont.)

X = ______________ σ = _______________

α = ________________ α/2 = _____________

Z0.025 = _____________ Z0.975 = ____________

Solution: 12.416 < μSTOP TIME < 13.984

Z Z

Z0.025 = -1.96 Z0.975 = 1.9613.2-(1.96)(2/sqrt(25)) = 12.416 13.2+(1.96)(2/sqrt(25)) = 13.984


Your turn …

What is the 90% C.I.? What does it mean?

-5 -4 -3 -2 -1 0 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.

90% 5%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:

2. If σ 2 is unknown and n is not “large”, we must use the t-statistic.

n

szX 2/


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:

n

stX

n

stX nn 1,2/1,2/

-5 -4 -3 -2 -1 0 1 2 3 4 5


Recall Our ExampleA 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?


Small Sample Example (cont.)

n = _______ df = _______ X = ______ s = _______

α = _______ α/2 = ____ t0.025,24 = _______

_______________ < μ < ________________

-5 -4 -3 -2 -1 0 1 2 3 4 5

13.2 - (2.064)(2/sqrt(25)) = 13.374 13.2 + (2.064)(2/sqrt(25)) = 14.026


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.


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


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

n

n

s


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

nzXx

nzX

11

11 2/02/

nstXx

nstX nn

11

11 1,2/01,2/


9.7: Tolerance Limits

For a normal distribution of unknown mean μ, and unknown standard deviation σ, tolerance limits are given by

x + kswhere 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.


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.


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 Fall 2011 9th edition Slide 1 Chapter 9: One- and Two- Sample Estimation ...

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