Simple Comparative Experiments

8/10/2019 Simple Comparative Experiments

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Simple Comparative Experiments

Simple Comparative Experiments

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Compare two conditions (sometimes called treatments)

Illustration

The tension bond strength of Portland cement mortar is an important

characteristic of the product. An engineer is interested in comparing

the strength of a modified formulation in which polymer latex

emulsions have been added during mixing to the strength of the

unmodified mortar. The experimenter has collected 10 observations

on strength for the modified formulation and another 10 observations

for the unmodified formulation. The data are shown in the table.

The two different formulations are referred to as two treatments or as

two levels of the factor formulations


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Tension Bond Strength of Portland Cement

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Modified Mortar Unmodified Mortar

16.85 17.50

16.40 17.63

17.21 18.25

16.35 18.00

16.52 17.86

17.04 17.75

16.96 18.22

17.15 17.90

16.59 17.96

16.57 18.15

Basic Concepts

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Each observation in the Portland cement experiment

would be called a run

The individual runs differ, so there is fluctuation, or noise,

in the results.

This noise is usually called experimental error or simply

error.

It is a statistical error, meaning that it arises from variation

that is uncontrolled and generally unavoidable.

The presence of error or noise implies that the response

variable, tension bond strength, is a random variable.

A random variable may be either discrete or continuous.


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Graphical Description Dot Plot

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Mean Strength

Modified Mortar 16.764

Unmodified Mortar 17.922

Do the two samples differ by a

non-trivial amount?

Shows central tendency as

well as dispersion.

Graphical Description - Boxplot

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Basic Concepts Expected Value & Variance

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Basic Concepts

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Basic Concepts

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If y1 and y2 are independent,

But, in general

Sampling and Estimators

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Most statistical methods assume that

random samples are used

If every element in the population has

an equal probability of being chosen,

then the procedure employed is called

random sampling

Estimators

An estimator of an unknown parameter

is a statistic that corresponds to that

parameter

Note that a point estimator is a random

variable

is a point es timatorof and s is a

point estimator of

Properties of estimators

The point estimator should be unbiased.

The long-run average or expected value of

the point estimator should be the

parameter that is being estimated.

An unbiased estimator should have

minimum variance.

This property states that the minimum

variance point estimator has a variance

that is smaller than the variance of any

other estimator of that parameter.

The probability distribution of a

statistic is called a sampling

distribution


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The Normal Distribution

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Sample runs that differ as a result of experimental error

often are well described by the normal distribution

The Central Limit Theorem

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This result states essentially that the sum of n independent and identically

distributed random variables is approximately normally distributed

We think of the error in an experiment as arising in an additive manner from

several independent sources; consequently, the normal distribution

becomes a plausible model for the combined experimental error


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The

distribution

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If z1, z2,,zkare normally and independently distributed

random variables with mean 0 and variance 1, then the

random variable

follows a chi-squared distribution with kdegrees of

freedom with

The distribution

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The t distribution

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If z and are independent standard normal and chi-

square random variables, respectively, the random

variable

follows the t distribution with k degrees of freedom,

denoted tk

Mean and variance of t are = 0 and =

for k > 2

The t distribution

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is distributed as t with n - 1 degrees of freedom.


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The F distribution

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If and

are two independent chi-square random

variables with u and v degrees of freedom, respectively,

then the ratio

follows the Fdistribution with u numerator degrees of

freedom and v denominator degrees of freedom

The F distribution

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Randomized Designs Inferences About Means

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Assume that a completely randomized experimental design

is used

In such a design, the data are viewed as if they were a random

sample from a normal distribution.

Recall the Portland cement data

A Model for the Data

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Often describe the results of an experiment with a model

yij is thejth observation from factor level i

is the mean of the response at the ith factor level

is a normal random variable associated with the ijth

observation

We assume that are NID(0,), i= 1, 2

is the random error componentof the model

Because the means and are constants, yijare NID(,)


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Hypotheses

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A statement either about the parameters of a probability distribution or theparameters of a model

Reflects some conjecture about the problem situation

In the Portland cement experiment, we may think that

The mean tension bond strength of the modified mortar formulation is equal to acertain value

The mean tension bond strengths of the two mortar formulations are equal

Null and alternative hypotheses

Type I ( is the significance level) and type II errors

Power

The Portland Cement Example Summary Statistics

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How the Two-Sample t-Test Works:

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How the Two-Sample t-Test Works

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How the Two-Sample t-Test Works:

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Values of t0 that are near zero are consistent with the null hypothesis

Values of t0 that are very different from zero are consistent with the

alternative hypothesis

t0 is a distance measure-how far apart the averages are expressed

in standard deviation units

Notice the interpretation of t0 as a signal-to-noise ratio

The Two-Sample (Pooled) t-Test

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So far, we havent really doneany statistics

We need an objective basis fordeciding how large the teststatistic t0 really is

In 1908, W. S. Gosset derivedthe referencedistributionfor t0 called the tdistribution

(tables of the tdistribution inthe textbook.)

t0 = -2.20


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A value of t0 between -2.101and +2.101 is consistent withequality of means

It is possible for the means tobe equal and t0 to exceedeither 2.101 or 2.101, but it

would be a rareevent leads to the conclusion thatthe means are different

Could also use the P-valueapproach

t0 = -2.20


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The P-value is the risk of wrongly rejecting the null hypothesisof equal means (it measures rareness of the event)

The P-value in our problem is P= 0.042

t0 = -2.20

Minitab Two-Sample t-Test Results

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Two-Sample T-Test and CI: Modified Mortar, Unmodified Mortar

Two- sampl e T f or Modi f i ed Mort ar vs Unmodi f i ed Mor t ar

N Mean StDev SE MeanModi f i ed Mort ar 10 16. 764 0. 316 0. 10Unmodi f i ed Mor t ar 10 17. 922 0. 248 0. 078

Difference = (Modified Mortar) - (Unmodified Mortar)Est i mat e f or di f f erence: - 1. 15895% CI f or di f f erence: ( - 1. 426, - 0. 890)T- Test of difference = 0 (vs ): T- Val ue = - 9.11 P-Val ue = 0.000 DF = 17


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Assumptions of the t-test

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Both samples are drawn from independent populations

Populations can be described by a normal distribution

The standard deviation or variances of both populations

are equal

The observations are independent random variables

The assumption of independence is critical, and if the run order

is randomized (and, if appropriate, other experimental units

and materials are selected at random) this assumption will

usually be satisfied

The equal-variance and normality assumptions are easy to

check using a normal probability plot

Probability Plot

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Observations in the sample are first ranked from smallest to

largest

The ordered observations Y(j)are then plotted against their

observed cumulative frequency (j - 0.5)/n.

If the hypothesized distribution adequately describes the data,

the plotted points will fall approximately along a straight line

Usually, the determination of whether or not the data plot as a

straight line is subjective

The assumption of equal population variances can be checked

by simply comparing the slopes of the two straight lines


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Normal Probability Plot for the Cement Example

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What can you conclude?

Choice of sample size

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Suppose we are testing the following hypothesis

and that the means are not equal so that = The probability of type II error depends on the true difference

in means

A graph of versus for a particular sample size is called theoperating characteristic curve, or O.C. curve for the test

The error is also a function of sample size.

For a given value of , the error decreases as the sample sizeincreases

A specified difference in means is easier to detect for larger samplesizes than for smaller ones


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Operating characteristic curves for the two-sided t-test

with = 0.05

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The greater the difference inmeans, , the smallerthe probability of type II errorfor a given sample size and .

That is, for a specified samplesize and , the test will detectlarge differences more easilythan small ones.

As the sample size gets larger,the probability of type II errorgets smaller for a givendifference in means and .

That is, to detect a specified

difference , we may make thetest more powerful byincreasing the sample size.

=| |

2 =

||

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Example 1

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Analysis of a random sample consisting of n1 = 20 specimens

of cold rolled steel to determine yield strengths resulted in a

sample average strength of 29.8 ksi. A second random

sample of n2 = 25 two-sided galvanized steel specimens gave

a sample average strength of 34.7 ksi. Assuming that the two

yield strength distributions are normal with s1 = 4.0 and s2 =5.0, does the data indicate that the corresponding average

yield strengths 1 and 2 are different? Use a significance

level of = 0.01.

Table of the t distribution


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Example 2

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A hardness testing machine presses a rod with a pointed tipinto a metal specimen with known force. By measuring thedepth of the depression caused by the tip, the hardness ofthe specimen is determined. Two different tips are availablefor the machine, and although the precision (variability) ofthe measurements seem to be the same, is suspected thatone tip produces different hardness readings from the other.

Ten specimens were tested and the hardness readings were

obtained as shown in the table.

What do you conclude?

Example 2 Data

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Example 2 (contd.)

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The metal specimens chosen for testing are from different

bar stock that are produced in different heats and are not

exactly homogenous in some other way that affects

hardness. This lack of homogeneity will contribute to the

variance and will inflate the experimental error.

Alternative Experimental Design for Example 2

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Assume that each specimen is large enough so that two

hardness determinations may be made on it.

This alternative design would consist of dividing each

specimen into two parts, then randomly assigning one tip

to one-half of each specimen and the other tip to the

remaining half.

The order in which the tips are tested for a particular

specimen would also be randomly selected.

The mathematical model would be as follows


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The Paired Comparison Design (The Paired t-test)

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Testing := is equivalent to testing

The test statistic would be

H0 would be rejected if > ,

Paired t-test Data

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The Paired Comparison Design

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Special case of a more general type of design called the randomized block design

Block refers to a relatively homogeneous experimental unit (in our case, the metal specimens

are the blocks)

The block represents a restriction on complete randomization because the treatment

combinations are only randomized within the block

Note that, although 2n = 2(10) = 20 observations have been taken, only n - 1 = 9

degrees of freedom are available for the t statistic

As the degrees of freedom for t increase the test becomes more sensitive

By blocking or pairing, we have effectively lost n 1 degrees of freedom, but we

hope we have gained a better knowledge of the situation by eliminating anadditional source of variability (the difference between specimens).

Minitab Output

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Two-Sample T-Test and CI: Tip 1, Tip 2

Two- sampl e T f or Ti p 1 vs Ti p 2

N Mean St Dev SE MeanTi p 1 10 4. 80 2. 39 0. 76Ti p 2 10 4. 90 2. 23 0. 71

Di f f erence = (Ti p 1) - (Ti p 2)Esti mate f or di f f erence: - 0. 1095% CI f or di f f erence: (- 2. 28, 2.08)

T- Test of difference = 0 (vs ): T- Val ue = - 0. 10 P- Val ue = 0. 924 DF = 18Both use Pool ed StDev = 2. 3154

Paired T-Test and CI: Tip 1, Tip 2

Pai red T f or Ti p 1 - Tip 2

N Mean StDev SE MeanTi p 1 10 4. 800 2. 394 0. 757Ti p 2 10 4. 900 2. 234 0. 706Di f f erence 10 - 0. 100 1. 197 0.379

95% CI f or mean di f f erence: ( - 0. 956, 0. 756)T- Test of mean difference = 0 (vs 0): T - Val ue = - 0.26 P-Val ue = 0. 798


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Inferences About Variances

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Sensitive to the normality assumption

The single sample test

Test statistic

The null hypothesis is rejected if

> ,

or 0.01. Use an = 0.05

level of significance. What assumptions are required for

this test?

Table of the Chi Sq distribution


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Two Sample Test for Variance

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Assumes normality of both populations

Test statistic

The null hypothesis is rejected if

> ,, or < ,,

Note that

Illustration

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A chemical engineer is investigating the inherent

variability of two types of test equipment that can be used

to monitor the output of a production process. He

suspects that the old equipment, type 1, has a larger

variance than the new one. Two random samples of n1 =

12 and n2 = 10 observations are taken, and the samplevariances are S1 = 14.5 and S2 = 10.8. What do you

conclude?

F0.05,11,9 = 3.10

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Simple Comparative Experiments