stat review -keller

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Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc

!hat is "tatistics#

Statistics is a way to get information from data

Data

Statistics

Information

Data: Facts, especially

numerical facts, collected

together for reference or

information.

Definitions: Oxford English Dictionary

Information: Knowledge

communicated concerning

some particular fact.

Statistics is atoolfor creatingnew understandingfrom a set of

numbers.


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$ey "tatistical Concepts%

Population

apopulationis the group of allitems of interest toa statistics practitioner.

frequently very large; sometimes infinite.

E.g. All 5 million Florida voters, per Example 12.5

Sample

Asampleis a set of data drawn from thepopulation.

Potentially very large, but less than the population.

E.g. a sample of 765 voters exit polled on election day.


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$ey "tatistical Concepts%

Populations have Parameters,

Samples have Statistics.

&arameter

&op'lation "ample

"tatistic

"'(set


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)escriptive "tatistics%

aremethodsof organizing, summarizing, and presenting

data in a convenient and informative way. These methodsinclude:

Graphical Techniques (Chapter 2), and

Numerical Techniques (Chapter 4).

The actual method used depends on what informationwe

would like to extract. Are we interested in

measure(s) of central location? and/or

measure(s) of variability (dispersion)?

Descriptive Statistics helps to answer these questions


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"tatistical Inference%

Statistical inferenceis theprocessof making an estimate,

prediction, or decision about a population based on a sample.

&arameter

&op'lation

"ample

"tatistic

Inference

What can we inferabout a Populations Parameters

based on a Samples Statistics?


7/209Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc

)e*nitions%

A variableis some characteristic of a population or sample.

E.g. student grades.

Typically denoted with a capital letter: X, Y, Z

The valuesof the variable are the range of possible valuesfor a variable.

E.g. student marks (0..100)

Dataare theobserved valuesof a variable.

E.g. student marks: {67, 74, 71, 83, 93, 55, 48}



Interval )ata%

Intervaldata

Real numbers, i.e. heights, weights, prices, etc.

Also referred to as quantitative or numerical.

Arithmetic operations can be performed on Interval Data,thus its meaningful to talk about 2*Height, or Price + $1,

and so on.


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rdinal )ata%

OrdinalDataappear to be categorical in nature, but their

values have anorder; a ranking to them:

E.g. College course rating system:

poor = 1, fair = 2, good = 3, very good = 4, excellent = 5

While its still not meaningful to do arithmetic on this data

(e.g. does 2*fair = very good?!), we can say things like:

excellent > poor or fair < very good

That is, order is maintained no matter what numeric values

are assigned to each category.



-raphical . Ta('lar Techni'es for +ominal)ata%The only allowable calculation on nominal data is to count

the frequency of each value of the variable.

We can summarize the data in a table that presents the

categories and their counts called afrequency distribution.

Arelative frequency distributionlists the categories and the

proportion with which each occurs.

Refer to Example 2.1
http://e/TT%20PowerPoint%20slides/References/Xm02-01.xlshttp://e/TT%20PowerPoint%20slides/References/Xm02-01.xls



+ominal )ata Ta('lar "'mmary1



+ominal )ata re'ency1

Bar Charts are often used to displayfrequencies



+ominal )ata

It all the same information,(based on the samedata).

Just differentpresentation.



-raphical Techni'es for Interval)ataThere are several graphical methods that are used when the

data are interval(i.e. numeric, non-categorical).

The most important of these graphical methods is the

histogram.

The histogram is not only a powerful graphical technique

used tosummarizeinterval data, but it is also used to help

explainprobabilities.



B'ilding a 3istogram%

1) Collect the Data

2) Create a frequency distribution for the data. 3) Draw the Histogram.



3istogram and "tem . Leaf%



give%

Is a graph of acumulativefrequency distribution.

We create an ogive in three steps

1) Calculate relative frequencies.

2) Calculatecumulative relative frequenciesby adding thecurrent class relative frequency to the previous classcumulative relative frequency.

(For the first class, its cumulative relative frequency is just its relative frequency)


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C'm'lative 4elative re'encies%

first class

next class: .355+.185=.540

last class: .930+.070=1.00

:

:


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give%

The ogive can (e 'sedto anser 'estions

like6

!hat telephone (ill

val'e is at the 50thpercentile#

(Refer also to Fig. 2.13 in your textbook)7aro'nd 895:


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"catter )iagram%

Example 2.9A real estate agent wanted to know to what

extent the selling price of a home is related to its size

1) Collect the data

2) Determine the independent variable (X house size) andthe dependent variable (Y selling price)

3) Use Excel to create a scatter diagram


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"catter )iagram%

It appears that in fact there is a relationship, that is, the

greater the house size the greater the selling price


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&atterns of "catter )iagrams%

Linearity and Direction are two concepts we are interested in

&ositive Linear 4elationship +egative Linear 4elationship

!eak or +on;Linear 4elationship


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Time "eries )ata%

Observations measured at the same point in time are called

cross-sectionaldata.

Observations measured at successive points in time are

calledtime-seriesdata.

Time-series data graphed on a line chart, which plots the

value of the variable on the vertical axis against the time

periods on the horizontal axis.


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+'merical )escriptive Techni'es%

Measures of Central Location

Mean, Median, Mode

Measures of Variability

Range, Standard Deviation, Variance, Coefficient of Variation

Measures of Relative Standing

Percentiles, Quartiles

Measures of Linear Relationship

Covariance, Correlation, Least Squares Line


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>rithmetic


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"tatistics is a pattern lang'age%

Population Sample

Size N n

Mean


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The >rithmetic


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4ange%

Therangeis the simplest measure of variability, calculated

as:

Range = Largest observation Smallest observation

E.g.

Data: {4, 4, 4, 4, 50} Range = 46

Data: {4, 8, 15, 24, 39, 50} Range = 46The range is the same in both cases,

but the data sets have very different distributions


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"tatistics is a pattern lang'age%

Population Sample

Size N n

Mean

Variance


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?ariance%

The variance of a populationis:

The variance of a sampleis:

pop'lation mean

sample mean

+oteA the denominator is sample sie n1 min's one A

pop'lation sie


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>pplication%

Example 4.7. The following sample consists of the number

of jobs six randomly selected students applied for: 17, 15,23, 7, 9, 13.

Finds its mean and variance.

What are we looking to calculate?

The following sampleconsists of the number of jobs six

randomly selected students applied for: 17, 15, 23, 7, 9, 13.

Finds its meanand variance.

as opposed to or 2


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"ample


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"tandard )eviation%

The standard deviation is simply the square root of the

variance, thus:

Population standard deviation:

Sample standard deviation:


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"tandard )eviation%

Consider Example 4.8where a golf club manufacturer has

designed a new club and wants to determine if it is hit moreconsistently (i.e. with less variability) than with an old club.

UsingTools )ata >nalysisDmay need to 7add in:% )escriptive

"tatisticsin Excel, we produce the following tables for

interpretation

You get moreconsistent

distance with the

new club.


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The Empirical 4'le% If the histogram isbellshaped

>ppro@imately FGH of all o(servations fall

ithin onestandard deviation of the mean

>ppro@imately 5H of all o(servations fall

ithin twostandard deviations of the mean

>ppro@imately JH of all o(servations fall

ithin threestandard deviations of the mean

Ch ( h K Th + t f d ( i l i


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Che(ysheKs Theorem%+otoften 'sed (eca'se interval isvery ide

A more general interpretation of the standard deviation is

derived from Chebysheffs Theorem, which applies to allshapes of histograms (not just bell shaped).

The proportion of observations in any sample that lie

within kstandard deviations of the mean is at least:

or k=2 say1, the theoremstates that at least9/M of all

o(servations lie ithin 2standard deviations of themean This is a 7loer (o'nd:compared to Empirical 4'lesappro@imation 5H1

l


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Bo@ &lots%

These box plots are based on

data in Xm04-15.

Wendys service time is

shortest and least variable.

Hardees has the greatest

variability, while Jack-in-

the-Box has the longest

service times.

< h d f C ll i )


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"ampling%

Recall that statistical inference permits us to draw

conclusions about a population based on a sample.

Sampling (i.e. selecting a sub-set of a whole population) is

often done for reasons ofcost(its less expensive to sample

1,000 television viewers than 100 million TV viewers) and

practicality(e.g. performing a crash test on every

automobile produced is impractical).

In any case, thesampled populationand thetarget

populationshould be similarto one another.

" li &l


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"ampling &lans%

Asampling planis just a method or procedure for

specifying how a sample will be taken from a population.

We will focus our attention on these three methods:

Simple Random Sampling,

Stratified Random Sampling, and

Cluster Sampling.

"i l 4 d " li


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"imple 4andom "ampling%

Asimple random sampleis a sample selected in such a way

that every possible sample of the same size is equally likelyto be chosen.

Drawing three names from a hat containing all the names of

the students in the class is an example of a simple random

sample: any group of three names is as equally likely as

picking any other group of three names.

"t ti* d 4 d " li


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"trati*ed 4andom "ampling%

After the population has been stratified, we can usesimple

random samplingto generate the complete sample:

e only have s'Ncient reso'rces to sample M00 people total,

e o'ld dra O00 of them from the lo income gro'p%

%if e are sampling O000 people, ed dra50 of them from the high income gro'p

Cl t " li


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Cl'ster "ampling%

Acluster sampleis a simple random sample of groups or

clusters of elements (vs. a simple random sample ofindividual objects).

This method is useful when it is difficult or costly to develop

a complete list of the population members or when the

population elements are widely dispersed geographically.

Cluster sampling may increase sampling error due tosimilarities among cluster members.

" li E


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"ampling Error%

Sampling errorrefers to differences between the sample and

the population that exist only because of the observationsthat happened to be selected for the sample.

Another way to look at this is: the differences in results for

different samples (of the same size) is due to sampling error:

E.g. Two samples of size 10 of 1,000 households. If we

happened to get the highest income level data points in ourfirst sample and all the lowest income levels in the second,

this delta is due to sampling error.

+ li E


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+onsampling Error%

Nonsampling errorsare more serious and are due to

mistakes made in the acquisition of data or due to the sampleobservations being selected improperly.Three types of

nonsampling errors:

Errors in data acquisition,

Nonresponse errors, and

Selection bias.

Note: increasing the sample size will notreduce this type of

error.

> h t > i i


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>pproaches to >ssigning&ro(a(ilities%There are three ways to assign a probability, P(Oi), to an

outcome, Oi, namely:

Classical approach: make certain assumptions (such asequally likely, independence) about situation.

Relative frequency: assigning probabilities based onexperimentation or historical data.

Subjective approach: Assigning probabilities based on theassignors judgment.

Interpreting &ro(a(ility


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Interpreting &ro(a(ility%

One way to interpret probability is this:

If a random experiment is repeated an infinitenumber of

times, the relative frequency for any given outcome is the

probability of this outcome.

For example, the probability of heads in flip of a balanced

coin is .5, determined using the classical approach. The

probability is interpreted as being the long-term relativefrequency of heads if the coin is flipped an infinite number

of times.

Conditional &ro(a(ility


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Conditional &ro(a(ility%

Conditional probabilityis used to determine how two events

are related; that is, we can determine the probability of oneeventgiventhe occurrence of another related event.

Conditional probabilities are written as P(A | B)and read as

the probability of A givenB and is calculated as:

Independence


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Independence%

One of the objectives of calculating conditional probability

is to determine whether two events are related.

In particular, we would like to know whether they areindependent, that is, if the probability of one event isnot

affectedby the occurrence of the other event.

Two events A and B are said to be independentif

P(A|B) = P(A)or

P(B|A) = P(B)

Complement 4'le


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Complement 4'le%

The complement of an event A is the event that occurs when

A does not occur.

Thecomplement rulegives us the probability of an event

NOT occurring. That is:

P(AC) = 1 P(A)

For example, in the simple roll of a die, the probability of the

number 1 being rolled is 1/6. The probability that some

number other than 1 will be rolled is 1 1/6 = 5/6.


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ddition 4'le


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>ddition 4'le%

Recall: theaddition rulewas introduced earlier to provide a

way to compute the probability of event AorBorboth Aand B occurring; i.e. the union of A and B.

P(A or B) = P(A) + P(B) P(A and B)

Why do we subtract the joint probability P(A and B) from

the sum of the probabilities of A and B?

P(A or B) = P(A) + P(B) P(A and B)

>ddition 4'le for


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>ddition 4'le for


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To Types of 4andom ?aria(les%

DiscreteRandom Variable

one that takes on acountablenumber of values E.g. values on the roll of dice: 2, 3, 4, , 12

ContinuousRandom Variable one whose values arenot discrete, not countable

E.g. time (30.1 minutes? 30.10000001 minutes?)

Analogy:

Integers are Discrete, while Real Numbers are Continuous

Las of E@pected ?al'e


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Las of E@pected ?al'e%

1. E(c) = c

The expected value of a constant (c) is just the value of theconstant.

2. E(X + c) = E(X) + c

3. E(cX) = cE(X)

We can pull a constant out of the expected value expression

(either as part of a sum with a random variable X or as a coefficient

of random variable X).

Las of ?ariance


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Las of ?ariance%

1. V(c) = 0

The variance of a constant (c) is zero.

2. V(X + c) = V(X)

The variance of a random variable and a constant is just the

variance of the random variable (per 1 above).

3. V(cX) = c2V(X)

The variance of a random variable and a constant coefficient isthe coefficient squared times the variance of the random variable.

Binomial )istri('tion


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Binomial )istri('tion%

Thebinomial distributionis the probability distribution that

results from doing a binomial experiment. Binomialexperiments have the following properties:

1. Fixed number of trials, represented as n.

2. Each trial has two possible outcomes, a success and a

failure.

3. P(success)=p (and thus: P(failure)=1p), for all trials.

4. The trials are independent, which means that the

outcome of one trial does not affect the outcomes of any

other trials.

Binomial 4andom ?aria(le


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Binomial 4andom ?aria(le%

The binomial random variablecountsthe number of

successes in ntrials of the binomial experiment. It can takeon values from 0, 1, 2, , n. Thus, its adiscreterandom

variable.

To calculate the probability associated with each value we

use combintorics:

for x=0, 1, 2, , n

Binomial Ta(le


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Binomial Ta(le%

What is the probability that Pat fails the quiz?

i.e. what is P(X 4), given P(success) = .20and n=10 ?

P(X 4) = .967

Binomial Ta(le


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Binomial Ta(le%

What is the probability that Pat getstwo answerscorrect?

i.e. what is P(X = 2), given P(success) = .20and n=10 ?

P(X = 2) = P(X2) P(X1) = .678 .376 = .302

remem(er, the ta(le shos cumulativepro(a(ilities%

=BI+


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=BI+


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=BI+


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Binomial )istri('tion%

As you might expect, statisticians have developed general

formulas for the mean, variance, and standard deviation of abinomial random variable. They are:

&oisson )istri('tion


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&oisson )istri('tion%

Named for Simeon Poisson, thePoisson distributionis a

discrete probability distribution and refers to the number ofevents (a.k.a. successes) within a specific time period or

region of space. For example:

The number of cars arriving at a service station in 1 hour. (The

interval of time is 1 hour.)The number of flaws in a bolt of cloth. (The specific region is a

bolt of cloth.)

The number of accidents in 1 day on a particular stretch of

highway. (The interval is defined by both time, 1 day, and space,the particular stretch of highway.)

The &oisson E@periment


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The &oisson E@periment%

Like a binomial experiment, aPoisson experimenthas four

defining characteristic properties:1. The number of successes that occur in any interval is

independent of the number of successes that occur in any

other interval.

2. The probability of a success in an interval is the same for

all equal-size intervals

3. The probability of a success is proportional to the size of

the interval.4. The probability of more than one success in an interval

approaches 0 as the interval becomes smaller.



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ThePoisson random variableis the number of successes

that occur in a period of time or an interval of space in aPoisson experiment.

E.g. On average, 96trucks arrive at a border crossing

every hour.

E.g. The number of typographic errors in a new textbook

edition averages 1.5per 100 pages.

s'ccesses

timeperiod

s'ccesses#A1

interval

&oisson &ro(a(ility )istri('tion


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&oisson &ro(a(ility )istri('tion%

The probability that a Poisson random variable assumes a

value of xis given by:

and eis the natural logarithm base.

FYI:

E@ample J O2


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E@ample JO2%

The number of typographical errors in new editions of

textbooks varies considerably from book to book. Aftersome analysis he concludes that the number of errors is

Poisson distributed with a mean of 1.5 per 100 pages. The

instructor randomly selects 100 pages of a new book. What

is the probability that there are no typos?

That is, what is P(X=0) given that = 1.5?

There is about a 22% chance of finding zero errors



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As mentioned on the Poisson experiment slide:

The probability of a success is proportional to the size of

the interval

Thus, knowing an error rate of 1.5 typos per 100 pages, we

can determine a mean value for a 400 page book as:

=1.5(4) = 6 typos / 400 pages.

E@ample JO9%


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E@ample JO9%

For a 400 page book, what is the probability that there are

notypos?

P(X=0) =

there is a very small chance there are no typos

E@ample JO9%


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E@ample JO9%

Excelis an even better alternative:

&ro(a(ility )ensity 'nctions%
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&ro(a(ility )ensity 'nctions%

Unlike a discrete random variable which we studied in

Chapter 7, acontinuous random variableis one that canassume an uncountablenumber of values.

We cannot list the possible values because there is an

infinite number of them.

Because there is an infinite number of values, the

probability of each individual value is virtually 0.

&oint &ro(a(ilities are Sero


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&oint &ro(a(ilities are Sero

Because there is an infinite number of values, the

probability of each individual value is virtually 0.

Thus, we can determine the probability of arange of values

only.

E.g. with a discreterandom variable like tossing a die, it is

meaningful to talk about P(X=5), say.

In a continuoussetting (e.g. with time as a random variable), theprobability the random variable of interest, say task length, takes

exactly5 minutes is infinitesimally small, hence P(X=5) = 0.

It is meaningful to talk about P(X 5).

&ro(a(ility )ensity 'nction%


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&ro(a(ility )ensity 'nction%

A function f(x) is called aprobability density function(over

the range a

x

bif it meets the followingrequirements:

1) f(x) 0 for all xbetween aand b, and

2) The total area under the curve between aand bis 1.0

f(x)

xba

area=1

The +ormal )istri('tion%


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Thenormal distributionis the most important of all

probability distributions. The probability density function ofanormal random variableis given by:

It looks like this:

Bell shaped,

Symmetrical around the mean



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e o a s (' o

Important things to note:

The normal distribution is fully defined by two parameters:its standard deviation andmean

Unlike the range of the uniform distribution (a x b)

Normal distributionsrange from minus infinity to plus infinity

The normal distribution is bell shaped andsymmetrical about the mean

"tandard +ormal )istri('tion%


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A normal distribution whose mean is zeroand standard

deviation is oneis called thestandard normal distribution.

As we shall see shortly, any normal distribution can be

convertedto a standard normal distribution with simple

algebra. This makes calculations much easier.

0

O

O

Calc'lating +ormal &ro(a(ilities%


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g

We can use the following function to convert any normal

random variable to a standardnormal random variable

"ome advice6alays dra a

pict'reA

0



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g

Example: The time required to build a computer isnormally

distributedwith a mean of 50 minutes and a standarddeviation of 10 minutes:

What is the probability that a computer is assembled in a

time between 45 and 60 minutes?

Algebraically speaking, what is P(45 < X < 60)?

0



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g

P(45 < X < 60)?

0

mean of 50 minutes and a

standard deviation of 10 minutes



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g

We can use Table 3in

Appendix B to look-upprobabilities P(0 < Z < z)

We can break up P(.5 < Z < 1)into:

P(.5 < Z < 0)+ P(0 < Z < 1)

The distribution issymmetricaround zero, so we have:P(.5 < Z < 0) = P(0 < Z < .5)

Hence: P(.5 < Z < 1) = P(0 < Z < .5)+ P(0 < Z < 1)

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g

How to use Table 3

This table gives probabilitiesP(0 < Z < z)

First column = integer + first decimal

Top row = second decimal place

P(0 < Z < 0.5)

P(0 < Z < 1)

P(.5 < Z < 1) = .1915 + .3414 = .5328

sing the +ormal Ta(le Ta(le 91%
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g

What is P(Z > 1.6)?

0 OF

&0 U S U OF1 = MM52

&S OF1 = 5 V &0 U S U OF1= 5 V MM52

= .0548

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g

What is P(Z < -2.23)?

0 229

&0 U S U 2291

&S U ;2291 = &S 2291= 5 V &0 U S U 2291

= .0129

;229

&S 2291&S U ;2291



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g

What is P(Z < 1.52)?

0 O52

&S U 01 = 5

&S U O521 = 5 W &0 U S U O521= 5 W M95J

= .9357

&0 U S U O521



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g

What is P(0.9 < Z < 1.9)?

0 0

&0 U S U 01

&0 U S U O1 = &0 U S U O1 V &0 U S U 01=MJO9 V 9O5

= .1554

O

&0 U S U O1

inding ?al'es of S%
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Other Z values are

Z.05= 1.645Z.01= 2.33

sing the val'es of S


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Because z.025= 1.96 and - z.025= -1.96, it follows that we can

state

P(-1.96 < Z < 1.96) = .95

Similarly

P(-1.645 < Z < 1.645) = .90

ther Contin'o's )istri('tions%


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Three other important continuous distributions which will be

used extensively in later sections are introduced here:

StudenttDistribution,

Chi-Squared Distribution, and

FDistribution.

"t'dent t)istri('tion%


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Here the lettertis used to represent the random variable,

hence the name. The density function for the Studenttdistribution is as follows

(nu) is calledthe degrees of freedom, and

(Gamma function) is (k)=(k-1)(k-2)(2)(1)

"t'dent t)istri('tion%


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In much the same way that and define the normal

distribution, , the degrees of freedom, defines the StudenttDistribution:

As the number of degrees of freedom increases, thet

distribution approaches the standard normal distribution.

ig're G2M

)etermining "t'dent t?al'es%


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The studenttdistribution is used extensively in statistical

inference. Table 4in Appendix B lists values of

That is, values of a Studenttrandom variable with degrees

of freedom such that:

The values for A are pre-determined

critical values, typically in the

10%, 5%, 2.5%, 1% and 1/2% range.

sing the tta(le Ta(le M1 for
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val'es%For example, if we want the value oftwith 10 degrees of

freedom such that the area under the Studenttcurve is .05:>rea 'nder the c'rve val'e t>1 6 CL


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TheFdensity function is given by:

F > 0. Two parameters define this distribution, and like

weve already seen these are againdegrees of freedom.

is the numerator degrees of freedom and

is the denominator degrees of freedom.

)etermining ?al'es of F%


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For example, what is the value of F for 5% of the area under

the right hand tail of the curve, with a numerator degree offreedom of 3 and a denominator degree of freedom of 7?

Solution: use the F look-up (Table 6)

+'merator )egrees of reedom 6 CL


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For areas under the curve on the left hand side of the curve,

we can leverage the following relationship:

Pay close attention to the order of the terms!


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OO00Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc

Chapter 9

Samplin !istributions

"ampling )istri('tion of the


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" O 2 9 M 5 F

P#"$ O/F O/F O/F O/F O/F O/F

A fair dieis thrown infinitely many times,

with the random variable X = # of spots on any throw.

The probability distribution of X is:

and the mean and variance are calculated as well:

"ampling )istri('tion of To )ice


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A sampling distribution is created by looking at

all samples of size n=2 (i.e. two dice) and their means

While there are 36 possible samples of size 2, there are only

11 values for , and some (e.g. =3.5) occur more

frequently than others (e.g. =1).

"ampling )istri('tion of To )ice%


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Thesampling distributionof is shown below:

1.0 1/36

1.5 2/36

2.0 3/36

2.5 4/36

3.0 5/36

3.5 6/36

4.0 5/36

4.5 4/36

5.0 3/36

5.5 2/36

6.0 1/36

P( )

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

F/9F

5/9F

M/9F

9/9F

2/9F

O/9F

P

(

)

Compare%


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Compare the distribution of X

with the sampling distribution of .

As well, note that:

1 2 3 4 5 6 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

Central Limit Theorem%


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The sampling distribution of the mean of a random sample

drawn from any population isapproximately normalfor asufficiently large sample size.

The larger the sample size, the more closely the sampling

distribution of X will resemble a normal distribution.

Central Limit Theorem%


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If the population is normal, then X is normally distributed

for all values of n.

If the population is non-normal, then X is approximately

normal only for larger values of n.

In many practical situations, a sample size of 30 may be

sufficiently large to allow us to use the normal distribution

as an approximation for the sampling distribution of X.

"ampling )istri('tion of the "ample


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108/209


The foreman of a bottling plant has observed that the amount

of soda in each 32-ounce bottle is actually a normallydistributed random variable, with a mean of 32.2 ounces and

a standard deviation of .3 ounce.

If a customer buys one bottle, what is the probability that thebottle will contain more than 32 ounces?

E@ample Oa1%


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We want to find P(X > 32), where X is normally distributed

and =32.2 and =.3

there is about a 75% chance that a single bottle of soda

contains more than 32oz.

E@ample O(1%


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The foreman of a bottling plant has observed that the amount

of soda in each 32-ounce bottle is actually a normally

distributed random variable, with a mean of 32.2 ounces and

a standard deviation of .3 ounce.

If a customer buys a carton of fourbottles, what is theprobability that themean amount of the four bottleswill be

greater than 32 ounces?

E@ample O(1%


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We want to find P(X > 32), where X is normally distributed

with =32.2 and =.3

Things we know:

1) X is normally distributed, therefore so will X.

2) = 32.2 oz.

3)

E@ample O(1%


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If a customer buys a carton of fourbottles, what is the

probability that themean amount of the four bottleswill be

greater than 32 ounces?

There is about a 91% chance the mean of the four bottles

will exceed 32oz.

-raphically "peaking%mean=92


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hat is the pro(a(ility that one(ottle ill contain more than 92

o'nces#

hat is the pro(a(ility that themean of fo'r (ottles ill e@ceed 92

o#

2

"ampling )istri('tion6 )iKerence of tomeans


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meansThe final sampling distribution introduced is that of the

differencebetween two sample means. This requires:

independentrandom samples be drawn from each of twonormalpopulations

If this condition is met, then the sampling distribution of thedifferencebetween the two sample means, i.e.

will be normally distributed.

(note: if the two populations are notboth normallydistributed, but the sample sizes are large (>30), thedistribution of is approximatelynormal)

"ampling )istri('tion6 )iKerence of tomeans


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meansThe expected valueand varianceof the sampling

distribution of are given by:

mean:

standard deviation:

(also called the standard error if the difference between two

means)

Estimation%


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There are two types of inference: estimation and hypothesis

testing; estimationis introduced first.

The objective of estimation is to determine theapproximate

valueof a population parameter on the basis of a sample

statistic.

E.g., the sample mean ( ) is employed to estimatethe

population mean ( ).

Estimation%


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The objective of estimation is to determine theapproximate

valueof a population parameter on the basis of a sample

statistic.

There are two types of estimators:

Point Estimator

Interval Estimator

&oint . Interval Estimation%


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For example, suppose we want to estimate the mean summer

income of a class of business students. For n=25 students,

is calculated to be 400 $/week.

point estimate interval estimate

An alternative statement is:

The mean income isbetween380 and 420 $/week.

Estimating hen is knon%the con*dence


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We established in Chapter 9:

Thus, theprobabilitythat the interval:

contains the population mean is 1 . This is aconfidence interval estimator for .

the sample meanis in the center of

the interval%

the con*denceinterval

o'r commonly 'sed con*dencelevels


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levels%Confidence Level

c't . keep handyA

Ta(le O0O

E@ample O0O%


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A computer company samples demand during lead time over

25 time periods:

Its is known that the standard deviation of demand over lead

time is 75 computers. We want to estimate themeandemand

over lead time with 95% confidence in order to set inventorylevels

235 374 309 499 253

421 361 514 462 369

394 439 348 344 330

261 374 302 466 535

386 316 296 332 334

E@ample O0O% C%&C'&%()


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In order to use our confidence interval estimator, we need the

following pieces of data:

therefore:

The lowerand upperconfidence limits are 340.76and 399.56.

370.16

1.96

75

n 25 Given

Calculated from the data

E@ample O0O% *+(),P,)(


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The estimation for the mean demand during lead time liesbetween 340.76 and 399.56 we can use this as input indeveloping an inventory policy.

That is, we estimated that the mean demand during lead time

falls between 340.76 and 399.56, and this type of estimatoris correct 95% of the time. That also means that 5% of thetime the estimator will be incorrect.

Incidentally, the media often refer to the 95% figure as 19times out of 20, which emphasizes the long-runaspect ofthe confidence level.

Interval !idth%


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A wide interval provides little information.

For example, suppose we estimate with 95% confidence thatan accountants average starting salary is between $15,000

and $100,000.

Contrastthis with: a 95% confidence interval estimate of

starting salaries between $42,000 and $45,000.

The second estimate is much narrower, providing accountingstudents more precise information about starting salaries.

Interval !idth%


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The width of the confidence interval estimate is a function of

the confidence level,the populationstandard deviation, and

the sample size

"electing the "ample "ie%


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We can control the width of the interval by determining the

sample size necessary to produce narrow intervals.

Suppose we want to estimate the mean demand to within 5

units; i.e. we want to the interval estimate to be:

Since:

It follows that

Solve for nto get requisite sample size!

"electing the "ample "ie%

S l i h i


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Solving the equation

that is, to produce a 95% confidence interval estimate of the

mean (5 units), we need to sample 865 lead time periods

(vs. the 25 data points we have currently).

"ample "ie to Estimate a


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The general formula for the sample size needed to estimate a

population mean with an interval estimate of:

Requires a sample size of at least this large:

E@ample O02%

A l b i h di f


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A lumber company must estimate the mean diameter of trees

to determine whether or not there is sufficient lumber to

harvest an area of forest. They need to estimate this to within

1 inch at a confidence level of 99%. The tree diameters are

normally distributed with a standard deviation of 6 inches.

How many trees need to be sampled?

E@ample O02%

Thi k


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Things we know:

Confidence level = 99%, therefore =.01

We want , hence W=1.

We are given that = 6.

1

E@ample O02%

W t


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We compute

That is, we will need to sample at least 239 trees to have a

99% confidence interval of 1


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+onstatistical 3ypothesis Testing%

Th t ibl


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There are two possible errors.

A Type I error occurs when we reject a true null hypothesis.That is, a Type I error occurs when the jury convicts an

innocent person.

A Type II error occurs when we dont reject a false nullhypothesis. That occurs when a guilty defendant is acquitted.


Th b bilit f T I i d t d (G k


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The probability of a Type I error is denoted as(Greek

letter alpha). The probability of a type II error is(Greek

letter beta).

The two probabilities are inversely related. Decreasing one

increases the other.


The critical concepts are theses:


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The critical concepts are theses:

1. There are two hypotheses, the null and the alternative

hypotheses.

2. The procedure begins with the assumption that the null

hypothesis is true.

3. The goal is to determine whether there is enough evidence toinfer that the alternative hypothesis is true.

4. There are two possible decisions:

Conclude that there is enough evidence to support the

alternative hypothesis.

Conclude that there is not enough evidence to support the

alternative hypothesis.


5 T o possible errors can be made


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5. Two possible errors can be made.

Type I error: Reject a true null hypothesisType II error: Do not reject a false null hypothesis.

P(Type I error) =

P(Type II error) =

Concepts of 3ypothesis Testing O1%

There are two hypotheses One is called the null hypothesis


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There are twohypotheses. One is called thenull hypothesis

and the other thealternativeorresearch hypothesis. The

usual notation is:

H0: the null hypothesis

H1: the alternative or research hypothesis

The null hypothesis (H0) will always state that theparameter

equals the valuespecified in the alternative hypothesis (H1)

prono'nced

3 7no'ght:

Concepts of 3ypothesis Testing%

Consider Example 10 1 (mean demand for computers during


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Consider Example 10.1 (mean demand for computers during

assembly lead time) again. Rather than estimate the mean

demand, our operations manager wants to know whether the

mean is different from 350 units. We can rephrase this

request into a test of the hypothesis:

H0: X = 350

Thus, our research hypothesis becomes:H1: X 350

This is hat e areinterested in

determining%

Concepts of 3ypothesis Testing M1%

There are two possible decisions that can be made:


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There are twopossible decisions that can be made:

Conclude that there isenough evidenceto support thealternative hypothesis

(also stated as: rejecting the null hypothesis in favor of the

alternative)

Conclude that there is notenough evidenceto support thealternative hypothesis

(also stated as: notrejecting the null hypothesis in favor ofthe alternative)

NOTE: we do notsay that we acceptthe null hypothesis

Concepts of 3ypothesis Testing%

Once the null and alternative hypotheses are stated the next


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Once the null and alternative hypotheses are stated, the next

step is to randomly sample the population and calculate atest

statistic(in this example, the sample mean).

If the test statistics value is inconsistent with the null

hypothesis we reject the null hypothesisand infer that thealternative hypothesis is true.

For example, if were trying to decide whether the mean is

not equal to 350, a large value of (say, 600) would provide

enough evidence. If is close to 350 (say, 355) we could not

say that this provides a great deal of evidence to infer that the

population mean is different than 350.

Types of Errors%

A Type I error occurs when we reject a true null hypothesis


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A Type I error occurs when werejectatruenull hypothesis

(i.e. Reject H0when it is TRUE)

A Type II error occurs when wedont rejectafalsenull

hypothesis (i.e. Do NOT reject H0when it is FALSE)

30 T

4eYect I

4eYect II


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143/209

E@ample OOO%

The system will be cost effective if the mean account balance


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The system will be cost effective if the mean account balance

for all customers is greater than $170.

We express this belief as a our research hypothesis, that is:

H1: > 170 (this is what we want to determine)

Thus, our null hypothesis becomes:

H0: = 170 (this specifies a single value for the

parameter of interest)


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E@ample OOO%

To test our hypotheses we can use two different approaches:


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To test our hypotheses, we can use two different approaches:

Therejection regionapproach (typically used when

computing statistics manually), and

Thep-valueapproach (which is generally used with a

computer and statistical software).

We will explore both in turn


147/209

E@ample OOO%

All thats left to do is calculate and compare it to 170


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All that s left to do is calculate and compare it to 170.

we can calculate this based on any level of

significance ( ) we want


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E@ample OOO% The Big &ict're%


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=175.34

=178

H1: > 170

H0: = 170

Reject H0in favor of

"tandardied Test "tatistic%

An easier method is to use the standardized test statistic:


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An easier method is to use the standardized test statistic:

and compare its result to : (rejection region: z > )

Since z = 2.46 > 1.645 (z.05), we reject H0in favor of H1


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p;?al'e

The p-value of a test is the probability of observing a test


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Thep valueof a test is the probability of observing a test

statistic at least as extreme as the one computed given that

the null hypothesis is true.

In the case of our department store example, what is the

probabilityof observing a sample meanat least as extremeas the one already observed (i.e. = 178), given that the null

hypothesis (H0: = 170) is true?

p;val'e

Interpreting the p;val'e%

The smaller the p-value, the more statistical evidence exists


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The smaller the p value, the more statistical evidence existsto support the alternative hypothesis.

If the p-value is less than 1%, there isoverwhelmingevidencethat supports the alternative hypothesis.

If the p-value is between 1% and 5%, there is astrongevidencethat supports the alternative hypothesis.

If the p-value is between 5% and 10% there is a weakevidencethat supports the alternative hypothesis.

If the p-value exceeds 10%, there isno evidencethat

supports the alternative hypothesis.We observe a p-value of .0069, hence there is

overwhelming evidenceto support H1: > 170.

Interpreting the p;val'e%

Compare the p-value with the selected value of the


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Co pa e t e p va ue w t t e se ected va ue o t e

significance level:

If the p-value is less than , we judge the p-value to be

small enough to reject the null hypothesis.

If the p-value is greater than , we do not reject the null

hypothesis.

Since p-value = .0069


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j ymean payment period. Thus, the parameter to be tested is the

population mean. We want to know whether there is enoughstatistical evidence to show that the population mean is lessthan 22 days. Thus, the alternative hypothesis is

H1:< 22

The null hypothesis is

H0:= 22

Chapter;pening E@ample%

The test statistic is


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We wish to reject the null hypothesis in favor of the

alternative only if the sample mean and hence the value of

the test statistic is small enough. As a result we locate therejection region in the left tail of the sampling distribution.

We set the significance level at 10%.

n

x

z

/

=


Rejection region: 28.110

==< zzz


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From the data in SSAwe compute

and

p-value = P(Z < -.91) = .5 - .3186 = .1814

28.110.< zzz

63.21220

759,4

220===

ixx

91.

220/6

2263.21

/

=

=

=

n

x

z


Conclusion: There is not enough evidence to infer that the
http://e/TT%20PowerPoint%20slides/References/SSA.xlshttp://e/TT%20PowerPoint%20slides/References/SSA.xls


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g

mean is less than 22.

There is not enough evidence to infer that the plan will be

profitable.

Since Z(- .91) > -Z.10(-1.28)

We fail to reject Ho: X 22

at a O0H level of signi*cance

&LT &!E4 C4?E


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4ight;Tail Testing%

Calculate the critical value of the mean ( ) and compare


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( ) p

against the observed value of the sample mean ( )

Left;Tail Testing%

Calculate the critical value of the mean ( ) and compare


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p

against the observed value of the sample mean ( )

ToVTail Testing%

Two tail testing is used when we want to test a research


163/209


g

hypothesis that a parameter is not equal () to some value

E@ample OO2%

AT&Ts argues that its rates are such that customers wont


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see a difference in their phone bills between them and their

competitors. They calculate the mean and standard deviationfor all their customers at $17.09 and $3.87 (respectively).

They then sample 100 customers at random and recalculate amonthly phone bill based on competitors rates.

What we want to show is whether or not:H1: 17.09. We do this by assuming that:

H0: = 17.09

E@ample OO2%

The rejection region is set up so we can reject the null


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hypothesis when the test statistic is largeorwhen it is small.

That is, we set up a two-tail rejection region. The total areain the rejection region must sum to , so we divide this

probability by 2.

stat is small stat is large

E@ample OO2%

At a 5% significance level (i.e. = .05), we have


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/2 = .025. Thus, z.025= 1.96 and our rejection region is:

z < 1.96 -or- z > 1.96

z-z.025 +z.0250

E@ample OO2%

From the data, we calculate = 17.55


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Using our standardized test statistic:

We find that:

Since z = 1.19 is not greater than 1.96, nor less than 1.96

we cannot reject the null hypothesis in favor of H1. That isthere is insufficient evidence to infer that there is a

difference between the bills of AT&T and the competitor.

&LT &!E4 C4?E


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"'mmary of ne; and To;TailTests%


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ne;Tail Test

left tail1

To;Tail Test ne;Tail Test

right tail1

Inference >(o't > &op'lation%D"I-+$+!+Z

&op'lation


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We will develop techniques to estimate and test three

population parameters:

Population Mean

Population Variance

Population Proportion p

&arameter

"ample

"tatistic

Inference

Inference !ith ?ariance nknon%

Previously, we looked at estimating and testing the


171/209


population mean when the population standard deviation ( )

was known or given:

But how often do we know the actual population variance?

Instead, we use the Student t-statistic, given by:

Testing hen is 'nknon%

When the population standard deviation is unknown and the


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population is normal, the test statistic for testing hypotheses

about is:

which is Studenttdistributed with = n1 degrees offreedom. The confidence interval estimator of is given

by:

E@ample O2O%

Will new workers achieve 90% of the level of experienced


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workers within one week of being hired and trained?

Experienced workers can process 500 packages/hour, thus if

our conjecture is correct, we expect new workers to be able

to process .90(500) = 450 packages per hour.

Given the data, is this the case?

E@ample O2O%

Our objective is todescribethe population of the numbers of

*!)+(*-


174/209


packages processed in 1 hour by new workers, that is we

want to know whether the new workers productivity is morethan 90% of that of experienced workers. Thus we have:

H1: > 450

Therefore we set our usual null hypothesis to:

H0: = 450

E@ample O2O%

Our test statistic is:

C/P'()


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With n=50 data points, we have n1=49 degrees of freedom.

Our hypothesis under question is:

H1: > 450

Our rejection region becomes:

Thus we will reject the null hypothesis in favor of the

alternative if our calculated test static falls in this region.

E@ample O2O%

From the data, we calculate = 460.38, s=38.83 and thus:

C/P'()


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Since

we reject H0in favor of H1, that is, there is sufficient

evidence to conclude that the new workers are producing at

more than 90% of the average of experienced workers.

E@ample O22%

Can we estimate the return on investment for companies that

li d ?

*!)+(*-


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won quality awards?

We are given a random sampleof n = 83 such companies.

We want to construct a 95% confidence interval for the mean

return, i.e. what is: ??

E@ample O22%

From the data, we calculate:

C/P'()


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For this term

and so:

Check 4e'isite Conditions%

The Student t distribution isrobust, which means that if the

l i i l h l f h d


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population is nonnormal, the results of the t-test and

confidence interval estimate are still valid provided that thepopulation is not extremelynonnormal.

To check this requirement,draw a histogramof the data andsee how bell shaped the resulting figure is. If a histogram

is extremely skewed (say in the case of an exponential

distribution), that could be considered extremely

nonnormal and hence t-statistics would be not be valid inthis case.

Inference >(o't &op'lation?ariance%If we are interested in drawing inferences about a

l ti i bilit h d


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populations variability, the parameter we need to

investigate is the population variance:

The sample variance (s2)is an unbiased, consistent andefficient point estimator for . Moreover,

the statistic, , has a chi-squared distribution,

with n1 degrees of freedom.

Testing . Estimating &op'lation?arianceCombining this statistic:


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With the probability statement:

Yields theconfidence interval estimator for :

loer con*dencelimit

'pper con*dencelimit

E@ample O29%

Consider a container filling machine. Management wants a

hi t fill 1 lit (1 000 ) th t th t i f th

*!)+(*-


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machine to fill 1 liter (1,000 ccs) so that that variance of the

fills is less than 1 cc2. A random sampleof n=25 1 liter fillswere taken. Does the machine perform as it should at the 5%

significance level?

We want to show that:

H1: < 1

(so our null hypothesis becomes: H0

: = 1). We will usethis test statistic:

?ariance is less than O cc2

E@ample O29%

Since our alternative hypothesis is phrased as:

C/P'()


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H1: < 1

We will reject H0in favor of H1if our test statistic falls into

this rejection region:

We computer the sample variance to be: s2=.8088

And thus our test statistic takes on this value

com

pare

E@ample O2M%

As we saw, we cannot reject the null hypothesis in favor of

th lt ti Th t i th i t h id t i f


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the alternative. That is, there is not enough evidence to infer

that the claim is true.

Note:the result does not say that the variance is greater than

1, rather it merely states that we are unableto show that the

variance is less than 1.

We could estimate (at 99% confidence say) the variance of

the fills

E@ample O2M%

In order to create a confidence interval estimate of the

i d th f l

C/P'()


185/209


variance, we need these formulae:

we know (n1)s2= 19.41 from our previous calculation, and

we have from Table 5 in Appendix B:

loer con*dencelimit

'pper con*dencelimit

Comparing To &op'lations%

Previously we looked at techniques to estimate and test

parameters for one population:


186/209


parameters for one population:

Population Mean , Population Variance

We will still consider these parameters when we are looking

attwo populations, however our interest will now be:

Thedifferencebetween two means.Theratioof two variances.


187/209

"ampling )istri('tion of

1. is normally distributed if the original populations

are normal or approximately normal if the populations are


188/209


are normal or approximately normal if the populations are

nonnormal and the sample sizes are large (n1, n2> 30)

2. The expected value of is

3. The variance of is

and the standard error is:

(o't

Since isnormally distributedif the original

populations are normal or approximately normal if the


189/209


populations are normal orapproximately normalif the

populations are nonnormal and the sample sizes are large (n1,n2> 30), then:

is a standard normal (or approximately normal) random

variable. We could use this to build test statistics orconfidence interval estimators for

(o't

except that, in practice, the z statistic is rarely used since

the population variances are unknown


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the population variances are unknown.

Instead we use a t-statistic. We consider two cases for the

unknown population variances: when we believe they areequaland conversely when they arenot equal.

!hen are variances e'al#

How do we know when the population variances are equal?


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Since the population variances are unknown, we cant know

for certain whether theyre equal, but we can examine the

sample variancesand informally judgetheir relative values

to determine whether we can assume that the populationvariances are equal or not.

Test "tatistic for e'alvariances11) Calculate thepooled variance estimatoras


192/209


2) and use it here:

degrees of freedom

CI Estimator for e'alvariances1The confidence interval estimator for when the

population variances are equal is given by:


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population variances are equal is given by:

degrees of freedompooled variance estimator

Test "tatistic for 'ne'alvariances1The test statistic for when the population variances

are unequal is given by:


194/209


are unequalis given by:

Likewise, the confidence interval estimator is:

degrees of freedom

E@ample O92%

Two methods are being tested for assembling office chairs.

Assembly times are recorded (25 times for each method) At

*!)+(*-


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Assembly times are recorded (25 times for each method). At

a 5% significance level, do the assembly times for the twomethodsdiffer?

That is, H1:

Hence, our null hypothesis becomes: H0:

Reminder: This is atwo-tailed test.

E@ample O92%

The assembly timesfor each of the two methods are

recorded and preliminary data is prepared

C/P'()


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recordedand preliminary data is prepared

The sample variances are similar, hence e ill ass'me thatthe pop'lation variances are e'al%

E@ample O92%

Recall, we are doing a two-tailed test, hence the rejection

region will be:

C/P'()


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region will be:

The number of degrees of freedom is:

Hence our critical values of t (and our rejection region)becomes:

E@ample O92%

In order to calculate our t-statistic, we need to first calculate

the pooled variance estimator followed by the t-statistic

C/P'()


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thepooled variance estimator, followed by the t statistic

E@ample O92% *+(),P,)(


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Since our calculated t-statisticdoes notfall into the rejection

region, we cannot reject H0in favor of H1, that is, there is not

sufficient evidence to infer that the mean assembly timesdiffer.

E@ample O92%

Excel, of course, also provides us with the information

*+(),P,)(


200/209


Compare%

%or look at p;val'e

Con*dence Interval%

We can compute a 95% confidence interval estimate for the

difference in mean assembly times as:


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difference in mean assembly times as:

That is, we estimate the mean difference between the twoassembly methods between .36 and .96 minutes. Note: zero

is included in this confidence interval


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independent samples.

If, however, an observation in one sample ismatchedwith

an observation in a second sample, this is called amatched

pairs experiment.

To help understand this concept, lets consider example 13.4

Identifying actors%

Factors that identify the t-test and estimator of :


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Inference a(o't the ratio of tovariancesSo far weve looked at comparing measures of centrallocation, namely themeanof two populations.


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, y p p

When looking at two population variances, we consider the

ratioof the variances, i.e. the parameter of interest to us is:

The sampling statistic: isFdistributed with

degrees of freedom.

Inference a(o't the ratio of tovariancesOur null hypothesis is always:


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H0:

(i.e. the variances of the two populations will be equal, hence

their ratio will be one)

Therefore, our statistic simplifies to:

df1 = n1- 1

df2 = n2- 1

E@ample O9F%

In example 13.1, we looked at the variances of the samples

of people who consumed high fiber cereal and those who did

*!)+(*-


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p p g

not and assumed they were not equal. We can use the ideasjust developed to test if this is in fact the case.

We want to show: H1:(the variances are not equal to each other)

Hence we have our null hypothesis: H0:

E@ample O9F%

Since our research hypothesis is: H1:

We are doing a two-tailed test and our rejection region is:

C%&C'&%()


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We are doing a two-tailed test, and our rejection region is:

F

E@ample O9F%Our test statistic is:

C%&C'&%()


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Hence there is sufficient evidence to reject the nullhypothesis in favor of the alternative; that is, there is a

difference in the variance between the two populations.

F.58 1.61

E@ample O9F%We may need to work with the Excel output before drawing

conclusions

*+(),P,)(


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'r research hypothesis3O6

re'ires to;tail testing,

('t E@cel only gives 's val'esfor one;tail testing%

stat review -keller

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