1. science
2. mathematics
3. statistics

What is Statistics?
“Statistics is a way to get information from data”
Statistics
Data
Data: Facts, especially
numerical facts, collected
together for reference or
information.
Information
Information: Knowledge
communicated concerning
some particular fact.
Statistics is a tool for creating new understanding from a set of numbers.
Definitions: Oxford English Dictionary
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
1.1
Key Statistical Concepts…
Population
— a population is the group of all items of interest to
a statistics practitioner.
— frequently very large; sometimes infinite.
E.g. All 5 million Florida voters, per Example 12.5
Sample
— A sample is a set of data drawn from the
population.
— Potentially very large, but less than the population.
E.g. a sample of 765 voters exit polled on election day.
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1.2
Key Statistical Concepts…
Parameter
— A descriptive measure of a population.
Statistic
— A descriptive measure of a sample.
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1.3
Key Statistical Concepts…
Population
Sample
Subset
Parameter
Statistic
Populations have Parameters,
Samples have Statistics.
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1.4
Descriptive Statistics…
…are methods of organizing, summarizing, and presenting
data in a convenient and informative way. These methods
include:
Graphical Techniques (Chapter 2), and
Numerical Techniques (Chapter 4).
The actual method used depends on what information we
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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1.5
Statistical Inference…
Statistical inference is the process of making an estimate,
prediction, or decision about a population based on a sample.
Population
Sample
Inference
Statistic
Parameter
What can we infer about a Population’s Parameters
based on a Sample’s Statistics?
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1.6
Definitions…
A variable is some characteristic of a population or sample.
E.g. student grades.
Typically denoted with a capital letter: X, Y, Z…
The values of the variable are the range of possible values
for a variable.
E.g. student marks (0..100)
Data are the observed values of a variable.
E.g. student marks: {67, 74, 71, 83, 93, 55, 48}
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1.7
Interval Data…
Interval data
• 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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1.8
Nominal Data…
Nominal Data
• The values of nominal data are categories.
E.g. responses to questions about marital status, coded
as:
Single = 1, Married = 2, Divorced = 3, Widowed = 4
Because the numbers are arbitrary arithmetic operations
don’t make any sense (e.g. does Widowed ÷ 2 = Married?!)
Nominal data are also called qualitative or categorical.
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1.9
Ordinal Data…
Ordinal Data appear to be categorical in nature, but their
values have an order; 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.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
1.10
Graphical & Tabular Techniques for Nominal Data…
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 a frequency distribution.
A relative frequency distribution lists the categories and the
proportion with which each occurs.
Refer to Example 2.1
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1.11
Nominal Data (Tabular Summary)
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1.12
Nominal Data (Frequency)
Bar Charts are often used to display frequencies…
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1.13
Nominal Data
It all the same information,
(based on the same data).
Just different presentation.
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1.14
Graphical Techniques for Interval Data
There 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 to summarize interval data, but it is also used to help
explain probabilities.
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1.15
Building a Histogram…
1) Collect the Data 
2) Create a frequency distribution for the data. 
3) Draw the Histogram. 
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1.16
Histogram and Stem & Leaf…
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1.17
Ogive…
Is a graph of a cumulative frequency distribution.
We create an ogive in three steps…
1) Calculate relative frequencies. 
2) Calculate cumulative relative frequencies by adding the
current class’ relative frequency to the previous class’
cumulative relative frequency.
(For the first class, its cumulative relative frequency is just its relative frequency)
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1.18
Cumulative Relative Frequencies…
first class…
next class: .355+.185=.540
:
:
last class: .930+.070=1.00
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1.19
Ogive…
The ogive can be used to
answer questions like:
What telephone bill value
is at the 50th percentile?
“around $35”
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
(Refer also to Fig. 2.13 in your textbook)
1.20
Scatter Diagram…
Example 2.9 A 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) and
the dependent variable (Y – selling price) 
3) Use Excel to create a “scatter diagram”…
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1.21
Scatter Diagram…
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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1.22
Patterns of Scatter Diagrams…
Linearity and Direction are two concepts we are interested in
Positive Linear Relationship
Negative Linear Relationship
Weak or Non-Linear Relationship
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1.23
Time Series Data…
Observations measured at the same point in time are called
cross-sectional data.
Observations measured at successive points in time are
called time-series data.
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.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
1.24
Numerical Descriptive Techniques…
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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1.25
Measures of Central Location…
The arithmetic mean, a.k.a. average, shortened to mean, is
the most popular & useful measure of central location.
It is computed by simply adding up all the observations and
dividing by the total number of observations:
Sum of the observations
Mean =
Number of observations
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1.26
Arithmetic Mean…
Population Mean
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Sample Mean
1.27
Statistics is a pattern language…
Size
Population
Sample
N
n
Mean
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1.28
The Arithmetic Mean…
…is appropriate for describing measurement data, e.g.
heights of people, marks of student papers, etc.
…is seriously affected by extreme values called “outliers”.
E.g. as soon as a billionaire moves into a neighborhood, the
average household income increases beyond what it was
previously!
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1.29
Measures of Variability…
Measures of central location fail to tell the whole story about
the distribution; that is, how much are the observations
spread out around the mean value?
For example, two sets of class
grades are shown. The mean
(=50) is the same in each case…
But, the red class has greater
variability than the blue class.
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1.30
Range…
The range is 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 = 46
The range is the same in both cases,
but the data sets have very different distributions…
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1.31
Statistics is a pattern language…
Size
Population
Sample
N
n
Mean
Variance
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1.32
Variance…
population mean
The variance of a population is:
population size
sample mean
The variance of a sample is:
Note! the denominator is sample size (n) minus one !
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1.33
Application…
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 sample consists of the number of jobs six
randomly selected students applied for: 17, 15, 23, 7, 9, 13.
Finds its mean and variance.
…as opposed to  or 2
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1.34
Sample Mean & Variance…
Sample Mean
Sample Variance
Sample Variance (shortcut method)
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1.35
Standard Deviation…
The standard deviation is simply the square root of the
variance, thus:
Population standard deviation:
Sample standard deviation:
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1.36
Standard Deviation…
Consider Example 4.8 where a golf club manufacturer has
designed a new club and wants to determine if it is hit more
consistently (i.e. with less variability) than with an old club.
Using Tools > Data Analysis [may need to “add in”… > Descriptive
Statistics in Excel, we produce the following tables for
interpretation…
You get more
consistent
distance with the
new club.
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1.37
The Empirical Rule…
If the histogram is bell shaped
Approximately 68% of all observations fall
within one standard deviation of the mean.
Approximately 95% of all observations fall
within two standard deviations of the mean.
Approximately 99.7% of all observations fall
within three standard deviations of the mean.
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1.38
Chebysheff’s Theorem…Not often used because interval is very wide.
A more general interpretation of the standard deviation is
derived from Chebysheff’s Theorem, which applies to all
shapes of histograms (not just bell shaped).
The proportion of observations in any sample that lie
within k standard deviations of the mean is at least:
For k=2 (say), the theorem states
that at least 3/4 of all observations
lie within 2 standard deviations of
the mean. This is a “lower bound”
compared to Empirical Rule’s
approximation (95%).
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1.39
Box Plots…
These box plots are based on
data in Xm04-15.
Wendy’s service time is
shortest and least variable.
Hardee’s has the greatest
variability, while Jack-inthe-Box has the longest
service times.
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1.40
Methods of Collecting Data…
There are many methods used to collect or obtain data for
statistical analysis. Three of the most popular methods are:
• Direct Observation
• Experiments, and
• Surveys.
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1.41
Sampling…
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 of cost (it’s 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, the sampled population and the target
population should be similar to one another.
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1.42
Sampling Plans…
A sampling plan is 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.
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1.43
Simple Random Sampling…
A simple random sample is a sample selected in such a way
that every possible sample of the same size is equally likely
to 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.
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1.44
Stratified Random Sampling…
After the population has been stratified, we can use simple
random sampling to generate the complete sample:
If we only have sufficient resources to sample 400 people total,
we would draw 100 of them from the low income group…
…if we are sampling 1000 people, we’d draw
50 of them from the high income group.
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1.45
Cluster Sampling…
A cluster sample is a simple random sample of groups or
clusters of elements (vs. a simple random sample of
individual 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 to
similarities among cluster members.
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1.46
Sampling Error…
Sampling error refers to differences between the sample and
the population that exist only because of the observations
that 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 our
first sample and all the lowest income levels in the second,
this delta is due to sampling error.
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1.47
Nonsampling Error…
Nonsampling errors are more serious and are due to
mistakes made in the acquisition of data or due to the sample
observations being selected improperly. Three types of
nonsampling errors:
Errors in data acquisition,
Nonresponse errors, and
Selection bias.
Note: increasing the sample size will not reduce this type of
error.
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1.48
Approaches to Assigning Probabilities…
There are three ways to assign a probability, P(Oi), to an
outcome, Oi, namely:
Classical approach: make certain assumptions (such as
equally likely, independence) about situation.
Relative frequency: assigning probabilities based on
experimentation or historical data.
Subjective approach: Assigning probabilities based on the
assignor’s judgment.
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1.49
Interpreting Probability…
One way to interpret probability is this:
If a random experiment is repeated an infinite number 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 relative
frequency of heads if the coin is flipped an infinite number
of times.
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1.50
Conditional Probability…
Conditional probability is used to determine how two events
are related; that is, we can determine the probability of one
event given the occurrence of another related event.
Conditional probabilities are written as P(A | B) and read as
“the probability of A given B” and is calculated as:
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1.51
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 are
independent, that is, if the probability of one event is not
affected by the occurrence of the other event.
Two events A and B are said to be independent if
P(A|B) = P(A)
or
P(B|A) = P(B)
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1.52
Complement Rule…
The complement of an event A is the event that occurs when
A does not occur.
The complement rule gives 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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1.53
Multiplication Rule…
The multiplication rule is used to calculate the joint
probability of two events. It is based on the formula for
conditional probability defined earlier:
If we multiply both sides of the equation by P(B) we have:
P(A and B) = P(A | B)•P(B)
Likewise, P(A and B) = P(B | A) • P(A)
If A and B are independent events, then P(A and B) = P(A)•P(B)
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1.54
Addition Rule…
Recall: the addition rule was introduced earlier to provide a
way to compute the probability of event A or B or both A
and 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)
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1.55
Addition Rule for Mutually Excusive Events
If and A and B are mutually exclusive the occurrence of one
event makes the other one impossible. This means that
P(A and B) = 0
The addition rule for mutually exclusive events is
P(A or B) = P(A) + P(B)
We often use this form when we add some joint probabilities
calculated from a probability tree
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1.56
Two Types of Random Variables…
Discrete Random Variable
– one that takes on a countable number of values
– E.g. values on the roll of dice: 2, 3, 4, …, 12
Continuous Random Variable
– one whose values are not discrete, not countable
– E.g. time (30.1 minutes? 30.10000001 minutes?)
Analogy:
Integers are Discrete, while Real Numbers are Continuous
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1.57
Laws of Expected Value…
1. E(c) = c
• The expected value of a constant (c) is just the value of the
constant.
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).
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1.58
Laws of Variance…
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 is
the coefficient squared times the variance of the random variable.
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1.59
Binomial Distribution…
The binomial distribution is the probability distribution that
results from doing a “binomial experiment”. Binomial
experiments 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)=1–p), 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.
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1.60
Binomial Random Variable…
The binomial random variable counts the number of
successes in n trials of the binomial experiment. It can take
on values from 0, 1, 2, …, n. Thus, its a discrete random
variable.
To calculate the probability associated with each value we
use combintorics:
for x=0, 1, 2, …, n
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1.61
Binomial Table…
“What is the probability that Pat fails the quiz”?
i.e. what is P(X ≤ 4), given P(success) = .20 and n=10 ?
P(X ≤ 4) = .967
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1.62
Binomial Table…
“What is the probability that Pat gets two answers correct?”
i.e. what is P(X = 2), given P(success) = .20 and n=10 ?
P(X = 2) = P(X≤2) – P(X≤1) = .678 – .376 = .302
remember, the table shows cumulative probabilities…
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1.63
=BINOMDIST() Excel Function…
There is a binomial distribution function in Excel that can
also be used to calculate these probabilities. For example:
What is the probability that Pat gets two answers correct?
# successes
# trials
P(success)
cumulative
(i.e. P(X≤x)?)
P(X=2)=.3020
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1.64
=BINOMDIST() Excel Function…
There is a binomial distribution function in Excel that can
also be used to calculate these probabilities. For example:
What is the probability that Pat fails the quiz?
# successes
# trials
P(success)
cumulative
(i.e. P(X≤x)?)
P(X≤4)=.9672
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1.65
Binomial Distribution…
As you might expect, statisticians have developed general
formulas for the mean, variance, and standard deviation of a
binomial random variable. They are:
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1.66
Poisson Distribution…
Named for Simeon Poisson, the Poisson distribution is a
discrete probability distribution and refers to the number of
events (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.)
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1.67
The Poisson Experiment…
Like a binomial experiment, a Poisson experiment has 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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1.68
Poisson Distribution…
The Poisson random variable is the number of successes
that occur in a period of time or an interval of space in a
Poisson experiment.
successes
E.g. On average, 96 trucks arrive at a border crossing
every hour.
time period
E.g. The number of typographic errors in a new textbook
edition averages 1.5 per 100 pages.
successes (?!)
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interval
1.69
Poisson Probability Distribution…
The probability that a Poisson random variable assumes a
value of x is given by:
and e is the natural logarithm base.
FYI:
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1.70
Example 7.12…
The number of typographical errors in new editions of
textbooks varies considerably from book to book. After some
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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1.71
Poisson Distribution…
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.
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1.72
Example 7.13…
For a 400 page book, what is the probability that there are
no typos?
P(X=0) =
“there is a very small chance there are no typos”
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1.73
Example 7.13…
…Excel is an even better alternative:
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1.74
Probability Density Functions…
Unlike a discrete random variable which we studied in
Chapter 7, a continuous random variable is one that can
assume an uncountable number 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.
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1.75
Point Probabilities are Zero
Because there is an infinite number of values, the
probability of each individual value is virtually 0.
Thus, we can determine the probability of a range of values
only.
E.g. with a discrete random variable like tossing a die, it is
meaningful to talk about P(X=5), say.
In a continuous setting (e.g. with time as a random variable), the
probability the random variable of interest, say task length, takes
exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
It is meaningful to talk about P(X ≤ 5).
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1.76
Probability Density Function…
A function f(x) is called a probability density function (over
the range a ≤ x ≤ b if it meets the following
requirements:
1) f(x) ≥ 0 for all x between a and b, and
f(x)
area=1
a
b
x
2) The total area under the curve between a and b is 1.0
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1.77
The Normal Distribution…
The normal distribution is the most important of all
probability distributions. The probability density function of
a normal random variable is given by:
It looks like this:
Bell shaped,
Symmetrical around the mean
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…
1.78
The Normal Distribution…
Important things to note:
The normal distribution is fully defined by two parameters:
its standard deviation and mean
The normal distribution is bell shaped and
symmetrical about the mean
Unlike the range of the uniform distribution (a ≤ x ≤ b)
Normal distributions range from minus infinity to plus infinity
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Standard Normal Distribution…
A normal distribution whose mean is zero and standard
deviation is one is called the standard normal distribution.
0
1
1
As we shall see shortly, any normal distribution can be
converted to a standard normal distribution with simple
algebra. This makes calculations much easier.
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Calculating Normal Probabilities…
We can use the following function to convert any normal
random variable to a standard normal random variable…
0
Some advice: always
draw a picture!
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Calculating Normal Probabilities…
Example: The time required to build a computer is normally
distributed with a mean of 50 minutes and a standard
deviation of 10 minutes:
0
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) ?
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Calculating Normal Probabilities…
P(45 < X < 60) ?
…mean of 50 minutes and a
standard deviation of 10 minutes…
0
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Calculating Normal Probabilities…
We can use Table 3 in
Appendix B to look-up
probabilities P(0 < Z < z)
We can break up P(–.5 < Z < 1) into:
P(–.5 < Z < 0) + P(0 < Z < 1)
The distribution is symmetric around 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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Calculating Normal Probabilities…
How to use Table 3…
This table gives probabilities P(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
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Using the Normal Table (Table 3)…
What is P(Z > 1.6) ?
P(0 < Z < 1.6) = .4452
z
0
1.6
P(Z > 1.6) = .5 – P(0 < Z < 1.6)
= .5 – .4452
= .0548
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Using the Normal Table (Table 3)…
What is P(Z < -2.23) ?
P(0 < Z < 2.23)
P(Z < -2.23)
P(Z > 2.23)
z
-2.23
0
2.23
P(Z < -2.23) = P(Z > 2.23)
= .5 – P(0 < Z < 2.23)
= .0129
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Using the Normal Table (Table 3)…
What is P(Z < 1.52) ?
P(0 < Z < 1.52)
P(Z < 0) = .5
z
0
1.52
P(Z < 1.52) = .5 + P(0 < Z < 1.52)
= .5 + .4357
= .9357
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Using the Normal Table (Table 3)…
What is P(0.9 < Z < 1.9) ?
P(0 < Z < 0.9)
P(0.9 < Z < 1.9)
z
0
0.9
1.9
P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9)
=.4713 – .3159
= .1554
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Finding Values of Z…
Other Z values are
Z.05 = 1.645
Z.01 = 2.33
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Using the values of Z
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
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Other Continuous Distributions…
Three other important continuous distributions which will be
used extensively in later sections are introduced here:
Student t Distribution,
Chi-Squared Distribution, and
F Distribution.
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Student t Distribution…
Here the letter t is used to represent the random variable,
hence the name. The density function for the Student t
distribution is as follows…
(nu) is called the degrees of freedom, and
(Gamma function) is (k)=(k-1)(k-2)…(2)(1)
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Student t Distribution…
In much the same way that
and
define the normal
distribution, , the degrees of freedom, defines the Student
t Distribution:
Figure 8.24
As the number of degrees of freedom increases, the t
distribution approaches the standard normal distribution.
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Determining Student t Values…
The student t distribution is used extensively in statistical
inference. Table 4 in Appendix B lists values of
That is, values of a Student t random variable with
of freedom such that:
degrees
The values for A are pre-determined
“critical” values, typically in the
10%, 5%, 2.5%, 1% and 1/2% range.
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Using the t table (Table 4) for values…
For example, if we want the value of t with 10 degrees of
freedom such that the area under the Student t curve is .05:
Area under the curve value (tA) : COLUMN
t.05,10
t.05,10=1.812
Degrees of Freedom : ROW
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F Distribution…
The F density function is given by:
F > 0. Two parameters define this distribution, and like
we’ve already seen these are again degrees of freedom.
is the “numerator” degrees of freedom and
is the “denominator” degrees of freedom.
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Determining Values of F…
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 of
freedom of 3 and a denominator degree of freedom of 7?
Solution: use the F look-up (Table 6)
There are different tables
for different values of A.
Make sure you start with
the correct table!!
F.05,3,7
F.05,3,7=4.35
Denominator Degrees of Freedom : ROW
Numerator Degrees of Freedom : COLUMN
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Determining Values of F…
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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Chapter 9
Sampling Distributions
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Sampling Distribution of the Mean…
A fair die is thrown infinitely many times,
with the random variable X = # of spots on any throw.
The probability distribution of X is:
x
P(x)
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
…and the mean and variance are calculated as well:
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Sampling Distribution of Two Dice
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).
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Sampling Distribution of Two Dice…
The sampling distribution of
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
5/36
)
6/36
4/36
P(
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
P( )
3/36
is shown below:
2/36
1/36
1.0
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1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
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Compare…
Compare the distribution of X…
1
2
3
4
5
6
1.0
1.5
…with the sampling distribution of
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
.
As well, note that:
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Central Limit Theorem…
The sampling distribution of the mean of a random sample
drawn from any population is approximately normal for a
sufficiently large sample size.
The larger the sample size, the more closely the sampling
distribution of X will resemble a normal distribution.
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Central Limit Theorem…
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.
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Sampling Distribution of the Sample Mean
1.
2.
3. If X is normal, X is normal. If X is nonnormal, X is
approximately normal for sufficiently large sample sizes.
Note: the definition of “sufficiently large” depends on the
extent of nonnormality of x (e.g. heavily skewed;
multimodal)
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Example 9.1(a)…
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 one bottle, what is the probability that the
bottle will contain more than 32 ounces?
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Example 9.1(a)…
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.”
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Example 9.1(b)…
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 four bottles, what is the
probability that the mean amount of the four bottles will be
greater than 32 ounces?
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Example 9.1(b)…
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)
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Example 9.1(b)…
If a customer buys a carton of four bottles, what is the
probability that the mean amount of the four bottles will be
greater than 32 ounces?
“There is about a 91% chance the mean of the four bottles
will exceed 32oz.”
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Graphically Speaking…
mean=32.2
what is the probability that one bottle will
contain more than 32 ounces?
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what is the probability that the mean of
four bottles will exceed 32 oz?
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Sampling Distribution:
Difference of two means
The final sampling distribution introduced is that of the
difference between two sample means. This requires:
 independent random samples be drawn from each of two
normal populations
If this condition is met, then the sampling distribution of the
difference between the two sample means, i.e.
will be normally distributed.
(note: if the two populations are not both normally
distributed, but the sample sizes are “large” (>30), the
distribution of
is approximately normal)
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Sampling Distribution:
Difference of two means
The expected value and variance of the sampling
distribution of
are given by:
mean:
standard deviation:
(also called the standard error if the difference between two
means)
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Estimation…
There are two types of inference: estimation and hypothesis
testing; estimation is introduced first.
The objective of estimation is to determine the approximate
value of a population parameter on the basis of a sample
statistic.
E.g., the sample mean (
population mean ( ).
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) is employed to estimate the
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Estimation…
The objective of estimation is to determine the approximate
value of a population parameter on the basis of a sample
statistic.
There are two types of estimators:
Point Estimator
Interval Estimator
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Point & Interval Estimation…
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 is between 380 and 420 $/week.
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Estimating
when
is known…
We established in Chapter 9:
Thus, the probability that the interval:
contains the population mean
is 1–
confidence interval estimator for .
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the confidence
interval
the sample mean is
in the center of the
interval…
. This is a
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Four commonly used confidence levels…
Confidence Level

cut & keep handy!

Table 10.1
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Example 10.1…
A computer company samples demand during lead time over
25 time periods:
235
421
394
261
386
374
361
439
374
316
309
514
348
302
296
499
462
344
466
332
253
369
330
535
334
Its is known that the standard deviation of demand over lead
time is 75 computers. We want to estimate the mean demand
over lead time with 95% confidence in order to set inventory
levels…
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Example 10.1…
CALCULATE
In order to use our confidence interval estimator, we need the
following pieces of data:
370.16
Calculated from the data…
1.96
75
n
Given
25
therefore:
The lower and upper confidence limits are 340.76 and 399.56.
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Example 10.1…
INTERPRET
The estimation for the mean demand during lead time lies
between 340.76 and 399.56 — we can use this as input in
developing 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 estimator
is correct 95% of the time. That also means that 5% of the
time the estimator will be incorrect.
Incidentally, the media often refer to the 95% figure as “19
times out of 20,” which emphasizes the long-run aspect of
the confidence level.
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Interval Width…
A wide interval provides little information.
For example, suppose we estimate with 95% confidence that
an accountant’s average starting salary is between $15,000
and $100,000.
Contrast this with: a 95% confidence interval estimate of
starting salaries between $42,000 and $45,000.
The second estimate is much narrower, providing accounting
students more precise information about starting salaries.
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Interval Width…
The width of the confidence interval estimate is a function of
the confidence level, the population standard deviation, and
the sample size…
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Selecting the Sample Size…
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 n to get requisite sample size!
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Selecting the Sample Size…
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).
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Sample Size to Estimate a Mean…
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:
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Example 10.2…
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?
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Example 10.2…
Things we know:
Confidence level = 99%, therefore
=.01
We want
1 , hence W=1.
We are given that
= 6.
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Example 10.2…
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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Nonstatistical Hypothesis Testing…
A criminal trial is an example of hypothesis testing without
the statistics.
In a trial a jury must decide between two hypotheses. The
null hypothesis is
H0: The defendant is innocent
The alternative hypothesis or research hypothesis is
H1: The defendant is guilty
The jury does not know which hypothesis is true. They must
make a decision on the basis of evidence presented.
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Nonstatistical Hypothesis Testing…
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 don’t reject a false null
hypothesis. That occurs when a guilty defendant is acquitted.
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Nonstatistical Hypothesis Testing…
The probability of a Type I error is denoted as α (Greek letter
alpha). The probability of a type II error is β (Greek letter
beta).
The two probabilities are inversely related. Decreasing one
increases the other.
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Nonstatistical Hypothesis Testing…
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
to infer 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.
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Nonstatistical Hypothesis Testing…
5. Two possible errors can be made.
Type I error: Reject a true null hypothesis
Type II error: Do not reject a false null hypothesis.
P(Type I error) = α
P(Type II error) = β
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Concepts of Hypothesis Testing (1)…
There are two hypotheses. One is called the null hypothesis
and the other the alternative or research hypothesis. The
usual notation is:
pronounced
H “nought”
H0: — the ‘null’ hypothesis
H1: — the ‘alternative’ or ‘research’ hypothesis
The null hypothesis (H0) will always state that the parameter
equals the value specified in the alternative hypothesis (H1)
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Concepts of Hypothesis Testing…
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: μ = 350
Thus, our research hypothesis becomes:
H1: μ ≠ 350
This is what we are interested
in determining…
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Concepts of Hypothesis Testing (4)…
There are two possible decisions that can be made:
Conclude that there is enough evidence to support the
alternative hypothesis
(also stated as: rejecting the null hypothesis in favor of the
alternative)
Conclude that there is not enough evidence to support the
alternative hypothesis
(also stated as: not rejecting the null hypothesis in favor of
the alternative)
NOTE: we do not say that we accept the null hypothesis…
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Concepts of Hypothesis Testing…
Once the null and alternative hypotheses are stated, the next
step is to randomly sample the population and calculate a test
statistic (in this example, the sample mean).
If the test statistic’s value is inconsistent with the null
hypothesis we reject the null hypothesis and infer that the
alternative hypothesis is true.
For example, if we’re 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.
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Types of Errors…
A Type I error occurs when we reject a true null hypothesis
(i.e. Reject H0 when it is TRUE)
H0
T
Reject
I
Reject
F
II
A Type II error occurs when we don’t reject a false null
hypothesis (i.e. Do NOT reject H0 when it is FALSE)
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Recap I…
1) Two hypotheses: H0 & H1
2) ASSUME H0 is TRUE
3) GOAL: determine if there is enough evidence to infer that
H1 is TRUE
4) Two possible decisions:
Reject H0 in favor of H1
NOT Reject H0 in favor of H1
5) Two possible types of errors:
Type I: reject a true H0 [P(Type I)= ]
Type II: not reject a false H0 [P(Type II)= ]
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Example 11.1…
A department store manager determines that a new billing
system will be cost-effective only if the mean monthly
account is more than $170.
A random sample of 400 monthly accounts is drawn, for
which the sample mean is $178. The accounts are
approximately normally distributed with a standard deviation
of $65.
Can we conclude that the new system will be cost-effective?
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Example 11.1…
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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Example 11.1…
What we want to show:
H1: > 170
H0: = 170 (we’ll assume this is true)
We know:
n = 400,
= 178, and
= 65
Hmm. What to do next?!
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Example 11.1…
To test our hypotheses, we can use two different approaches:
The rejection region approach (typically used when
computing statistics manually), and
The p-value approach (which is generally used with a
computer and statistical software).
We will explore both in turn…
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Example 11.1… Rejection Region…
The rejection region is a range of values such that if the test
statistic falls into that range, we decide to reject the null
hypothesis in favor of the alternative hypothesis.
is the critical value of
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to reject H0.
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Example 11.1…
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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Example 11.1…
At a 5% significance level (i.e.
=0.05), we get
Solving we compute
=175.34
Since our sample mean (178) is greater than the critical value we
calculated (175.34), we reject the null hypothesis in favor of H1, i.e.
that:
> 170 and that it is cost effective to install the new billing
system
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Example 11.1… The Big Picture…
H1:
H0:
> 170
= 170
=175.34
=178
Reject H0 in favor of
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Standardized Test Statistic…
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 H0 in favor of H1…
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PLOT POWER CURVE
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p-Value
The p-value of 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
probability of observing a sample mean at least as extreme
as the one already observed (i.e. = 178), given that the null
hypothesis (H0: = 170) is true?
p-value
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Interpreting the p-value…
The smaller the p-value, the more statistical evidence exists
to support the alternative hypothesis.
•If the p-value is less than 1%, there is overwhelming
evidence that supports the alternative hypothesis.
•If the p-value is between 1% and 5%, there is a strong
evidence that supports the alternative hypothesis.
•If the p-value is between 5% and 10% there is a weak
evidence that supports the alternative hypothesis.
•If the p-value exceeds 10%, there is no evidence that
supports the alternative hypothesis.
We observe a p-value of .0069, hence there is
overwhelming evidence to support H1: > 170.
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Interpreting the p-value…
Compare the p-value with the selected value of the
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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= .05, we reject H0 in favor of H1
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Chapter-Opening Example…
The objective of the study is to draw a conclusion about the
mean payment period. Thus, the parameter to be tested is the
population mean. We want to know whether there is enough
statistical evidence to show that the population mean is less
than 22 days. Thus, the alternative hypothesis is
H1:μ < 22
The null hypothesis is
H0:μ = 22
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Chapter-Opening Example…
The test statistic is
x 
z
/ n
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 the
rejection region in the left tail of the sampling distribution.
We set the significance level at 10%.
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Chapter-Opening Example…
z   z    z.10  1.28
Rejection region:
From the data in SSA we compute
and
x

x
4,759

 21 .63
220
220
z
x 
/ n
i

21 .63  22
 .91
6 / 220
p-value = P(Z < -.91) = .5 - .3186 = .1814
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Chapter-Opening Example…
Conclusion: There is not enough evidence to infer that the
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: μ > 22
at a 10% level of significance.
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PLOT POWER CURVE
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Right-Tail Testing…
Calculate the critical value of the mean ( ) and compare
against the observed value of the sample mean ( )…
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Left-Tail Testing…
Calculate the critical value of the mean ( ) and compare
against the observed value of the sample mean ( )…
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Two–Tail Testing…
Two tail testing is used when we want to test a research
hypothesis that a parameter is not equal (≠) to some value
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Example 11.2…
AT&T’s argues that its rates are such that customers won’t
see a difference in their phone bills between them and their
competitors. They calculate the mean and standard deviation
for all their customers at $17.09 and $3.87 (respectively).
They then sample 100 customers at random and recalculate a
monthly phone bill based on competitor’s rates.
What we want to show is whether or not:
H1: ≠ 17.09. We do this by assuming that:
H0: = 17.09
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Example 11.2…
The rejection region is set up so we can reject the null
hypothesis when the test statistic is large or when it is small.
stat is “small”
stat is “large”
That is, we set up a two-tail rejection region. The total area
in the rejection region must sum to , so we divide this
probability by 2.
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Example 11.2…
At a 5% significance level (i.e. = .05), we have
/2 = .025. Thus, z.025 = 1.96 and our rejection region is:
z < –1.96
-z.025
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-or-
0
z > 1.96
+z.025
z
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Example 11.2…
From the data, we calculate
= 17.55
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 is
“there is insufficient evidence to infer that there is a
difference between the bills of AT&T and the competitor.”
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PLOT POWER CURVE
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Summary of One- and Two-Tail Tests…
One-Tail Test
(left tail)
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Two-Tail Test
One-Tail Test
(right tail)
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Inference About A Population…[SIGMA UNKNOWN]
Population
Sample
Inference
Statistic
Parameter
We will develop techniques to estimate and test three
population parameters:
Population Mean
Population Variance
Population Proportion p
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Inference With Variance Unknown…
Previously, we looked at estimating and testing the
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:
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Testing
when
is unknown…
When the population standard deviation is unknown and the
population is normal, the test statistic for testing hypotheses
about is:
which is Student t distributed with = n–1 degrees of
freedom. The confidence interval estimator of
is given by:
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Example 12.1…
Will new workers achieve 90% of the level of experienced
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?
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Example 12.1…
IDENTIFY
Our objective is to describe the population of the numbers of
packages processed in 1 hour by new workers, that is we
want to know whether the new workers’ productivity is more
than 90% of that of experienced workers. Thus we have:
H1:
> 450
Therefore we set our usual null hypothesis to:
H0:
= 450
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Example 12.1…
COMPUTE
Our test statistic is:
With n=50 data points, we have n–1=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.
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Example 12.1…
From the data, we calculate
COMPUTE
= 460.38, s =38.83 and thus:
Since
we reject H0 in 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.
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Example 12.2…
IDENTIFY
Can we estimate the return on investment for companies that
won quality awards?
We are given a random sample of n = 83 such companies.
We want to construct a 95% confidence interval for the mean
return, i.e. what is:
??
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Example 12.2…
COMPUTE
From the data, we calculate:
For this term
and so:
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Check Requisite Conditions…
The Student t distribution is robust, which means that if the
population is nonnormal, the results of the t-test and
confidence interval estimate are still valid provided that the
population is “not extremely nonnormal”.
To check this requirement, draw a histogram of the data and
see 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 in
this case.
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Inference About Population Variance…
If we are interested in drawing inferences about a
population’s variability, the parameter we need to
investigate is the population variance:
The sample variance (s2) is an unbiased, consistent and
efficient point estimator for
. Moreover,
the statistic,
, has a chi-squared distribution,
with n–1 degrees of freedom.
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Testing & Estimating Population Variance
Combining this statistic:
With the probability statement:
Yields the confidence interval estimator for
lower confidence limit
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:
upper confidence limit
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Example 12.3…
IDENTIFY
Consider a container filling machine. Management wants a
machine to fill 1 liter (1,000 cc’s) so that that variance of the
fills is less than 1 cc2. A random sample of n=25 1 liter fills
were taken. Does the machine perform as it should at the 5%
significance level?
Variance is less than 1 cc2
We want to show that:
H1:
<1
(so our null hypothesis becomes: H0:
= 1). We will use
this test statistic:
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Example 12.3…
COMPUTE
Since our alternative hypothesis is phrased as:
H1:
<1
We will reject H0 in favor of H1 if 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…
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Example 12.4…
As we saw, we cannot reject the null hypothesis in favor of
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 unable to show that the
variance is less than 1.
We could estimate (at 99% confidence say) the variance of
the fills…
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Example 12.4…
COMPUTE
In order to create a confidence interval estimate of the
variance, we need these formulae:
lower confidence limit
upper confidence limit
we know (n–1)s2 = 19.41 from our previous calculation, and
we have from Table 5 in Appendix B:
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Comparing Two Populations…
Previously we looked at techniques to estimate and test
parameters for one population:
Population Mean , Population Variance
We will still consider these parameters when we are looking
at two populations, however our interest will now be:
 The difference between two means.
 The ratio of two variances.
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Difference of Two Means…
In order to test and estimate the difference between two
population means, we draw random samples from each of
two populations. Initially, we will consider independent
samples, that is, samples that are completely unrelated to one
another.
Because we are compare two population means, we use the
statistic:
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Sampling Distribution of
1.
is normally distributed if the original populations
are normal –or– approximately normal if the populations are
nonnormal and the sample sizes are large (n1, n2 > 30)
2. The expected value of
3. The variance of
is
is
and the standard error is:
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Making Inferences About
Since
is normally distributed if the original
populations are normal –or– approximately normal if 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 or
confidence interval estimators for
…
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Making Inferences About
…except that, in practice, the z statistic is rarely used since
the population variances are unknown.
??
Instead we use a t-statistic. We consider two cases for the
unknown population variances: when we believe they are
equal and conversely when they are not equal.
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When are variances equal?
How do we know when the population variances are equal?
Since the population variances are unknown, we can’t know
for certain whether they’re equal, but we can examine the
sample variances and informally judge their relative values
to determine whether we can assume that the population
variances are equal or not.
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Test Statistic for
1) Calculate
(equal variances)
– the pooled variance estimator as…
2) …and use it here:
degrees of freedom
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CI Estimator for
(equal variances)
The confidence interval estimator for
when the
population variances are equal is given by:
pooled variance estimator
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degrees of freedom
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Test Statistic for
(unequal variances)
The test statistic for
when the population variances
are unequal is given by:
degrees of freedom
Likewise, the confidence interval estimator is:
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Example 13.2…
IDENTIFY
Two methods are being tested for assembling office chairs.
Assembly times are recorded (25 times for each method). At
a 5% significance level, do the assembly times for the two
methods differ?
That is, H1:
Hence, our null hypothesis becomes: H0:
Reminder: This is a two-tailed test.
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Example 13.2…
COMPUTE
The assembly times for each of the two methods are
recorded and preliminary data is prepared…
The sample variances are similar, hence we will assume that the
population variances are equal…
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Example 13.2…
COMPUTE
Recall, we are doing a two-tailed test, hence the rejection
region will be:
The number of degrees of freedom is:
Hence our critical values of t (and our rejection region)
becomes:
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Example 13.2…
COMPUTE
In order to calculate our t-statistic, we need to first calculate
the pooled variance estimator, followed by the t-statistic…
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Example 13.2…
INTERPRET
Since our calculated t-statistic does not fall into the rejection
region, we cannot reject H0 in favor of H1, that is, there is
not sufficient evidence to infer that the mean assembly times
differ.
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Example 13.2…
INTERPRET
Excel, of course, also provides us with the information…
Compare…
…or look at p-value
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Confidence Interval…
We can compute a 95% confidence interval estimate for the
difference in mean assembly times as:
That is, we estimate the mean difference between the two
assembly methods between –.36 and .96 minutes. Note: zero
is included in this confidence interval…
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Matched Pairs Experiment…
Previously when comparing two populations, we examined
independent samples.
If, however, an observation in one sample is matched with
an observation in a second sample, this is called a matched
pairs experiment.
To help understand this concept, let’s consider example 13.4
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Identifying Factors…
Factors that identify the t-test and estimator of
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:
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Inference about the ratio of two variances
So far we’ve looked at comparing measures of central
location, namely the mean of two populations.
When looking at two population variances, we consider the
ratio of the variances, i.e. the parameter of interest to us is:
The sampling statistic:
is F distributed with
degrees of freedom.
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Inference about the ratio of two variances
Our null hypothesis is always:
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
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Example 13.6…
IDENTIFY
In example 13.1, we looked at the variances of the samples
of people who consumed high fiber cereal and those who did
not and assumed they were not equal. We can use the ideas
just 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:
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Example 13.6…
CALCULATE
Since our research hypothesis is: H1:
We are doing a two-tailed test, and our rejection region is:
F
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Example 13.6…
CALCULATE
Our test statistic is:
.58
1.61
F
Hence there is sufficient evidence to reject the null
hypothesis in favor of the alternative; that is, there is a
difference in the variance between the two populations.
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Example 13.6…
INTERPRET
We may need to work with the Excel output before drawing
conclusions…
Our research hypothesis
H1:
requires two-tail testing,
but Excel only gives us values
for one-tail testing…
If we double the one-tail p-value Excel gives us, we have the p-value of
the test we’re conducting (i.e. 2 x 0.0004 = 0.0008). Refer to the text
and CD Appendices for more detail.
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