BUS 308 Week 3 Lecture 1 Examining Differences - Continued.docx

BUS 308 Week 3 Lecture 1
Examining Differences - Continued
Expected Outcomes
After reading this lecture, the student should be familiar with:
1. Issues around multiple testing
2. The basics of the Analysis of Variance test
3. Determining significant differences between group means
4. The basics of the Chi Square Distribution.
Overview
Last week, we found out ways to examine differences between a
measure taken on two
groups (two-sample test situation) as well as comparing that
measure to a standard (a one-sample
test situation). We looked at the F test which let us test for
variance equality. We also looked at
the t-test which focused on testing for mean equality. We noted
that the t-test had three distinct
versions, one for groups that had equal variances, one for
groups that had unequal variances, and
one for data that was paired (two measures on the same subject,
such as salary and midpoint for
each employee). We also looked at how the 2-sample unequal t-
test could be used to use Excel
to perform a one-sample mean test against a standard or
constant value. This week we expand
our tool kit to let us compare multiple groups for similar mean

values.
A second tool will let us look at how data values are distributed
– if graphed, would they
look the same? Different shapes or patterns often means the
data sets differ in significant ways
that can help explain results.
Multiple Groups
As interesting as comparing two groups is, often it is a bit
limiting as to what it tells us.
One obvious issue that we are missing in the comparisons made
last week was equal work. This
idea is still somewhat hard to get a clear handle on. Typically,
as we look at this issue, questions
arise about things such as performance appraisal ratings,
education distribution, seniority impact,
etc.
Some of these can be tested with the tools introduced last week.
We can see, for
example, if the performance rating average is the same for each
gender. What we couldn’t do, at
this point however, is see if performance ratings differ by
grade, do the more senior workers
perform relatively better? Is there a difference between ratings
for each gender by grade level?
The same questions can be asked about seniority impact. This
week will give us tools to expand
how we look at the clues hidden within the data set about equal
pay for equal work.
ANOVA
So, let’s start taking a look at these questions. The first tool for

this week is the Analysis
of Variance – ANOVA for short. ANOVA is often confusing
for students; it says it analyzes
variance (which it does) but the purpose of an ANOVA test is to
determine if the means of
different groups are the same! Now, so far, we have considered
means and variance to be two
distinct characteristics of data sets; characteristics that are not
related, yet here we are saying that
looking at one will give us insight into the other.
The reason is due to the way the variance is analyzed. Just as
our detectives succeed by
looking at the clues and data in different ways, so does
ANOVA. There are two key variances
that are examined with this test. The first, called Within Group
variance, is the average variance
of the groups. ANOVA assumes the population(s) the samples
are taken from have the same
variation, so this average is an estimate of the population
variance.
The second is the variance of the entire group, Between Group
Variation, as if all the
samples were from the same group. Here are exhibits showing
two situations. In Exhibit A, the
groups are close together, in fact they are overlapping, and the
means are obviously close to each
other. The Between Group variation (which would be from the
data set that starts with the
orange group on the right and ends with the gray group on the
left) is very close to the Within
Group (the average) variation for the three groups.

So, if we divide our estimate of the Between Group (overall)
variation by the estimate of
our Within Group (average) variation, we would get a value
close to 1, and certainly less than
about 1.5. Recalling the F statistic from last week, we could
guess that there is not a significant
difference in the variation estimates. (Of course, with the
statistical test we do not guess but
know if the result is significant or not.)
Look at three sample distributions in Exhibit A. Each has the
same within group
variance, and the overall variance of the entire data set is not all
that much larger than the
average of the three separate groups. This would give us an F
relatively close to 1.00.
Exhibit A: No Significant Difference with Overall Variation
0
0.05
0.1
0.15
0.2
0.25
0.3

0.35
0.4
0.45
-5 -4 -3 -2 -1 0 1 2 3 4 5
Exhibit B: Significant Difference with Overall Variation
Now, if we look at exhibit B, we see a different situation. Here
the group distributions do
not overlap, and the means are quite different. If we were to
divide the Between Group (overall)
variance by the Within Group (average) variance we would get a
value quite a bit larger than the
value we calculated with the pervious samples, probably large
enough to indicate a difference
between the within and between group variation estimates.
And, again, we would examine this F
value for statistical significance.
This is essentially what ANOVA does; we will look at how and
the output in the next
lecture. If the F statistic is statistically significant (the null
hypothesis of no difference is
rejected), then we can say that the means are different. Neat!
So, why bother learning a new tool to test means? Why don’t
we merely use multiple t-
tests to test each pair separately. Granted, it would take more
time that doing a single test, but
with Excel that is not much of an issue. The best reason to use

ANOVA is to ensure we do not
reduce our confidence in our results. If we use an alpha of
0.05, it is essentially saying we are
95% sure we made the right decision in rejecting the null.
However, if we do even 3 t-tests on
related data, our confidence drops to the P(Decision 1 correct +
Decision 2 correct + Decision 3
correct). As we recall from week 1, the probability of three
events occurring is the product of
each event separately, or .95*.95*.95 = 0.857! And in
comparing means for 6 groups (such as
means for the different grade levels), we have 16 comparisons
which would reduce our overall
confidence that all decisions were correct to 44%. Not very
good. Therefore, a single ANOVA
test is much better for our confidence in making the right
decision than multiple T-tests.
The hypothesis testing procedure steps are set up in a similar
fashion to what we did in
with the t-tests. There is a single approach to wording the null
and alternate hypothesis
statements with ANOVA:
Ho: All means are equal
0
0.05
0.1
0.15
0.2

0.25
0.3
0.35
0.4
0.45
-10 -5 0 5 10
Ha: At least one mean differs.
The reason for this is simple. No matter how many groups we
are testing, if a single mean
differs, we will reject the null hypothesis. And, it can get
cumbersome listing all possible
outcomes of one or more means differing for the alternate.
One issue remains for us if we reject the null of no differences
among the mean, which
means are different? This is done by constructing what we can
call, for now, difference
intervals. A difference interval will give us a range of values
that the “real” difference between
two means could really be. Remember, since the means are
from samples, they are close
approximations to the actual population mean, which might be a
bit larger or smaller than any
given mean. These difference intervals will take into account
the possible sampling error we
have. (How we do this will be discussed in lecture 2 for this
week.).

A difference interval might be -2 to +1.8. This says that the
actual difference when we
subtract one mean from another could be any value between -2
to +1.8. Since this interval says
the difference could be 0 (meaning the means could be the
same), we would find this pair of
means to be not significantly different. If, however, our
difference range was, for example, from
+1.8 to + 3.8 (the same range but all positive values), we would
say the difference between the
means is significant as 0 is not within the range.
ANOVA is a very useful tool when we need to compare multiple
groups. For example,
this can be used to see if average shipping costs are the same
across multiple shippers. The
average time to fill open positions using different advertising
approaches, or the associated costs
of each, can also be tested with this technique. With our equal
pay issues, we can look at mean
equality across grades of variables such as compa-ratio, salary,
performance rating, seniority, and
even raise.
Chi Square Tests
The ANOVA test somewhat relies upon the shape of the
samples, both with our
assumption that each sample is normally distributed with an
equal variance and with their
relative relationship (how close or distant they are). In many
cases, we are concerned more with
the distribution of our variables than with other measures. In
some cases, particularly with
nominal labels, distribution is all we can measure.

In our salary question, one issue that might impact our analysis
is knowing if males and
females are distributed across the grades in a similar pattern. If
not, then whichever gender holds
more higher-level jobs would obviously have higher salaries.
While this might be an affirmative
action or possible discrimination issue, it is not an equal pay for
equal work situation.
So, again, we have some data that we are looking at, but are not
sure how to make the
decision if things are the same or not. And, just by examining
means we cannot just look at the
data we have and tell anything about how the variables are
distributed.
But, have no fear, statistics comes to our rescue! Examining
distributions, or shapes, or
counts per group (all ways of describing the same data) is done
using a version of the Chi Square
test; and, after setting up the data Excel does the work for us.
In comparing distributions, and we can do this with discrete
(such as the number of
employees in each grade) variables or continuous variables
(such as age or years of service
which can take any value within a range if measured precisely
enough) that we divide into
ranges, we simply count how many are in each group or range.
For something like the
distribution of gender by grades; simply count how many males
and females are in each grade,
simple even if a bit tedious. For something like compa-ratio,

we first set up the range values we
are interested in (such as .80 up to but not including .90, etc.),
and then count how many values
fall within each group range.
These counts are displayed in tables, such as the following on
gender distribution by
grade. The first is the distribution of employees by grade level
for the entire sample, and the
second is the distribution by gender. The question we ask is for
both kinds of tables is basically
the same, is the difference enough to be statistically significant
or meaningfully different from
our comparison standard?
A B C D E F
Overall 15 7 5 5 12 6
A B C D E F
Male 3 3 3 2 10 4
Female 12 4 2 3 2 2
The answer to the question of whether the distributions are
different enough, when using
the Chi Square test, depends with the group we are comparing
the distribution with. When we
are dealing with a single row table, we need to decide what our
comparison group or distribution
is. For example, we could decide to compare the existing
distribution or shape against a claim
that the employees are spread out equally across the 6 grades
with 50/6 = 8.33 employees in each
grade. Or we could decide to compare the existing distribution
against a pyramid shape - a more

typical organization hierarchy, with the most employees at the
lower grades (A and B) and fewer
at the top; for example, 17, 10, 8, 7, 5, 3. The expected
frequency per cell does not need to be a
whole number. What is important is having some justification
for the comparison distribution
we use.
When we have multi-row tables, such as the second example
with 2 rows, the comparison
group is known or considered to be basically the average of the
existing counts. We will get into
exactly how to set this up in the next lecture. In either case the
comparison (or “expected”)
distribution needs to have the row and column total sums to be
the same as the original or actual
counts.
The hypothesis claims for either chi square test are basically the
same:
Ho: Variable counts are distributed as expected (a claim of no
difference)
Ha: Variable counts are not distributed as expected (a claim that
a difference exists)
Comparing distributions/shapes has a lot of uses in business.
Manufacturing generally
produces parts that have some variation in key measures; we can
use the Chi Square to see if the
distribution of these differences from the specification value is
normally distributed, or if the
distribution is changing overtime (indicating something is

changing – such as machine
tolerances). The author used this approach to compare the
distribution/pattern of responses to
questions on an employee opinion survey between departments
and the overall division.
Different response patterns suggested the issue was a
departmental one while similar patterns
suggested that the division “owned” the results, indicating
which group should develop ways to
improve the results.
Summary
This week we looked at two different tests, one that looks for
mean differences among
two or more groups and one that looks for differences in
patterns, distributions, or shapes in the
data set.
The Analysis of Variance (ANOVA) test uses the difference in
variance between the
entire data set and the average variance of the groups to see if
at least one mean differs. If so, the
construction of difference intervals will tell us which of the
pairs of means actually differ.
The Chi Square tests look at patterns within data sets and lets us
compare them to a
standard or to each other.
Both tests are found in the Data Analysis link in Excel and
follow the same basic set-up
process as we saw with the F and t-tests last week.
If you have any questions on this material, please ask your

instructor.
After finishing with this lecture, please go to the first
discussion for the week, and engage
in a discussion with others in the class over the first couple of
days before reading the second
lecture.
BUS308 Week 3 Lecture 2
Examining Differences – ANOVA and Chi Square
Expected Outcomes
After reading this lecture, the student should be familiar with:
1. Conducting hypothesis tests with the ANVOA and Chi Square
tests
2. How to interpret the Analysis of Variance test output
3. How to interpret Determining significant differences between
group means
4. The basics of the Chi Square Distribution.
Overview
This week we introduced the ANOVA test for multiple mean
equality and the Chi Square
tests for distributions. This lecture will focus on interpreting
the outcomes of both tests. The
process of setting them up will be covered in Lecture 3 for this
week.

ANOVA
Hypothesis Test
The week 3 question 1 asks if the average salary per grade is
equal? While this might
seem like a no-brainer (we expect each grade to have higher
average salaries), we need to test all
assumed relationships. This is much like our detectives saying
“we need to exclude you from the
suspect pool; where were you last night?” This example will, of
course use the compa-ratio
instead of the salary values you will use in the homework.
The ANOVA test is found in the Data | Analysis tab.
Step 5 in the hypothesis testing process asks us to “Perform the
test.” Here is a screen
shot of the ANOVA output for a test of the null hypothesis: “All
grade compa-ratio means are
equal.” For this question we will be using the ANOVA-Single
Factor option as we are testing
mean equality for a single factor, Grades. We will briefly cover
the other ANOVA options in
Lecture 3 for this week.
Note that The ANOVA single factor output includes the test
name, a summary table, and
an ANOVA table. The summary table that gives us the count,
sum, average, and variance for the
compa-ratios by the analysis groups (in this case our grades).

Note that we are assuming equal
variances within the grades within the population for this
example, and your assignment. This
may not actually be true for this example (note the values in the
Variance column), but we will
ignore this for now. ANOVA is somewhat robust around
violations on the variance equality
assumption – means it may still produce acceptable results with
unequal variances. There is a
non-parametric alternate if the variances are too different, but
we do not cover it in this course.
Please note that the column and row values are present in this
screenshot. These will be needed
as references in question 2.
The next table is the meat of the test. While for all practical
purposes, we are only
interested in the highlighted p-value, knowing what the other
values are is helpful. When we
introduced ANVOA in lecture 1, we discussed the between and
within groups variation. As you
recall, the between groups focused on the data set as a single
group and not distinct groups. For
the Between Groups row, we have an Sum of Squares (SS)
value, which is a raw estimate of the
variation that exists. The degrees of freedom (df) for Between
Groups equals the number of
groups (k) we have minus 1 (k-1), which equals 5 for our 6
groups. The Mean Square variation
estimate equals the SS divided by the df.
The Within Group focuses on the average variation for all our
groups. SS gives us the
same raw estimate as for the BG row. The df for Within Groups

is the total count (N) minus the
number of groups (N-k), or 44 for our 50 employees in the 6
groups. MSwg equals SS/df.
The F statistic is calculated by dividing the MSbg by MSwg.
The next column gives us
our p-value followed by the critical value of F (when the p-
value would be exactly 0.05). The
total line is the sum of the SS values and the overall df which
equal the total count -1 (N – 1).
(As with the t and F tests, we could make our decision by
comparing the calculated F
value (in cell O20, with critical value of F in cell Q20. We
reject the null when the calculated F
is greater than the critical F. The critical value of F or any
statistic in an Excel output table is the
value that exactly provides a p-value equaling our selected
value for Alpha. However, we will
continue to use the P-value in our decisions.)
Now that we have our test results, we can complete step 6 of the
hypothesis testing
procedure.
Step 6: Conclusions and Interpretation
What is the p-value? Found in the table, it is 0.0186 (rounded).
(Side note: at times Excel will produce a p-value that looks
something like 3.8E-14. This
is called the scientific or exponential format. It is the same as
writing 3.8 * 10-14 and
equals 0.000000000000038. A simple way of knowing how
many 0s go between the
decimal point and the first non-zero number is to subtract 1

from the E value, so with E-
14, we have 13 zeros. At any rate, any Excel p-value using E-
xx format will always be
less than 0.05.)
Decision: Reject the null hypothesis.
Why? P-value is less than 0.05.
Conclusion: at least one mean differs across the grades.
Question 2: Group Comparisons
Now that we know at least one grade compa-ratio mean is not
equal to the rest, we need
to determine which mean(s) differ. We do this by creating
ranges of the possible difference in
the population mean values. Remember, that our sample results
are only a close approximation
of the actual population mean. We can estimate the range of
values that the population mean
actually equals (remember that discussion of the sampling
distribution of the mean from last
week). So, using the variation that exists in our groups, we
estimate the range of differences
between means (the possible outcomes of subtracting one mean
from another).
The following screen shot shows a completed comparison table
for the grade related
compa-ratio means.
Let’s look at what this table tells us before focusing on how to

develop the values
(covered in Lecture 3 for this week). Looking at the Groups
Compared Column, we see the
comparison groups listed, A-B for grades A and B, A-C for
grades A and C, etc. The next
column is the difference between the average compa-ratio
values for each pair of grades. The T
value column is the value for a 95% two tail test for the degrees
of freedom we have. (Lecture 3
discusses how to identify the correct value). Note that it is the
same value for all of our
comparison groups, the explanation comes in Lecture 3.
The next column, labeled the +/- term, is the margin of error
that exists for the mean
difference being examined. This is a function of sampling error
that exists within each sample
mean. These are all of the values we need to create a range of
values that represent, with a 95%
confidence, what the actual population mean differences are
likely to be. We subtract this value
from the mean (in column B) to get our low-end estimate (Low
column values), and we add it to
the mean to get our high-end estimate (High column values).
Now, we need to decide which of these ranges indicates a
significantly different pair of
means (within the population) and which ranges indicate the
likelihood of equal population
means (non-significant differences). This is fairly simple, if the
range contains a 0 (that is, one
endpoint is negative and the other is positive), then the
difference is not significant (since a mean
difference of 0 would never be significant). Notice in the table,
that the A-B, A-C, and A-D
range all contain 0, and the results are not significant different.

The A-E and A-F comparisons,
however have positive values for each end, and do not contain
0; these means are different in the
population.
We now know how to interpret an ANOVA table and an
accompanying table of
differences for significant mean differences between and among
groups.
Chi Square Tests
With the Chi Square tests, we are going to move from looking at
population parameters,
such as means and standard deviations, and move to looking at
patterns or distributions. The
shape or distribution of variables is often an important way of
identifying differences that could
be important. For example, we already suspect that males and
females are not distributed across
the grades in a similar manner. We will confirm or refute this
idea in the weekly assignment.
Generally, when looking at distributions and patterns we can
create groups within our
variable of interest. For example, the Grades variable is already
divided into 6 groups, making it
easy to count how many employees exist in each group. But
what about a continuous variable
such as Compa-ratio, where no such clear division into separate
groups exists. This is not a
problem as we can always divide any range of values into
groups such as quartiles (4 groups) or

any other number of distinct ranges. Most variables can be
subdivided this way.
The Chi Square test is actually a group of comparisons that
depend upon the size of the
table the data is displayed in. We will examine different tables
and tests in Lecture 3, for this
lecture we want to focus on how to interpret the outcome of a
Chi Square test – as outcomes are
the same regardless of the table size. The details of setting up
the data will be covered in Lecture
3.
Example – Question 3
The third question for this week asks about employee grade
distribution. We are
concerned here about the possible impact of an uneven
distribution of males and females in
grades and how this might impact average salaries. While we
are concerned about an uneven
distribution, our null hypothesis is always about equality, so the
null would respond to a question
such as are males and females distributed across the grades in a
similar pattern; that is, we are
either males or females more likely to be in some grades rather
than others.
A similar question can be asked about degrees, are graduate and
undergraduate degrees
distributed across grades in a similar pattern? If not, this might
be part of the cause for unequal
salary averages.
The step 5 output for a Chi Square test is very simple, it is the
p-value, the probability of

getting a chi square value as large or larger than what we see if
the null hypothesis is true.
That’s it – the data is set up, the Chi Square test function is
selected from the Fx statistical list,
and we have the p-value. There is not output table to examine.
So, for an examination of are degrees distributed across grades
in a similar manner, we
would have an actual distribution table (counts of what exists)
looking like this:
Place the actual distribution in the table below.
A B C D E F Total
UnderG 7 5 3 2 5 3 25
Grad 8 2 2 3 7 3 25
Total 15 7 5 5 12 6 50
This table would be compared to an expected table where we
show what we expect if the null
hypothesis was correct. (Setting up this table is discussed in
Lecture 3.) Then we just get our
answer.
So, steps 5 and 6 would look like:
Step 5: Conduct the test. 0.85 (the Chi Square p-value from the
Chisq.Test function
Step 6: Conclusion and Interpretation
What is the p-value? 0.85

Decision on rejecting the null: Do Not Reject the null
hypothesis.
Why? P-value is > 0.05.
Conclusion on impact of degrees? Degrees are distributed
equally across the grades
and do not seem to have any correlation with grades. This
suggests they are not an
important factor in explaining differing salary averages among
grades.
Of course, a bit more of getting the Chi Square result depends
on the data set up than
with the other tests, but the overall interpretation is quite
similar – does the p-value indicate we
should reject or not reject the null hypothesis claim as a
description of the population?
Summary
Both the ANOVA and Chi Square tests follow the same basic
logic developed last week
with the F and t-tests. The analysis is started with developing
the first four (4) hypothesis testing
steps which set-up the purpose and decision-making rules for
the analysis.
Running the tests (step 5) will be covered in the third lecture
for this week.
Step 6 (Interpretation) is also done in the same fashion as last
week. Look for the p-value
for each test and compare it to the alpha criteria. If the p-value
is less than alpha, we reject the

null hypothesis.
When the null is rejected in the ANOVA test, we then create
difference intervals to
determine which pair of means differs. If any of these intervals
contains the value 0 (meaning
one end is a negative value and the other is a positive value),
we can say that those means are not
significantly different within the population.
The Chi Square has two tests that were presented. One test
looks at a single group
compared to an expected distribution, which we provide. The
other version compares two or
more groups to an expected distribution which is generated by
the existing distributions. How
these “expected” tables are generated will be discussed in
Lecture 3 for this week.
Please ask your instructor if you have any questions about this
material.
When you have finished with this lecture, please respond to
Discussion thread 2 for this
week with your initial response and responses to others over a
couple of days.

BUS 308 Week 3 Lecture 1 Examining Differences - Continued.docx

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