Need discussion for BUS308 Statistics for Managers details & lecture below.

 Read Lecture 1. React to the material in the lecture. What is new? Is there anything you found to be unclear? How could you relate these ideas to issues and problems within your degree (business administration) area? 

BUS308 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.