Business Email Rubric Subject Line Subject line clearly states the main point of the email 5points Subject line is a bit long (5+ words) or a bit short (1 word) but states the point of the email 3points Subject line does not correspond to the main point of the email 2points Subject line is missing 0points Greeting Email includes a professional greeting that is appropriate for the audience; uses the person's first name 5points Email includes a professional greeting that is adequate for the audience but uses the person's first and last name or just the last name 3points Email includes a greeting but it is not personalized 2points Email lacks a greeting 0points Introductory Comment Email includes an introductory positive, relevant comment 5points it Email includes an introductory, positive comment but may be very general 3points Email includes an introduction only 2points Email lacks an introductory comment 0points Content Purpose of the email is clear, as is the outcome; content is succinct and well organized and does not include unnecessary information 5points Purpose of the email is clear, as is the outcome; content could be better organized 3points Purpose of the email is not clear and/or content is poorly organized; it took more than one reading to understand what the email is about 2points Email seems to be a collection of unrelated statements; it is difficult to figure out what the purpose is 0points Closing and Signature Email includes a complementary closing and signature with all required items (name, title/position, company name, phone number) 5points Email includes a complementary closing and signature but the signature is incomplete 3points Email may be missing either a complementary closing or signature 2points Email lacks a closing and signature 0points Writing Conventions Email looks professional and does not have any formatting or writing errors 5points Email is well presented with minimal (<3) formatting or writing errors 3points Email includes several (3+) formatting or writing errors 2points Email is poorly presented and has an accumulation of writing errors that interfere with readability 0points 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 distrib ...

1. Business Email Rubric
Subject Line
Subject line
clearly states the
main point of the
email
5points
Subject line is a
bit long (5+
words) or a bit
short (1 word) but
states the point of
the email
3points
Subject line does
not correspond to
the main point of

2. the email
2points
Subject line is
missing
0points
Greeting
Email includes a
professional
greeting that is
appropriate for the
audience; uses
the person's first
name
5points
Email includes a
professional
greeting that is
adequate for the

3. audience but uses
the person's first
and last name or
just the last name
3points
Email includes a
greeting but it is
not personalized
2points
Email lacks a
greeting
0points
Introductory
Comment
Email includes an
introductory
positive, relevant
comment

4. 5points
it Email includes
an introductory,
positive comment
but may be very
general
3points
Email includes an
introduction only
2points
Email lacks an
introductory
comment
0points
Content
Purpose of the
email is clear, as
is the outcome;

5. content is succinct
and well organized
and does not
include
unnecessary
information
5points
Purpose of the
email is clear, as
is the outcome;
content could be
better organized
3points
Purpose of the
email is not clear
and/or content is
poorly organized;
it took more than

6. one reading to
understand what
the email is about
2points
Email seems to be
a collection of
unrelated
statements; it is
difficult to figure
out what the
purpose is
0points
Closing and
Signature
Email includes a
complementary
closing and
signature with all

7. required items
(name,
title/position,
company name,
phone number)
5points
Email includes a
complementary
closing and
signature but the
signature is
incomplete
3points
Email may be
missing either a
complementary
closing or
signature

8. 2points
Email lacks a
closing and
signature
0points
Writing
Conventions
Email looks
professional and
does not have any
formatting or
writing errors
5points
Email is well
presented with
minimal (<3)
formatting or
writing errors

9. 3points
Email includes
several (3+)
formatting or
writing errors
2points
Email is poorly
presented and has
an accumulation
of writing errors
that interfere with
readability
0points
BUS308 Week 3 Lecture 2
Examining Differences – ANOVA and Chi Square

10. 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.

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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.