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2. BNAD 276: Statistical Inference in Management Spring 2016
Welcome
Green sheets

4. Schedule of readings Before our next exam (April 7th)
OpenStax Chapters 1 – 12
Plous (2, 3, & 4)
Chapter 2: Cognitive Dissonance
Chapter 3: Memory and Hindsight Bias
Chapter 4: Context Dependence

5. By the end of lecture today 3/24/16
Analysis of Variance (ANOVA)
Constructing brief, complete summary statements

6. “df” = degrees of freedom
Remember, you should know these two formulas by heart
“SS” = “Sum of Squares”
“SS” = “Sum of Squares”
= s2 =
Sample
Variance
Sample
Standard Deviation
= s =
“SS” = “Sum of Squares”
“df” = degrees of freedom

7. Study Type 3: One-way Analysis of Variance (ANOVA)
We are looking to compare two means
Study Type 2: t-test
Study Type 3: One-way Analysis of Variance (ANOVA)
Comparing more than two means

8. Review Study Type 3: One-way ANOVA
Single Independent Variable comparing more than two groups
Single Dependent Variable (numerical/continuous)
Used to test the effect of the IV on the DV
Ian was interested in the effect of incentives for girl scouts on the number
of cookies sold. He randomly assigned girl scouts into one of three groups.
The three groups were given one of three incentives and looked to see who
sold more cookies. The 3 incentives were 1) Trip to Hawaii, 2) New Bike
or 3) Nothing. This is an example of a true experiment
Dependent
variable is always quantitative
Sales per
Girl scout
Sales per
Girl scout
None
New
Bike
Trip
Hawaii
None
New
Bike
Trip
Hawaii
In an ANOVA, independent variable is qualitative
(& more than two groups)
Review

9. One-way ANOVA versus Chi Square
Be careful you are not designing a Chi Square
If this is just frequency you may have a problem
This is a
Chi Square
Total Number
of Boxes Sold
Sales per
Girl scout
This is an ANOVA
None
New
Bike
Trip
Hawaii
None
New
Bike
Trip
Hawaii
These are just frequencies
These are just frequencies
These are just frequencies
These are means
These are means
These are means

10. One-way ANOVA One-way ANOVAs test only one independent variable
Number of
cookies sold
One-way ANOVA
None
Bike
Hawaii trip
Incentives
One-way ANOVAs test only one independent variable
- although there may be many levels
“Factor” = one independent variable
“Level” = levels of the independent variable
treatment
condition
groups
“Main Effect” of independent variable = difference between levels
Note: doesn’t tell you which specific levels (means) differ from each other
A multi-factor experiment would be a multi-independent
variables experiment

11. Comparing ANOVAs with t-tests
Similarities still include:
Using distributions to make decisions about
common and rare events
Using distributions to make inferences about
whether to reject the null hypothesis or not
The same 5 steps for testing an hypothesis
Tells us generally about number of participants / observations
Tells us generally about number of groups / levels of IV
The three primary differences between t-tests and ANOVAS are:
1. ANOVAs can test more than two means
2. We are comparing sample means indirectly by
comparing sample variances
3. We now will have two types of degrees of freedom
t(16) = 3.0; p < F(2, 15) = 3.0; p < 0.05
Tells us generally about number of participants / observations

12. A girl scout troop leader wondered whether providing an
incentive to whomever sold the most girl scout cookies would
have an effect on the number cookies sold. She provided
a big incentive to one troop (trip to Hawaii), a lesser
incentive to a second troop (bicycle), and no incentive to
a third group, and then looked to see who sold more cookies.
How many levels of the Independent Variable?
What is Independent Variable?
Troop 1
(nada) 10 8 12 7 13
Troop 2
(bicycle) 12 14 10 11 13
Troop 3
(Hawaii) 14 9 19 13 15
What is
Dependent Variable?
How many groups?
n = 5
x = 10
n = 5
x = 12
n = 5
x = 14

13. Hypothesis testing: Step 1: Identify the research problem
Is there a significant difference in the number of
cookie boxes sold between the girlscout troops that were
given the different levels of incentive?
Describe the null and alternative hypotheses

14. Hypothesis testing: Decision rule = .05
Degrees of freedom (between) = number of groups - 1
= = 2
Degrees of freedom (within) = # of scores - # of groups
= (15-3) = 12*
Critical F (2,12) =
3.98
*or = (5-1) + (5-1) + (5-1) = 12.

15. Appendix
B.4 (pg.518)
F (2,12)
α= .05
Critical F(2,12)
= 3.89 15

16. ANOVA table Source df MS F SS Between 40 ? ? ? ? Within 88 ? ? ? Total
“SS” = “Sum of Squares” - will be given for exams - you can think of this as the numerator in a standard deviation formula
ANOVA table
Source df MS F SS
Between 40 ? ? ? ?
Within 88 ? ? ?
Total
128 ? ?

17. “df” = degrees of freedom
Remember, you should know these two formulas by heart
“SS” = “Sum of Squares”
“SS” = “Sum of Squares”
= s2 =
Sample
Variance
Sample
Standard Deviation
= s =
“SS” = “Sum of Squares”
“df” = degrees of freedom

18. Writing Assignment - ANOVA
1. Write formula for standard deviation of sample
2. Write formula for variance of sample
3. Re-write formula for variance of sample using the nicknames
for the numerator and denominator SS df
= MS
4. Complete this ANOVA table
ANOVA table
Source SS df MS F
Between 40 ? ? ?
Within 88 ? ?
Total
128 ?

19. ANOVA table Source df MS F SS Between 40 ? 2 ? ? ? Within ? 88 12 ? ?
“SS” = “Sum of Squares” - will be given for exams
ANOVA table
Source df MS F SS
Between 40 ? 2
# groups - 1 ? ? ?
3-1=2
15-3=12
Within ? 88 12 ? ?
# scores - number of groups
Total
128 ? ? 14
# scores - 1
15- 1=14

20. “SS” = “Sum of Squares” - will be given for exams
ANOVA table
SSbetween
dfbetween
“SS” = “Sum of Squares” - will be given for exams 40 2 40 2
=20
MSbetween
MSwithin
ANOVA table
Source df MS F SS 20
7.33
=2.73
Between 40 2 ? 20 ?
2.73
Within 88 12
7.33 ?
Total
128 14
SSwithin
dfwithin 88 12
=7.33 88 12

21. Make decision whether or not to reject null hypothesis
Observed F = 2.73
Critical F(2,12) = 3.89
2.73 is not farther out on the curve than 3.89
so, we do not reject the null hypothesis
F(2,12) = 2.73; n.s.
Conclusion: There appears to be no effect of type of
incentive on number of girl scout cookies sold
The average number of cookies sold for three different incentives
were compared. The mean number of cookie boxes sold for the
“Hawaii” incentive was 14 , the mean number of cookies boxes sold
for the “Bicycle” incentive was 12, and the mean number of cookies
sold for the “No” incentive was 10. An ANOVA was conducted and
there appears to be no significant difference in the number of
cookies sold as a result of the different levels of incentive
F(2, 12) = 2.73; n.s.

22. incentive then the means are significantly different from each other
Main effect of incentive: Will offering an incentive result in more girl scout cookies being sold?
If we have a “effect” of
incentive then the means
are significantly different
from each other
we reject the null
we have a significant F
p < 0.05
To get an effect we want:
Large “F” - big effect and small variability
Small “p” - less than 0.05 (whatever our alpha is)
We don’t know which means are different from which …. just that they are not all the same 22

23. Let’s do same problem Using MS Excel
A girlscout troop leader wondered whether providing an
incentive to whomever sold the most girlscout cookies would
have an effect on the number cookies sold. She provided
a big incentive to one troop (trip to Hawaii), a lesser
incentive to a second troop (bicycle), and no incentive to
a third group, and then looked to see who sold more cookies.
Troop 1
(Nada) 10 8 12 7 13
Troop 2
(bicycle) 12 14 10 11 13
Troop 3
(Hawaii) 14 9 19 13 15
n = 5
x = 10
n = 5
x = 12
n = 5
x = 14

24. Let’s do one Replication of study (new data)

25. Let’s do same problem Using MS Excel

26. Let’s do same problem Using MS Excel

27. ANOVA table SSbetween “Sum of Squares” 40 =20 dfbetween 2 MSbetween
MSwithin
# groups - 1 20
7.33
=2.73
3-1=2 88 12
=7.33
# scores - # of groups
SSwithin
dfwithin
15-3=12
# scores - 1
15- 1=14

28. No, so it is not significant Do not reject null
F critical (is observed F greater than critical F?)
P-value (is it less than .05?)
“Sum of Squares”

29. Make decision whether or not to reject null hypothesis
Observed F = 2.73
Critical F(2,12) = 3.89
2.7 is not farther out on the curve than 3.89
so, we do not reject the null hypothesis
Also p-value is not smaller than 0.05
so we do not reject the null hypothesis
Step 6: Conclusion: There appears to be no effect of type of
incentive on number of girl scout cookies sold

30. Make decision whether or not to reject null hypothesis
Observed F =
F(2,12) = 2.73; n.s.
Critical F(2,12) =
2.7 is not farther out on the curve than 3.89
so, we do not reject the null hypothesis
Conclusion: There appears to be no effect of type of
incentive on number of girl scout cookies sold
The average number of cookies sold for three different incentives
were compared. The mean number of cookie boxes sold for the
“Hawaii” incentive was 14 , the mean number of cookies boxes sold
for the “Bicycle” incentive was 12, and the mean number of cookies
sold for the “No” incentive was 10. An ANOVA was conducted and
there appears to be no significant difference in the number of
cookies sold as a result of the different levels of incentive
F(2, 12) = 2.73; n.s.

31. Thank you!
See you next time!!