BST 322 Week Three Slides
Revised March 2020
Brooks Ensign, M.B.A.
Week Three
• More hypothesis testing, with many of the
same assumptions, but:
• This week is EASIER
• One chapter, with the focus on one test
Search Youtube
• In Youtube: search for
• Chi Square Bozeman Science
• Excellent video !!
Have you watched this video?
Excellent introduction to Chi Square
• Youtube: Chi Square Bozeman Science
Types of Statistical Tests
We have two broad type of tests:
(last week): Parametric tests (ex: t tests) Dependent
variable is interval or ratio and there is a Normal
distribution in a population , etc….
This Week: esp. KAI-square
Non- Parametric tests (ex: Chi-Square) variables
are measured on a nominal or ordinal scale (no
assumption about normal distribution)
We are usually talking about NOMINAL DVs
Parametric tests are more powerful
• Parametric tests are more powerful
• But sometimes ( i.e., NOW )we cannot meet
the more stringent requirements for
parametric tests, so we use nonparametric
tests instead.
Week Three: Easier
• Inferential statistics (like last week)
• But this week: “non”-parametric ( i.e.,
nominal and ordinal) D.V., with last week
being parametric (interval and ratio)
• 70% of effort: one test: Chi Square, pages
170-179
• 30% of effort: the “other” nonparametric
tests, pages 180-192 (do not calculate)
Week Three (“Chi-squared”) Χ 2
• The easiest week of the course! (week four is
easy also)
• One chapter: chapter 8 (mostly the first half)
• One test: the Χ 2 Chi (Kai)-square test
• Focus on the first half of the chapter
• Skim the second half, lightly. Look for the
concepts explained in this slide deck. You
need to know “just a little bit” about these
“other nonparametric tests”
Review
Test statistic - computed from a formula to
determine critical region of the data
If the absolute value we find for the test statistic is >
than a certain critical value Then the null
hypothesis is rejected and the result is significant.
Statistical significance - the results seen are
“PROBABLY not” attributable to chance
P value–the probability that the result occurred by chance
“They Work Together:”
Think of the test statistic and the p value as the opposite ends of a seesaw.
They work in opposite directions. For statistical significance, we want a
large test statistic (larger than the table value) and a small p value (smaller
than 0.05, i.e., alpha).
KAI-squared
Chi-Square X2
Test statistic
greater than
table value
P value less
than alpha
(0.05), the level
of significance
Review – cont.
Interpreting your p value
-For small p values (usually < 0.05), reject Ho,
Your data don’t support Ho and your evidence is
beyond a reasonable doubt
-For large p values (usually > 0.05), you can’t
reject Ho, you don’t have enough evidence against it
-If your p value is close to 0.05 – your results
marginal (could go either way)
Selecting the Test (next slide)
• The Next Slide is Very Important! ( course
roadmap for hypothesis test selection)
• Select the column first: DV, output data, the
dependent data: Nominal, Ordinal, Interval or
Ratio?
• Last week: the tests in the far right column:
t test and ANOVA
• This week: look for the Small Blue Box
Selecting the Test (next slide)
• The Next Slide is Very Important!
• Select the column first: output data, the
dependent data: Nominal, Ordinal, Interval or
Ratio?
• Select the Row: 2 groups? 3 (or more)?
• Select the Row: Are groups independent or
dependent?
• Look in the middle column (“nominal DV”) –
look for the Blue Box (middle, top)
This Week
Non-Parametric Tests
Last Week,
Week 2
Level of Measurement of Dependent Variable
Number of
Groups
Nominal
Ordinal
Interval/Ratio
2
Independent
χ2 test or
Fisher’s Exact
MannWhitney U
t-test
3+
Independent
χ2 test
Kruskal-Wallis
ANOVA
2
Dependent/
repeated
measures
McNemar
Wilcoxon
signed-ranks
Paired t-test
3+
Dependent/
repeated
measures
Cochran’s Q
Friedman
RANOVA
Selecting the Test (next slide)
• The Next Slide is Very Important!
• Select the column first: output data, the
dependent data: Nominal, Ordinal, Interval or
Ratio?
• Select the Row: 2 groups? 3 (or more)?
• Select the Row: Are groups independent or
dependent?
• Look in the middle column (“nominal DV”) –
look for the Blue Box (middle, top)
Tests for Independent Groups
Level of Measurement
(Outcome Variable)
Nominal
Number of
Measures
Groups
Two Groups Chi-square test
Three or
More
Groups
Chi-square test
Ordinal
Measures
Mann-Whitney
U test
Kruskal-Wallis
test
“KAI” Chi Square Test: “ Χ2 “
Chi ( “kai” ) Square:
Χ2
Most important topic this week
Friendlier name (English, not Greek):
Comparison of Proportions (observed
vs. expected)
Military lesson:
• Military expression: “You don’t get what you
expect, you get what you inspect.”
• So, … ( Observed minus Expected )
• How different is what we observe (inspect)
from what we expect?
• Chi Square: ( observed minus expected)
• Null hypothesis: observed - expected = 0
(because observed = expected )
Null hypothesis: O = E
• What is the null hypothesis? O – E = 0
• The opposite of our experimental idea
(alternative hypothesis).
• NULL HYP: There is no (“ null “) relationship
between the independent and dependent
variables
• In Chi Square: the “expected proportionality”
will be distributed in each group, because
group membership does not matter.
Apply hypothesis tests to Chi-Square
Deciding to accept or reject the Null hypothesis
(Polit p. 105)
If the absolute value of the computed statistic is greater than
the tabled value, the null hypothesis can be rejected and the
result is said to be statistically significant at the specified
probability level
Apply to Chi-Square
(Polit p. 171)
Null Hypothesis here is ______________
A. The variables are not independent (they are related).
B. The variables are independent (unrelated).
C. The variables are moderately related.
Vote now!
Apply hypothesis tests to Chi-Square
Deciding to accept or reject the Null hypothesis
(Polit p. 105)
If the absolute value of the computed statistic is greater than
the tabled value, the null hypothesis can be rejected and the
result is said to be statistically significant at the specified
probability level
Apply to Chi-Square
(Polit p. 171)
Null Hypothesis here is ______________
B. The variables are independent (unrelated).
… so the distribution should be “proportionate / pro rata”
And the expected results should be the observed results!
If the null hypothesis is true?
• (O minus E): The Chi-squared ( Χ2 ) statistic
will be (close to zero) zero, because there will
be no large difference between the expected
values and the observed values.
• The null hypothesis says that the expected
values should be “proportionately” (evenly)
distributed ( so, take the proportion, the
percentage from the “overall” column, and
apply it to each subgroup = Expected Values)
Kai-squared Test Statistic Value
X2 = 0 for two variables that are completely
unrelated (see the distribution curve, p. 171)
TABLED VALUE: Page 416, Table A.4
If the X2 value is less than the tabled value, we
accept the null hypothesis: not statistically
significant
If the X2 value is greater than the tabled value, we
reject the null hypothesis: statistically
significant
Discussion Question 1
fig 8.2
From our textbook, in the first six pages of chapter 8
We compare two protocols: change the heparin lock
after 72 hours vs. after 96 hours. We suspect that the
72-hour protocol may result fewer complications
(infections? blockage?) We see a small difference in the
data,
But:
IS THE DIFFERENCE STATISTICALLY SIGNIFICANT?
What is the nature of the dependent variable? Nominal
data: Complications, Yes, or No (count the
complications) – counting is the only calculation we can
make with nominal data
Teaser: the critical values
•
•
•
•
•
•
Page 416
The critical values (table values) are easy
To find, on page 416
For 2X2 tables the critical value is 3.84
For 3X2: the critical value value is 5.99
Why? Stay tuned
The results (from the book)
• At the bottom of the next page:
• Chi-squared value is 0.25 (very low), < 3.84
• P value is 0.6 (very high), > 0.05
• These results are NOT SIGNIFICANT, probably
could have occurred randomly
Discussion Question 1
fig 8.2
Chi Sq
Use StatCrunch
StatCrunch is much easier to use, as
explained in the video (#9).
Most of the time: Use the
“WITH SUMMARY” method
StatCrunch this week, W3:
• Use “With Summary” for both DQs and for
first some homework questions. No data file
needed. (if you already have summarized
data, in a small table)
• Ask for help (my written guidance): THE WITH
SUMMARY METHOD
• For other HW questions : use “with data” – I
will help you. There is a data file in the
Blackboard course shell. The “with data”
approach is the same as W1 DQ2 (see video).
StatCrunch Results – remember
to add the “finishing touches” –
see the Cross-tabulation table
video in DQ-2, Week One;
Try to make your table look like
The table on page 61.
Contingency table results:
Rows: Complic
Columns: Group
Cell format
Count
(Row percent)
(Column percent)
(Total percent)
Expected count
72hr
no
yes
96hr
Total
41
(51.25%)
(82%)
(41%)
40
39
80
(48.75%)
(100.00%)
(78%)
(80%)
(39%)
(80%)
40
9
(45%)
(18%)
(9%)
10
11
20
(55%)
(100.00%)
(22%)
(20%)
(11%)
(20%)
10
50
50
100
(50%)
(50%) (100.00%)
Total
(100.00%) (100.00%) (100.00%)
(50%)
(50%) (100.00%)
Chi-Square test:
Statistic DF Value P-value
Chi-square
1 0.25 0.6171
Chi-squared value is 0.25
P value is 0.6
DQ2 Week Three
• For this second question - let's try to make a crosstabulation
table for a problem we don't have in the textbook.
• Using the data set provided above from a Pew research study
on dementia in 400 Nashville elderly residents (labeled
"Week3DiscussionQuestion2Data"), can you create a
StatCrunch crosstabulation table (copied and pasted, as well
as updated in Word) result for "used pot" and "Forgot date"?
And the Chi-Square analysis? What hypotheses did you test?
What can you conclude using marijuana and dementia in this
sample?
W3 DQ2
• can you create a StatCrunch crosstabulation
table (copied and pasted, as well as updated
in Word) result for "used pot" and "Forgot
date"? And the Chi-Square analysis? What
hypotheses did you test? What can you
conclude using marijuana and dementia in
this sample?
Contingency table results:
Rows: Forgot Date
Columns: Usedpot
Null hypothesis: There is no relationship between marijuana use
and forgetting the date
Alternative hypothesis: The two variables between marijuana use
and forgetting the date are related.
X2 =0.56017047, p-value= 0.7557, alpha=0.05, critical value=5.99,
df=2
The results are not statistically significant because the x2 is less than
the critical value and the p-value is greater than the level of
significance. Therefore, we must fail to reject the null hypothesis. It
shows that there is no relationship between marijuana use and
forgetting the date.
Start with O minus E
• O: the observed value
• E: the expected value
• Essence of Chi-Square (Χ2) is: Subtract E from
O: Χ2 = O minus E
• Then: calculate: Χ2 = ( O – E )2 / E
• The problem gives us the O values. How do
we find the E values? (stay tuned, - we apply
“expected proportionality”)
Contingency Table Example
Does this experimental therapy reduce incontinence?
ExperimentControl
al Group (E) Group (C)
Total
10
20.0%
20
40.0%
30.0%
Not
Incontinent
40
80.0%
30
60.0%
70
70.0%
Total
50
100.0%
50
100.0%
100
100.0%
Incontinent
•A
30
higher proportion of Cs than Es were incontinent—but is
this just random fluctuation?
Divide by E ??
• E cannot be zero … why not? What is the title
of this slide? … dividing by zero is not allowed
• E may also be a problem if it is too low … look
for the explanation about the “corrections”
(Yates and Fisher) on page 193 and later in
these slides
Start with O minus E
• O: the observed value
• E: the expected value
• Essence of Chi-Saquare is: Subtract E from O:
= O minus E
• Then: calculate: ( O – E )2 / E
• The problem gives us the O values. How do
we find the E values? (stay tuned, - we apply
“expected proportionality”)
Contingency Table Example
these are the Observed values
ExperimentControl
al Group (E) Group (C)
Total
10
20.0%
20
40.0%
30.0%
Not
Incontinent
40
80.0%
30
60.0%
70
70.0%
Total
50
100.0%
50
100.0%
100
100.0%
Incontinent
30
• A higher proportion of Cs than Es were incontinent—but is this just random
fluctuation?
Why can’t E be zero?
• The formula: For each cell in the table,
compute the following:
• Chi-squared: Χ2 = (O – E)2 ÷ E
• Χ2 ( say kai-squared, “Chi-squared,” not “ex”-squared
• Can you divide by zero? NO!!!
• We actually want the E, the expected value to be
larger than 10. If it is greater than zero, but less than
10, we need to use either the Yates Correction or the
Fisher Exact Test (stay tuned, and see p. 193)
Computation of Chi-Square
• Page 173: For each cell, compute the
following: Χ2 = (O – E)2 ÷ E
• Cell A in our example:
(10 - 15)2 ÷ 15 = 1.67
• Then add all the cell components together to
obtain χ2
• In our example with four cells:
χ2 = 1.67 + 1.67 + .71 + .71 = 4.46
Q: How Do We Get the Expected
Values? (proportionality)
even though StatCrunch does this for us !!!
A: We apply the “overall
proportion” (the expected
proportion) to each cell
But, How???
Expected Values (page 172)
χ2 = (∑ (Oij – Eij)2 ) / Eij
with df = (r-1)(c-1)
Where
Oij = observed cell frequencies
E = expected proportions
Eij = expected cell frequencies = row total X column total
Total count (N)
r = number of rows
c = number of columns
Calculating the Expected Value
Eij = expected cell frequencies =
row total X column total / Total count (N)
Or (preferred explanation, intuitive):
(row total / Total count (N) ) x times column total
Provides “proportionality”
Observed Versus Expected
Frequencies
E Group
C Group
Total
Incontinent
10 (20.0%)
(CELL A)
20 (40.0%)
30 (30.0%)
Not Incontinent
40 (80.0%)
30 (60.0%)
70 (70.0%)
Total
50 (100.0%)
50 (100.0%)
100 (100.0%)
• If null true, both Es and Cs would have 30%
incontinent (see Total Row %): 15 each
• Cell A, Observed (OA) = 10 Expected (EA) = 15
• EA = Row TotA (30) ÷ N (100) = 30% and
• 30% Col TotA (50) = 15 (E) expected value
“ phi “
in a 2X2
Required vocabulary
Know this definition from Chapter 8
• phi is the measure of strength of
relationship most commonly used with a
2X2 table.
Testing Significance of Chi-Square
• Why is your d.f. probably = 1 or 2
• ??? Why table value = 3.84 or 5.99 ??
• Page 416: Table of critical values requires knowing (1) df
and (2) significance criterion (e.g., .05)
• Can also use: =CHIINV(.05,df)
• df is “usually” just = 1 or = 2, because …
• In χ2 , df =(#Rows – 1) (#Columns – 1)
– Here: df = (2 – 1) (2 – 1) = 1
• If calculated χ2 > tabled value, results are significant
– Critical value for df = 1 and α = .05: 3.84
χ2 = 4.46, so null hypothesis is rejected
3.84 or 5.99 will often be the critical values – WHY?? Page 416
The critical value
• Next slide: Page 416: the “table” ( critical)
value
• Will usually be 3.84 or 5.99 ( Why?)
• Use shaded column ( α = 0.05 )
• Row: usually one or two, why?
• d.f. = ( r -1 ) x ( c – 1 ) = probably 1 or 2
• With a 2X2 table: d.f. = 1, so: 3.84
• With a 3X2 table: d.f. = 2, so: 5.99
Remember the Happy(?) Married
Men?
• We asked last week about “happiness,”
measured with a continuous variable that we
assumed to have a normal distribution (the
“happiness score”)
• Some of us questioned whether this score was
a valid measurement. Good question!
• Let’s use a nominal dependent variable: are
you, or are you not, on Prozac?
Another Example (2 X 3 table)
(same as Assignment question #1)
Happiness scores for 3 groups of men:
Married: 42 45 43 48 45 47 48 46 35 50
Single: 49 50 45 38 47 34 49 44 41 42
Divorced: 37 47 50 44 41 41 42 46 45 50
The same 3 group of men (10 in each group) were asked if
they are on Prozac or not:
Data:
2 of the married men and 2 of the single men and 7 of the
divorced men were on Prozac
Is the chi-square value statistically significant?
Married
Prozac: Yes
Prozac: No
Total
Prozac: Yes
Prozac: No
Total
Single
2
8
10
Expected Frequencies
Married
Single
3.7
6.3
10
Divorced
2
8
10
Total
7
3
10
Divorced
3.7
6.3
10
Excel solution to the Prozac question –
look for the StatCrunch solution in the
Discussion Questions…
11
19
30
Total
3.7
6.3
10
11
19
30
Test Results
0 Correction
7.177 χ2
2 Rows
3 Columns
2 df
0.028 p (χ2)
0.489 V (or φ)
This example is shown in the video by Dr. Myers, with
StatCrunch; watch this video!
Prozac example: See the Discussion Questions (#1)
SOLVED IN THE DISCUSSION
QUESTION ONE POSTING
Homework: final HW-Question
Chi-Square “Corrections”
• Smallest expected value:
• Homework Question guidance Page 193:
• When an expected frequency is low, in a 2X2
table (only relevant for 2X2 tables):
• Look at the “lowest cell value” in the expected
table
• If less than 10: Yates Correction
• If less than 5: Fisher’s Exact test
• If ten or greater: Chi-Square with no correction
• If expected value is zero: CAN’T USE CHI-SQUARE
The “Other” Nonparametrics
SKIM THE SECOND HALF OF
CHAPTER EIGHT, FOLLOWING THE
SLIDES IN THIS PRESENTATION
Page 180: Ready for the easier
stuff? The “other” tests
Introducing (just introducing) the
“Other Nonparametric Tests” (harder, but easier;
why?)
( pages 180-190)
More advanced tests, used much less frequently
Easier: In this class, we do not “use” these tests, we
just “talk about them”
Selecting the Statistical Test:
1. Identify your dependent variable: is it “continuous / scale”
(i.e. interval or ratio ?) – if so, then use t test or ANOVA, the
parametric tests from Week Two
2. With your dependent variable: choose the column in the
following table
3. How many groups in the independent variable? Two or…
“Three or more”
4. Are the groups in the independent variable related or
unrelated?
See the following table, from Chapter 8 (page182)
Also: see the inside cover of the textbook:
- Week Two tests are on the left side
- Week Three tests are on the right side
Selecting the Test (next slide)
• The Next Slide is Very Important!
• Select the column first: output data, the
dependent data: Nominal, Ordinal, Interval or
Ratio?
• Select the Row: 2 groups? 3 (or more)?
• Select the Row: Are groups independent or
dependent?
Selecting the Statistical Test
Level of Measurement of Dependent Variable
Number of
Groups
Nominal
Ordinal
Interval/Ratio
2
Independent
χ2 test or
Fisher’s Exact
MannWhitney U
t-test
3+
Independent
χ2 test
Kruskal-Wallis
ANOVA
2
Dependent/
repeated
measures
McNemar
Wilcoxon
signed-ranks
Paired t-test
3+
Dependent/
repeated
measures
Cochran’s Q
Friedman
RANOVA
Ready for the “easy” stuff?
• The rest of chapter 8 is “harder” because the
tests seem obscure and technical, BUT
• EASIER: Back away !!! : we just want to
“SKIM”
• Seek a “light / conceptual” (no math)
understanding of the “other nonparametric
tests” (from these slides)
• Pages 180-192: a “light skim” will do…
Let’s skim the second half of
Chapter Eight (the “other” tests)
HOW FAST DO YOU WANT TO GO?
“Analogs” / “Counterparts:”
Tests that are in the same row,
because they share some of the
same inputs
Parametric-Nonparametric Analogs
(analogs are in same row)
Parametric Test
Nonparametric Test
Independent groups
t-test
Dependent groups ttest
One-way ANOVA
Mann-Whitney U test
RM-ANOVA
Friedman test
Wilcoxon signedranks test
Kruskal-Wallis test
Tests for Independent Groups
Level of Measurement
(Outcome Variable)
Nominal
Number of
Measures
Groups
Two Groups Chi-square test
Three or
More
Groups
Chi-square test
Ordinal
Measures
Mann-Whitney
U test
Kruskal-Wallis
test
After mastering Chi-Square tests, what do you need to know
about the “Other” Non-Parametric Tests (after page 182)?
[ “not much” ] Learn “WHY / WHEN” (not HOW)
What is the nature of the Dependent Variable?
In the IV: 2 groups or 3 (or more?)
Are the groups in the IV independent or related?
Find your place in the master table, from page 178…
What is the name of the “other” non-parametric test?
What is the name of the variable?
What information is on the following slides?
Just learn the
information in these slides
No math is required! -
Mann-Whitney U test
• Non-parametric tests that involve ordinal
dependent variables use some form of
ranking. The Mann-Whitney U test is
popularly used when the dependent variable
is ordinal and is the analog of the independent
groups t-test.
Kruskal-Wallis Test
• Tests the null hypothesis that three or
more population distributions are identical:
– The nonparametric analog of one-way ANOVA
• Compares the ranks of the values for the
groups
• Test statistic is H, which follows chi-square
distribution
Tests for Dependent Groups
Number of
Groups (or
Measurement
Periods)
Two
Three or More
Level of Measurement
(Outcome Variable)
Nominal
Measures
Ordinal
Measures
McNemar Test Wilcoxon SignedRanks Test
Cochran’s Q
Test
Friedman Test
McNemar Test
• Tests differences in proportions for the
same people measured twice (or for
paired groups, like mothers/ daughters)
• Yields a statistic distributed as a chisquare, with df = 1
Wilcoxon Signed-Ranks Test
• Tests differences in ordinal-level measures for
the same people measured twice (or for
paired groups, like Sibling A/Sibling B)
– The nonparametric analog of a paired t-test
• Another example of a rank test
• For n > 10, it follows a normal distribution, so
the test statistic is z
Cochran’s Q Test
• Tests differences in proportions for the
same people measured three or more
times (or correlated groups)
• Yields a statistic distributed as a chisquare, with df = 1
• Not many applications in the nursing
literature
Friedman Test
• Tests differences in ordinal-level measures for
the same people measured three or more
times (or for correlated groups)
– The nonparametric analog of an RM-ANOVA
• Another example of a rank test
• Test statistic is a chi-square with (k – 1)
degrees of freedom (k = number of
measurements)
Independent Project
• Let’s talk about:
– The Independent Project !!
– Not due until the very end of the course, but let’s
understand what is required.
– Data file is in the course shell
– I will send specific guidance … ( you know I will!)
– Chi Square IS REQUIRED in the project (Q3)
Independent Project
Use the data set supplied in docsharing and using StatCrunch,
provide the following for the variable(s) of your choice:
1.Frequency distribution of a variable and bar graph of the same
variable
2.Descriptives of a continuous variable : mean, median, mode,
skewness, kurtosis, standard deviation
3.Cross tabulation of two variables
4.Comparison of the effect of three or more groups (single variable)
on a single continuous variable
5.Scatterplot of two continuous variables
6.Correlation between the two continuous variables (from #5)
Think carefully about what kind of variables to choose for the given
tasks.
Ask questions now – we can discuss. I will send you a lot of
information to help you.
Independent Project:
hints in purple
1.Frequency distribution of a NOMINAL variable and bar graph of
the same variable
2.Descriptives of a continuous variable : mean, median, mode,
skewness, kurtosis, standard deviation – define terms, provide
histogram
3.Cross tabulation of two variables with Chi Square test and
analysis
4.Comparison of the effect of three or more groups (single variable)
on a single continuous variable One-Way Anova
5.Scatterplot of two continuous variables
6.Correlation between the two continuous variables (from #5) with
significance testing ( page 199), easy in StatCrunch
Think carefully about what kind of variables to choose for the given
tasks.
Ask questions now – we can discuss. I will send you a lot of
information to help you.
Independent Project (cont)
REQUIRED WRITTEN PARAGRAPH: A descriptive written
paragraph should accompany each of the above including a
description of the variables used and any meaning that may be
attached to the results. The student must show that she or he is
able to synthesize and apply the materials learned in class. Part of
the class computer time is expected to be spent on this project.
Submit to the appropriate Drop Box in a Word document.
Grading on this project is as follows:
3 points for each task 1-6: 1 point each for variable choice,
appropriate display/test, description of result.
2 points for overall format/readability/construct (so make it neat and
tidy)
Project guidelines, More guidance:
• ANOVA cannot use “discrete quantitative” variable as the
dependent variable; example: number of pregnancies is not
approximately normal, not continuous, same problem for #
miscarriages
• Write a paragraph for each question, define your terms
• Variable selection is critical; what is appropriate /
inappropriate ??

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