1. Lecture #27 Tuesday, November 29, 2016 Textbook: 15.1
Statistics 200
Lecture #27 Tuesday, November 29, 2016
Textbook: 15.1
Objectives:
• Review hypothesis testing for a relationship between two categorical variables in a population.
• Extend techniques beyond the 2x2 table case.
• Predict how larger samples will influence chi-square statistic.

2. Motivating Example: Have you ever missed class because of alcohol?
1. Have you ever missed class because you were drinking the night before?
We asked a random sample of PSU students the following questions:
What kind of variables are these?
Quantitative
Categorical
2. What year are you in school?
a. Freshman b. Sophomore
c. Junior d. Senior

3. 2 Way Contingency Tables
Summarizing responses:
Generic notation:
(explanatory)×(response) table
Example:
(4 × 2) table implies:
explanatory variable: ___ levels
response variable: ____ levels Yr
Yes No
Total Fr 15 85
100 So 25 75 Jr 30 70 Se 50 4 2

4. Have you ever missed class because of alcohol?
PSU Students
n1 = 100 Fr
n2 = 100 So
n3 = 100 Jr
n4 = 100 Se
Which is the explanatory variable?
Year in school.
Missed class or not
Student Yr
Yes No
Total Fr 15 85
100 So 25 75 Jr 30 70 Se 50
Response Variable:
_______ answer
k = _____ samples
Samples are…
Dependent
Independent
Y / N 4
___ x ____ table

5. Null Hypothesis: H0 Starting Position: __________ is happening Nothing
Null: With (2 way Tables):
In the ______________ there is:
____ relationship between the two variables
____ difference in the groups
population No No
Statistical Hypotheses:
never include statements about ________
sample
Never trying to prove the ______ is true
null

6. Alternative Hypothesis: Ha
Challenging Position: __________ is happening
Something
Goal: hope the _____ support the alternative
data
Alternative: With (2 × way Tables):
In the population, there is:
_____ relationship between the two variables;
_____ difference in the groups
some
some

7. We can state hypotheses two ways
Null Hypothesis:
H0: In the population, no relationship between school year and the likelihood of missing class because of alcohol.
H0:
p1 = p2 = p3 = p4
Alternative Hypothesis:
Ha: In the population, a relationship exists between school year and the likelihood of missing class because of alcohol
Ha:
at least one pi differs

8. Do descriptive results show support for Ha?
Row Percents Yr
Yes No
Total Fr
15 (15%) 85
100 So
25 (25%) 75 Jr
30 (30%) 70 Se
50 (50%) 50
120
280
400
Yes – those percents look different for the 4 groups

9. We quantify evidence for the alternative using…
Chi-Square statistic
Next step: Find chi-square statistic using sample data
This statistic measures the difference in the __________ and ___________ counts in the contingency table.
Calculate using sample data
observed
expected

10. Chi-square Contribution =
Expected count = Yr
Yes No
Total Fr
15 (15%) E:
Chi: 85
100 So
25 (25%) 75 Jr
30 (30%) 70 Se
50 (50%) 50
120
280
400
Chi-square Contribution =
Get Chi-square statistic by summing contributions from all the cells

11. Check the necessary conditions for using a chi-square test (page 603)Yr
Yes No
Total Fr
15 (15%) E:
Chi: 85
100 So
25 (25%) 75 Jr
30 (30%) 70 Se
50 (50%) 50
120
280
400
Are all expected counts at least 1?
Are 80% or more of the expected counts at least 5?
If the answer is yes to both, we’re safe; the sample is large enough to make the test work well.

12. How many df for the chi-Square statistic?Yr
Yes No
Total Fr
15 (15%) 85
100 So
25 (25%) 75 Jr
30 (30%) 70 Se
50 (50%) 50
120
280
400
The chi-square statistic follows a chi-square distribution with a specific number of degrees of freedom (df ).
df = (r – 1)(c – 1)
r = # row variable categories
d = # column variable categories
For our example, df =
1 B. 2
C D. 4

13. Chi-Square Results df = (r – 1)x (c – 1)
Rows: SchoolYr Columns: Missed_Class
Yes No All Fr So Jr Se
All
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = , DF = 3, P-Value = 0.000
df = (r – 1)x (c – 1)

14. Analysis & Conclusion one differs H0: p1 = p2 = p3 = p4
Ha: at least one pi differs
Chi-Sq = & p-value = 0.000
Conclusions: We can claim that
a relationship exists between school year and the likelihood of having missed class because of alcohol in the population.
at least _____ population proportion of yeses _________ when comparing the four school years
one
differs

15. New Example:
Was PSU your first choice?
* no
* yes
How many college applications did you submit?
* (1-2)
* (3-5)
* (6-8)
* (≥ 9)

16. Example 2: Two Different Categorical Variables
Variable 4 levels
Not a binomial
Identify Variables:
explanatory: PSU first choice
response: number of applications submitted
This is an example of a ___ x ___ table
2 4
PSU First Choice
(1 – 2)
(3 – 5)
(6 – 8)
(≥ 9)
Total No 12 60 29 19
120
Yes
100
128 41 11
280
112
188 70 30
401

17. Row percents for each level of explanatory variable
(1 – 2)
(3 – 5)
(6 – 8)
(≥ 9)
Total No
(10%)
(50%)
(24%)
(16%)
(100%)
Yes
(36%)
(46%)
(15%)
(4%)
Descriptively:
what do the
row percents
suggest?
They suggest a difference

18. State Hypotheses Null Hypothesis:
Ho: no relationship between PSU choice and number of college applications submitted in the population
H0: distributions are _____ the same for the two PSU choices
both
Alternative Hypothesis:
Ha: relationship exists between PSU choice and number of college applications submitted in the population
Ha: distributions are ___ ____the same for two the PSU choices
not both

19. Minitab Output: Stat > Tables > Cross tabulation & Chi-square
Is there a statistically significant relationship found?
________
_________ distributions are found
Rows: PSU_Choice Columns: College_Applied
(1-2) (3-5) (6-8) (=>9) All No
Yes
All
Cell Contents: Count
Expected count
Pearson Chi-Square = , DF = 3,
P-Value = 0.000
Yes!
Different

20. Example 2: Two Different Categorical Variables
PSU First Choice
(1 – 2)
(3 – 5)
(6 – 8)
(≥ 9)
Total No 12 60 29 19
120
Yes
100
128 41 11
280
112
188 70 30
401
Pearson chi-square =
Double all counts to get:
PSU First Choice
(1 – 2)
(3 – 5)
(6 – 8)
(≥ 9)
Total No 24
120 58 38
240
Yes
200
256 82 22
560
224
376
140 60
802
Pearson chi-square = ,
which is exactly doubled!

21. Yet another pair of categorical variables:
How satisfied are you with your overall appearance?
• (very)
• (somewhat)
• (not much)
What do you best like to wear when going to and from classes?
• shorts
• athletic wear
• jeans
• other
Explanatory Variable:
Response Variable:
Table: __ x __
Satisfaction
Which clothes

22. What does the pattern of row percents show us?
Very similar?
Very different?
Look for significance using Chi-square statistic to decide!

23. Inferentially no support for a relationship in the population
H0: No relationship, or no difference among the three distributions.
Ha: Relationship exists, or some difference exists among the three distributions.
Pearson Chi-Square = 6.34, DF = 6, P-Value = 0.386
NOT
Can _____ claim that there is a statistically significant relationship or that different distributions are found
We cannot rule out randomness as a plausible explanation for what has happened.

24. If you understand today’s lecture…
15.1, 15.3, 15.11, 15.17, 15.19, 15.47, 15.49, 15.59
Objectives:
• Review hypothesis testing for a relationship between two categorical variables in a population.
• Extend techniques beyond the 2x2 table case.
• Predict how larger samples will influence chi-square statistic.