1. Discrete (Categorical) Data Analysis
TOPIC 10
Discrete (Categorical) Data Analysis

2. Discrete Random Variables
Recall that discrete random variables may take only discrete values.
For example,
Number of errors in a software product:
0, 1, 2, 3, 4, …
Categories of a product’s quality level”
High, medium, or low
Characteristics of a machine breakdown:
Mechanical failure, electrical failure, or operator misuse.

3. Sample Proportions
Recall that the success probability p can be estimated by the sample proportion
For large enough values of n the sample proportion can be taken to have approximately the normal distribution
This expression may be written in terms of a standard normal distribution as
= Standard Error

4. Confidence Interval Estimation for p
Since the probability of p is unknown then we replace p with its estimated
Assumptions:

5. Example
You’re a production manager for a newspaper. You want to find the % defective. Of 200 newspapers, 35 had defects. What is the 90% confidence interval estimate of the population proportion defective?

6. Example Solution 84

7. Sample Size for Estimating p
I don’t want to sample too much or too little!
SE = Sampling Error
If no estimate of p is available, use p = 1 – p = 0.5 89

8. Example
What sample size is needed to estimate p with 90% confidence and a width L of .03? 91

9. Exercises
Exercise:
In an election poll a random sample of 500 people showed that 42 preferred voting for a particular candidate. Set up a 90% confidence interval estimate for the population proportion, p of the particular candidate.
Suppose that the auditing procedures require you to have 95% confidence in estimating the population proportion of sales invoices with errors within ± The results from the past months indicate that the largest proportion has been no more than Find the sample size needed to satisfy the requirements of the company.

10. Z Test of Hypothesis for the Proportion
One sample Z test for the proportion
where
Hypothesized proportion of successes in the population
Sample proportion of successes
Number of items having the characteristic of interest
Sample size

11. Example
You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level of significance?

12. Example Solution Test Statistic: p = .04 H0: Ha:  = , /2 = 0.025
Decision:
Conclusion:
p = .04
p  .04
H0:
Ha:
 = , /2 = 0.025
n =
Critical Value(s):
.05
500
Do not reject H0 at  = .05
Reject H
Reject H
.025
.025
There is evidence proportion is 4%
-1.96
1.96 Z

13. Exercise
A fast-food chain has developed a new process to ensure that orders at the drive-through are filled correctly. The previous process filled orders correctly 85% of the time. Based on a sample of 100 orders using the new process, 94 were filled correctly. At a 0.01 level of significance, can you conclude that the new process has increased the proportion of orders filled correctly?

14. Large-Sample Inference about p1 – p2
Assumptions:
Independent, random samples
Normal approximation can be used if

15. Large-Sample Inference about p1 – p2
(1 – α)100% Confidence Interval for ( p1 – p2)
where

16. Example
As personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. Find a 99% confidence interval for the difference in perceptions.
To check assumptions, use sample proportions as estimators of population proportion:
n1·p1 = 78·63/78 = 63
n1·q1 = 78·(1-63/78) = 15
n2·p2 = 82·49/82 = 49
n2·q2 = 82·(1-49/82) = 33

17. Example Solution
To check assumptions, use sample proportions as estimators of population proportion:
n1·p1 = 78·63/78 = 63
n1·q1 = 78·(1-63/78) = 15
n2·p2 = 82·49/82 = 49
n2·q2 = 82·(1-49/82) = 33

18. Hypothesis Testing for Two Proportions
Large-Sample Inference about p1 – p2
Hypothesis Testing for Two Proportions
No Difference
Pop 1
Pop 2
Pop 1
Pop 2
Hypothesis
Any Difference
Pop 1 < Pop 2
Pop 1 > Pop 2 H0 Ha
Z – Test Statistic:
Hypothesized difference
where
The rejection region follows the way similar to that in the one sample tests

19. Example
As personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. At the .01 level of significance, is there a difference in perceptions?
To check assumptions, use sample proportions as estimators of population proportion:
n1·p1 = 78·63/78 = 63
n1·q1 = 78·(1-63/78) = 15
n2·p2 = 82·49/82 = 49
n2·q2 = 82·(1-49/82) = 33

20. Example Solution 12

21. Example Solution .005 z H0: Ha:  = n1 = n2 = Critical Value(s):
p1 - p2 = 0
p1 - p2  0
Test Statistic:
Decision:
Conclusion:
H0:
Ha:
 =
n1 = n2 =
Critical Value(s):
Z = +2.90
.01 78 82
Reject H0 at  = .01 z
2.58
-2.58
Reject H
.005
There is evidence of a difference in proportions 11

22. Chi-Square Tests for k Proportions
This topic extends hypothesis testing to analyze differences between population proportions based on two or more samples.
Qualitative data that fall in more than two categories often result from a multinomial experiment.
Some of the characteristics of the multinomial experiment are
The probabilities of the k outcomes, denoted p1, p2, … , pk, remain the same from trial to trial, where p1 + p2 + … + pk = 1
The trials are independent
Recall, binomial experiment is a multinomial experiment with k = 2

23. Chi-Square (2) Tests Populations p1 = p2 = p3 = p4 = ….. pk
Evidence to accept/reject our claim
Populations
2 Test for equality of proportions
p1 = p2 = p3 = p4 = ….. pk
Observed and expected frequencies
x , e
Draw Sample

24. Road Map Decision Making One/Two Samples Analysis of Variance χ2 Tests
One-Way Table
Two-Way Table

25. Multinomial Experiment
n identical and independent trials
k outcomes to each trial
Constant outcome probability, pk
Random variable is count, nk
Example: ask 100 people (n) which of 3 candidates (k) they will vote for
Uses one-way contingency table: Shows number of observations in k independent groups (outcomes or variable levels)

26. One Way Contingency Table
Outcomes (k = 3)
Candidate
Tom
Bill
Mary
Total 35 20 45
100
Number of responses

27. 2 Test Basic Idea
Compares observed frequency (xi) to expected frequency [ei] assuming null hypothesis is true
Closer observed frequency is to expected frequency, the more likely the H0 is true
Measured by squared difference relative to expected frequency
Reject large values
Assumptions:
A multinomial experiment has been conducted
The sample size n is large: ei is greater than or equal to 5 for every cell ( i = 1, 2, 3, …, k)

28. 2 Test for k Proportions
Hypothesized probability
1. Hypotheses
H0: p1 = p1,0, p2 = p2,0, ..., pk = pk,0
Ha: At least one pi is different from above
2. Test Statistic
Observed frequency
Expected frequency: ei = npi,0
3. Degrees of Freedom: k – 1
Number of outcomes 24

29. c Finding Critical Value Example Reject H0 5.991 Upper Tail Area df
What is the critical 2 value if k = 3, and  =.05?
If xi = ei, 2 = Do not reject H0 c 2
Upper Tail Area df
.995 …
.95
.05 1
...
0.004
3.841
0.010
0.103
5.991
2 Table (Portion)
Reject H0
 = .05
df = k - 1 = 2
5.991 26

30. 2 Test for k Proportions Example
As personnel director, you want to test the perception of fairness of three methods of performance evaluation. Of 180 employees, 63 rated Method 1 as fair, 45 rated Method 2 as fair, 72 rated Method 3 as fair. At the .05 level of significance, is there a difference in perceptions?
To check assumptions, use sample proportions as estimators of population proportion:
n1·p = 78·63/78 = 63
n1·(1-p) = 78·(1-63/78) = 15 10

31. 2 Test for k Proportions Solution
x1 = 63
x2 = 45
x3 = 72 12

32. 2 Test for k Proportions Solution
H0:
Ha:
 =
n1 = n2 = n3 =
Critical Value(s):
p1 = p2 = p3 = 1/3
At least 1 is different
Test Statistic:
Decision:
Conclusion:
2 = 6.3
.05 63 45 72
Reject H0 at  = .05 c 2
Reject H0
There is evidence of a difference in proportions
5.991
 = .05 11

33. Road Map Decision Making One/Two Samples Analysis of Variance χ2 Tests
One-Way Table
Two-Way
Table
Test of Independence

34. 2 Test of Independence Multinomial experiment has been conducted
Shows if a relationship exists between two qualitative (categorical) variables
One sample is drawn
Does not show causality
Uses two-way contingency table
Assumptions:
Multinomial experiment has been conducted
The sample size, n, is large: eij is greater than or equal to 5 for every cell

35. Two-Way Contingency Table
Shows number of observations from 1 sample jointly in 2 qualitative variables
Levels of variable 2
Levels of variable 1 40

36. Degrees of Freedom: (r – 1)(c – 1)
2 Test of Independence
Hypotheses
H0: Variables are independent
Ha: Variables are related (dependent)
Test Statistic
Observed frequency
Expected frequency
Degrees of Freedom: (r – 1)(c – 1)
Rows Columns 41

37. 2 Test of Independence Expected Frequencies
Statistical independence means joint probability equals product of marginal probabilities
Compute marginal probabilities and multiply for joint probability
Expected frequency is sample size times joint probability e =
Column Tot al
Sample Siz
Row Total a f   f a f

38. Expected Frequency Example
Joint probability =
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total Ri Cj
Expected freq. = 160×
= 54.6
Marginal probability = 43

39. Expected Frequency Calculation
ri: Total frequency in row i-th
cj: Total frequency in column j-th
112×
54.6
House Location
112×
57.4 = =
Urban
Rural
House Style
Obs.
Exp.
Obs.
Exp.
Total
Split
Level 63 49
112
Ranch
48×
23.4 15 33
48×
24.6 48
Total 78 78 82 82
160 = = 43

40. Example
As a realtor you want to determine if house style and house location are related. At the .05 level of significance, is there evidence of a relationship? 44

41.  Example Solution 112×78 160 112×82 160 = = 48×78 160 48×82 160 = =
eij  5 in all cells
112×
112× = =
48×
48× = = 45

42. Example Solution
Test Statistic: 12

43. Example Solution c Reject H0  = .05 3.841 H0: Ha:  = df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
2 = 8.41
.05
(2 – 1) (2 – 1) = 1
Reject H0 at  = .05 c 2
Reject H0
There is evidence of a relationship
3.841
 = .05 11

44. Exercise 1
You’re a marketing research analyst. You ask a random sample of 286 consumers if they purchase Diet Pepsi or Diet Coke. At the .05 level of significance, is there evidence of a relationship?
Diet Pepsi
Diet Coke No
Yes
Total No 84 32
116
Yes 48
122
170
Total
132
154
286 44

45.  Exercise 1 Solution eij  5 in all cells 116×132 286 154×116 286 = =
116×
154× = =
170×
170× = = 45

46. Exercise 1 Solution
Test Statistic: 12

47. Exercise 1 Solution c Reject H0  = .05 3.841 H0: Ha:  = df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
2 = 54.29
.05
(2 – 1) (2 – 1) = 1
Reject H0 at  = .05 c 2
Reject H0
There is evidence of a relationship
3.841
 = .05 11

48. Exercise 2
There is a statistically significant relationship between purchasing Diet Coke and Diet Pepsi. So what do you think the relationship is? Aren’t they competitors?
Diet Pepsi
Diet Coke No
Yes
Total No 84 32
116
Yes 48
122
170
Total
132
154
286 48

49. You Re-Analyze the Data
High Income
Diet Pepsi
Diet Coke No
Yes
Total No 4 30 34
Yes 40 2 42
There is a spurious relationship between purchasing Diet Coke & Diet Pepsi. Income is an intervening or control variable & is the true cause.
The analysis here uses only descriptive statistics.
For low income, consumers are price conscious. Either they can’t afford to buy either or they buy whatever is on sale.
For high income, consumers buy depending on preference regardless of price.
Total 44 32 76
Low Income
Diet Pepsi
Diet Coke No
Yes
Total No 80 2 82
Yes 8
120
128
Total 88
122
210 49

50. True Relationships* Underlying causal relation Apparent relation
Diet Coke
There is a spurious relationship between purchasing Diet Coke & Diet Pepsi. Income is an intervening or control variable & is the true cause.
The analysis here uses only descriptive statistics.
For low income, consumers are price conscious. Either they can’t afford to buy either or they buy whatever is on sale.
For high income, consumers buy depending on preference regardless of price.
Underlying causal relation
Apparent relation
Control or intervening variable (true cause)
Diet Pepsi 50

51. Numbers don’t think - People do!
Moral of the Story*
Numbers don’t think - People do! 51

52. Any Questions ?