STA 2023 FIU Event Combinations and Statistical Independence Mathab Task Please help me solve my lab 3 questions. Excel Lab attached with pdf instructions.

STA 2023 FIU Event Combinations and Statistical Independence Mathab Task Please help me solve my lab 3 questions. Excel Lab attached with pdf instructions. Will need answers to questions in a word doc. LAB 3: Event Combinations and Statistical Independence
Learning Objectives
• Identify combinations of events (intersection and union) on a sample space.
• Apply the Additive Law of Probability to compute the probability of the union of two
events.
• Invoke the Law of Complements to compute the complement of an event.
• State the property of statistical independence.
• Assess whether two variables in a categorical data-set are statistically independent.
Key Concepts
Event Combinations
Previously, we computed probabilities for single events on a finite sample space using
the Classical Method. We can also compute probabilities for combinations of events
using the Classical Method. Consider a random experiment of rolling two six-sided dice
and define the following events:
• Event A: The sum of the two dice equals five.
• Event B: The maximum value of the two dice equals three.
The sample space consisting of all possible combinations of two six-sided dice is depicted below:
Sample Space of a Roll of Two Six-Sided Dice
Event A includes four sample points (Die 1/Die 2): One/Four, Two/Three, Three/Two and
Four/One. Event B includes five sample points (Die 1/Die 2): One/Three, Two/Three,
4
Three/Two, Three/One, and Three/Three. Applying the Classical Method, P(A)= 36
and
5
P(B)= 36 .
The union of event A and event B (A ? B) includes any sample point included in event
A OR event B. A ? B includes seven sample points: One/Four, Two/Three, Three/Two,
Four/One, One/Three, Three/One, and Three/Three. Applying the Classical Method,
7
P(A ? B) = 36
The intersection of event A and event B (A ? B) includes any sample point that is
included in event A AND event B. A ? B includes two sample points: Two/Three and
2
.
Three/Two. Applying the Classical Method, P(A ? B) = 36
Additive Law of Probability
The third Axiom of Probability states: ”Suppose there are two collections of outcomes
(A and B) that do not overlap. Then: P(A OR B) = P(A) + P(B).” When two events, A
and B, are non-overlapping, then A ? B = ? (the empty set) which implies that P(A ? B)
= 0. However, events often do overlap as depicted in the Venn Diagram below:
Venn Diagram of Generalized Events A and B
If we were to compute P(A ? B) = P(A) + P(B) for overlapping events, the sample
point probabilities mapped to the intersection of event A and event B would be doublecounted. If the overlap is large enough, P(A?B) could exceed one, an invalid probability.
The Additive Law of Probability applies to any events A and B and states:
For any two events A and B defined on a sample space, P(A ? B) is given by:
P(A ? B) = P(A) + P(B) ? P(A ? B).
4
5
2
For the events A and B defined above, P(A)= 36
, P(B)= 36
and P(A ? B) = 36
. Applying
4
5
2
7
the Additive Law of Probability, P(A ? B) = 36 + 36 ? 36 = 36 , confirming our previous
calculation.
Law of Complements
Consider the random experiment of rolling two six-sided dice and define:
Event C: The maximum of the two dice is greater than two.
The complement of event C (C c ) is the set of sample points that are not included
in event C. In this example, C c includes four sample points (Die 1/Die 2): One/One,
4
One/Two, Two/One and Two/Two. Applying the Classical Method, P(C c )= 36
.
The Law of Complements states that for any event C, defined on a sample space:
P(C c ) = 1 – P(C)
4
. It follows from the Law of Complements
In the example above, we found that P(C c ) = 36
32
that P(C)= 36 . Typically, the Law of Complements is used when it is simpler to compute
the probability of the event complement than the event itself.
Statistical Independence
Two events A and B are statistically independent if and only if: P(A ? B) = P(A)P(B).
Does statistical independence hold for event A and event B defined above? P(A)P(B) =
4
× 365 6= 362 . No, events A and B are not statistically independent!
36
Are Disjoint Events Statistically Independent?
Students often assume that non-overlapping (disjoint) events are statistically independent. Consider the random experiment of rolling two six-sided dice and define event
D: The maximum value of the two dice equals five. Event D contains nine sample
points (Die 1/Die 2): One/Five, Two/Five, Three/Five, Four/Five, Five/One, Five/Two,
9
.
Five/Three, Five/Four and Five/Five. Applying the Classical Method, P(F ) = 36
Recall event A: The sum of the two dice equals five. Event A and event C share no
sample points in common: A ? C = ? (the empty set) which implies P(A ? C) = 0.
4
Therefore, P(A)P(C) = 36
× 369 6= 0. Event A and event C are not independent!
Statistical Independence in a Categorical Data-Set
Oftentimes, we record information about multiple categorical variables simultaneously.
Are these categorical variables statistically independent of one another? The crosstabulation data table below reflects a particular student’s daily habits as well as the
weather over a 350-day period:
Observed Cross-Tabulated Data Table for 350 Days of School
If the decision to eat lunch in the cafeteria is statistically independent of the weather,
what would the cross-tabulated data table look like? The expectation under statistical independence is: P(Cafeteria AND Rain) = P(Cafeteria)*P(Rain). We can estimate
and P(Rain) by 138
. Therefore, the expectation for P(Cafeteria AND
P(Cafeteria) by 113
350
350
113 138
Rain) = 350 * 350 =0.127. Under statistical independence, the expected number of days
the student ate in the cafeteria and it rained equals 0.127*350 = 44.55 days.
Under statistical independence, the cross-tabulated data table would appear as follows:
Expected Cross-Tabulated Data Table for 350 Days of School (Under Statistical
Independence)
The observed and expected cross-tabulated data tables appear very similar. The decision to eat lunch in the cafeteria and the weather appear to be statistically independent.
Comparison of Cross-Tabulated Data
There is a more mathematical approach for comparing the observed and expected
cross-tabulated data tables, in order to evaluate statistical independence. Consider
the following test statistic (T):
T =
P
i
(Observedi ?Expectedi )2
Expectedi
where i is an indicator for each cell in the cross-tabulated data table. If T is greater than
3.84, it suggests that the data diverge significantly from the expectation under statistical
independence. For the data above, the test statistic (T) equals:
T =
(46?45)2
45
+
(92?93)2
93
+
(67?68)2
68
+
(145?144)2
144
= 0.055.
A value of T=0.055 suggests that the data is not inconsistent with statistical independence. You will learn more details about this calculation if you continue your study of
Statistics!
Assignment
Question 1: Combinations of Events
Watch the Lab 3 video tutorial entitled ”Computing Probabilities of Event Combinations”.
Watch the Lab 3 video tutorial entitled ”Statistical Independence”.
Two five-sided dice are rolled. Consider the following events:
• Event A: The sum of the two dice equals five.
• Event B: The maximum value of the two dice equals five.
• Event C: At least one of the two dice equals five.
Note: If both Die1 and Die 2 equal five, then the maximum value is five.
(a) Compute P(A), P(B) and P(C) using the Classical Method.
(b) Do event A and event B share any sample points in common? If so, identify those
sample points.
(c) Compute P(A ? B) using the Classical Method.
(d) Are events A and B statistically independent? Explain your answer.
(e) Compute P(A ? B) using the Additive Law of Probability.
(f) Do event A and event C share any sample points in common? If so, identify those
sample points.
(g) Compute P(A ? C) using the Classical Method.
(h) Are events A and C statistically independent? Explain your answer.
(i) Compute P(A ? C) using the Additive Law of Probability.
(j) Do event B and event C share any sample points in common? If so, identify those
sample points.
(k) Compute P(B ? C) using the Classical Method.
(l) Are events B and C statistically independent? Explain your answer.
(m) Compute P(B ? C) using the Additive Law of Probability.
Question 2: Statistical Independence
Watch the Lab 3 video entitled ”Assessing Statistical Independence for a Dice Simulation”.
The data set labeled ”Dice Simulation” reports the outcome of a computer simulation in
which two five-sided dice (numbered from 1-5) are rolled repeatedly (N=5,000 trials).
(a) For each of the pairwise events defined in Question 1 (A-B, A-C, B-C), complete
the following steps i-iii.
i: Create a cross-tabulation table.
ii: Fix the marginal counts and predict the expected cross-tabulation table under statistical independence.
P
2
i ?Expectedi )
.
iii: Compute the test statistic: T = i (Observed
Expected
i
(b) Evaluate each of the pair-wise comparisons (A-B, A-C, B-C) for consistency with
statistical independence.
(c) Are your findings consistent with your results in Question 1? Explain your answer.
Question 3: Statistical Independence
Watch the Lab 3 video tutorial entitled ”Assessing Statistical Independence for a
Categorical Data Set”.
The data-set labeled ”Jasmine” reports on the last 250 days that a particular student
(Jasmine) attended class. During that time, Jasmine either drove to campus or took the
bus, packed lunch or ate in the cafeteria. The weather is also reported (Rain/No Rain).
Consider the following events:
• Event A: Jasmine drove to campus.
• Event B: Jasmine packed a lunch.
• Event C: It rained.
(a) For each of the pairwise events (A-B, A-C, B-C), complete the following steps i-iii.
i: Create a cross-tabulation table.
ii: Fixing the marginal counts, predict the cross-tabulation table that you would expect
under statistical independence.
iii: Compute the test statistic: T =
P
i
(Observedi ?Expectedi )2
.
Expectedi
(b) Evaluate each of the pair-wise comparisons (A-B, A-C, B-C) for consistency with
statistical independence.
Day
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
Rain
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
RAIN
NO RAIN
RAIN
RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
Pack Lunch
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
Drive/Bus
DROVE
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
BUS
DROVE
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
DROVE
BUS
DROVE
BUS
DROVE
BUS
BUS
DROVE
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
DROVE
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
DROVE
DROVE
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
RAIN
RAIN
NO RAIN
NO RAIN
RAIN
RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
RAIN
RAIN
NO RAIN
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
BUS
BUS
BUS
BUS
BUS
DROVE
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
DROVE
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
NO RAIN
CAFETERIA
CAFETERIA
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
CAFETERIA
PACK LUNCH
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
BUS
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
NO RAIN
RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
RAIN
NO RAIN
RAIN
NO RAIN
NO RAIN
NO RAIN
RAIN
NO RAIN
RAIN
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
CAFETERIA
PACK LUNCH
PACK LUNCH
PACK LUNCH
BUS
DROVE
BUS
BUS
DROVE
BUS
BUS
BUS
BUS
DROVE
DROVE
BUS
BUS
BUS
BUS
DROVE
Trial #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
Die 1
2
4
1
1
2
3
5
3
4
1
5
5
1
5
3
2
3
4
5
4
4
4
4
4
2
3
3
1
1
5
4
1
1
1
1
1
5
4
5
2
5
5
1
2
5
1
Die 2
5
2
5
1
2
1
4
2
5
1
5
5
3
5
2
5
1
1
2
4
5
4
3
5
1
4
4
4
5
5
4
5
5
1
1
3
1
3
1
1
2
2
5
4
3
2
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
3
4
3
1
2
1
2
4
4
2
3
1
3
5
2
3
2
3
3
4
5
2
3
2
3
3
5
5
4
1
3
5
2
1
1
1
3
2
1
4
3
1
2
5
3
3
5
4
5
1
5
2
4
2
5
2
4
5
3
4
1
2
2
5
5
5
3
3
4
1
3
5
3
1
1
2
4
3
2
1
3
5
5
2
1
1
3
5
4
2
4
4
5
3
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
5
5
2
5
4
1
5
1
3
3
2
2
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