One thousand voters selected at random were asked whether or not they agreed wit
ID: 3295194 • Letter: O
Question
One thousand voters selected at random were asked whether or not they agreed with a certain proposal by the government. The results of opinion are cross-tabulated by age in the following "contingency table": a) Give the marginal distribution of the age groups. Check your distribution... does it add to 1? b) Find the conditional distribution of the age groups given each opinion? Based on these distributions, do you think that the opinion might depend on age group or not? Why? c) What is the probability that a randomly chosen voter is 50+ and agrees with the proposal? To what type of distribution does this situation belong? d) What is the probability that a randomly chosen 50+ aged person does agree?Explanation / Answer
Solution
Let
O1j = observed frequency of voters (i.e., number of voters) in the jth age-group who agree with the proposal, j = 1, 2, 3, 4
O2j = observed frequency of voters (i.e., number of voters) in the jth age-group who disagree with the proposal, j = 1, 2, 3, 4
O1. = total of O1j = Total number of voters who agree with the proposal;
O2. = total of O2j = Total number of voters who disagree with the proposal;
N = O.. = O1. + O2. = Total number of voters
Given dataset incorporating the above total are given below:
Opinion
Age group (j)
Total
18 – 29 (1)
30 – 39 (2)
40 – 49 (3)
50 + (4)
Agree
250
100
40
90
480
Disagree
100
100
60
260
520
Total
350
200
100
350
1000 (N)
Part (a)
Marginal Distribution of age-group
Age-group
18 – 29 (1)
30 – 39 (2)
40 – 49 (3)
50 + (4)
Total
Probability
0.35
0.20
0.10
0.35
1 (Check)
Probability = (O1j + O2j)/N = column total/N ANSWER
Part (b)
Conditional Distribution of age-group given opinion
(1) Agree
Age-group
18 – 29 (1)
30 – 39 (2)
40 – 49 (3)
50 + (4)
Total
Probability
0.521
0.208
0.083
0.188
1 (Check)
Probability = O1j /[j = 1,4](O1j) = cell frequency/row 1 total ANSWER 1
(2) Disagree
Age-group
18 – 29 (1)
30 – 39 (2)
40 – 49 (3)
50 + (4)
Total
Probability
0.1923
0.1923
0.1154
0.5
1 (Check)
Probability = O2j /[j = 1,4](O2j) = cell frequency/row 2 total ANSWER 2
Part (c)
Probability that a randomly chosen voter is 50 +aged and agree with the proposal = Number of voters 50 +aged and agreeing with the proposal/N = 90/1000 = 0.009 ANSWER1
The distribution involved is Binomial ANSWER 2
Part (d)
Probability that a randomly chosen 50 +aged voter does agree with the proposal
= Number of 50 +aged voters and agreeing with the proposal/total number of 50 +aged voters = 90/350 = 0.2571 ANSWER
[Note that the divisor is NOT N.]
Opinion
Age group (j)
Total
18 – 29 (1)
30 – 39 (2)
40 – 49 (3)
50 + (4)
Agree
250
100
40
90
480
Disagree
100
100
60
260
520
Total
350
200
100
350
1000 (N)
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