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Tatiana [17]
3 years ago
11

An exit poll in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that

news organizations can try to figure out as soon as possible who won the election, before the votes are officially counted. This has been notoriously inaccurate in various elections, sometimes because of selection bias: the sample of people who are invited to and agree to participate in the survey may not be similar enough to the overall population of voters.
Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let A be the event that they voted for A, and W be the event that they are willing to participate in the exit poll. Suppose that P(W|A) = 0.7 but P(W|A^c ) = 0.3. In the exit poll, 60% of the respondents say they voted for A (assume that they are all honest), suggesting a comfortable victory for A.

Required:
Find P(A), the true proportion of people who voted for A.
Mathematics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

P(A) = 0.39

Step-by-step explanation:

We are given;

P(W|A) = 0.7

P(W|A^c ) = 0.3

We are told that 60% of the respondents said they voted for A. Thus;

P(A|W) = 60% = 0.6

Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).

Thus;

P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]

Now, P(A^c) can be expressed as 1 - P(A)

Thus, we now have;

P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]

Plugging in the relevant values gives;

P(W) = 0.7P(A) + 0.3(1 - P(A))

P(W) = 0.7P(A) + 0.3 - 0.3P(A)

P(W) = 0.3 + 0.4P(A)

Now,using Baye's theorem, we can find an expression for P(A|W)

Thus;

P(A|W) = [P(A ∩ W)]/P(W)

This can be further expressed as;

P(A|W) = [P(A) × P(W|A)]/P(W)

Plugging in relevant values, we have;

0.6 = 0.7P(A)/(0.3 + 0.4P(A))

Cross multiply to get;

0.6(0.3 + 0.4P(A)) = 0.7P(A)

0.18 + 0.24P(A) = 0.7P(A)

0.18 = 0.7P(A) - 0.24P(A)

0.46P(A) = 0.18

P(A) = 0.18/0.46

P(A) = 0.39

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12 ÷ 2 = 6

Using Pythagoras theorem,

\text{Base}^2+\text{Opposite}^2=\text{Hypotenuse}^2

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In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt
ankoles [38]

Answer:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

df=n-1=44-1=43  

p_v =P(t_{(43)}>1.436)=0.079  

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

\bar X=13.88 represent the sample mean

s=3.14 represent the sample standard deviation

n=44 sample size  

\mu_o =13.2 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:  

Null hypothesis:\mu \leq 13.2  

Alternative hypothesis:\mu > 13.2  

The statistic for this case is:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Calculating the statistic

Replacing the info given we got:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

P-value

The degrees of freedom are:

df=n-1=44-1=43  

Since is a right tailed test the p value is:  

p_v =P(t_{(43)}>1.436)=0.079  

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0
3 years ago
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