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goblinko [34]
3 years ago
12

Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-

handed. They find the point estimate is 0.12. Each independently constructed a confidence interval based on the point​ estimate, but​ Jaime's interval has a lower bound of 0.078 and an upper bound of 0.193​, while​ Mariya's interval has a lower bound of 0.051 and an upper bound of 0.189. Which interval is​ wrong? Why?
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

Jaime's. Interval not centered around the point estimate.

Step-by-step explanation:

When constructing a confidence interval based on a point estimate, the obtained point estimate must be the central value of the interval.

For Jaime's interval

Lower bound = 0.078

Upper Bound = 0.193

Central = \frac{0.078+0.193}{2} =0.1355

For Mariya's interval

Lower bound = 0.051

Upper Bound = 0.189

Central = \frac{0.051+0.189}{2} =0.12

For a point estimate of 0.12, only Mariya's interval is adequate since Jaime's is not centered around the point estimate.

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Read 2 more answers
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
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Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

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P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

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c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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