# Confidence interval help needed

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Let θ denote the true ability of a student in a certain subject, representing the probability of astudent answering correctly an arbitrary question and let X1, . . . , Xn be the marks to n questions the student answered, taking values either 0 or 1. We model them as n i.i.d. Bernoulli(θ) randomvariables. Then, we approximate a student’s ability by its ML estimate.

a) We can approximate the distribution of

(θ(X) - θ) / ((θ(X)(1−θ(X)))/n)

(Where θ(X) is the Maximum likelihood estimate)

by a standard gaussian N(0,1) distribution, under distribution P

b) How big should n be so that we can estimate the true ability within 0.01 marks, with probabilityat least 95%? Justify your answer.

I have no idea where to start here. Why are we approximating that scary looking distribution in the first place, and what does it mean by 'under distribution P

a) We can approximate the distribution of

(θ(X) - θ) / ((θ(X)(1−θ(X)))/n)

^{0.5}(Where θ(X) is the Maximum likelihood estimate)

by a standard gaussian N(0,1) distribution, under distribution P

_{0. }Using this result, construct the 95 percent confidence interval for θ.b) How big should n be so that we can estimate the true ability within 0.01 marks, with probabilityat least 95%? Justify your answer.

I have no idea where to start here. Why are we approximating that scary looking distribution in the first place, and what does it mean by 'under distribution P

_{0}? How do we start to construct an interval?
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Let θ denote the true ability of a student in a certain subject, representing the probability of astudent answering correctly an arbitrary question and let X1, . . . , Xn be the marks to n questions the student answered, taking values either 0 or 1. We model them as n i.i.d. Bernoulli(θ) randomvariables. Then, we approximate a student’s ability by its ML estimate.

a) We can approximate the distribution of

(θ(X) - θ) / ((θ(X)(1−θ(X)))/n)

(Where θ(X) is the Maximum likelihood estimate)

by a standard gaussian N(0,1) distribution, under distribution P

b) How big should n be so that we can estimate the true ability within 0.01 marks, with probabilityat least 95%? Justify your answer.

I have no idea where to start here. Why are we approximating that scary looking distribution in the first place, and what does it mean by 'under distribution P

**pineapplechemist**)Let θ denote the true ability of a student in a certain subject, representing the probability of astudent answering correctly an arbitrary question and let X1, . . . , Xn be the marks to n questions the student answered, taking values either 0 or 1. We model them as n i.i.d. Bernoulli(θ) randomvariables. Then, we approximate a student’s ability by its ML estimate.

a) We can approximate the distribution of

(θ(X) - θ) / ((θ(X)(1−θ(X)))/n)

^{0.5}(Where θ(X) is the Maximum likelihood estimate)

by a standard gaussian N(0,1) distribution, under distribution P

_{0. }Using this result, construct the 95 percent confidence interval for θ.b) How big should n be so that we can estimate the true ability within 0.01 marks, with probabilityat least 95%? Justify your answer.

I have no idea where to start here. Why are we approximating that scary looking distribution in the first place, and what does it mean by 'under distribution P

_{0}? How do we start to construct an interval?is approximately distributed as a standard normal variable. I've changed your notation of to the more common . The point here is that is your estimate of and

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

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What it is saying here is that, if is the maximum likelihood estimator of , then

is approximately distributed as a standard normal variable. I've changed your notation of to the more common . The point here is that is your estimate of and

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

**Gregorius**)What it is saying here is that, if is the maximum likelihood estimator of , then

is approximately distributed as a standard normal variable. I've changed your notation of to the more common . The point here is that is your estimate of and

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

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**Gregorius**)

What it is saying here is that, if is the maximum likelihood estimator of , then

is approximately distributed as a standard normal variable. I've changed your notation of to the more common . The point here is that is your estimate of and

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

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So, is my interval the estimator +/- 1.96 ? (Sorry for large text, I copied and pasted). Do I need to do anything else?

**pineapplechemist**)So, is my interval the estimator +/- 1.96 ? (Sorry for large text, I copied and pasted). Do I need to do anything else?

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(Original post by

That's it.

**Gregorius**)That's it.

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#7

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Thanks. They didn't actually teach us how to do it. For the second part of the question, is it similar? Do I need to rearrange the interval for n?

**pineapplechemist**)Thanks. They didn't actually teach us how to do it. For the second part of the question, is it similar? Do I need to rearrange the interval for n?

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Yes, you have to choose n so that the confidence interval has width 0.01.

**Gregorius**)Yes, you have to choose n so that the confidence interval has width 0.01.

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My methods therefore is upper interval - lower interval <= 0.01. I rearranged for n and got 153664(theta)(1-theta)<= n. It doesn't seem right to have it in terms of my estimator, is it ok?

**pineapplechemist**)My methods therefore is upper interval - lower interval <= 0.01. I rearranged for n and got 153664(theta)(1-theta)<= n. It doesn't seem right to have it in terms of my estimator, is it ok?

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(Original post by

If you are not actually given any information that allows you to calculate an actual value of then you need to think about what the "worst" value of could be, from the point of view of the width of the confidence interval. As a hint, consider the graph of the function f(x)=x(1-x).

**Gregorius**)If you are not actually given any information that allows you to calculate an actual value of then you need to think about what the "worst" value of could be, from the point of view of the width of the confidence interval. As a hint, consider the graph of the function f(x)=x(1-x).

Thanks for your help by the way, really appreciate it. Finding this module very hard!

edit: So I get n needs to be at least 38416. Seems huge!

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