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A diagnostic test for a certain disease is applied to n individuals known to not have the disease. Let X = the number among the n test results that are positive (indicating presence of the disease, so X is the number of false positives) and p = the probability that a disease-free individual's test result is positive (i.e., p is the true proportion of test results from disease-free individuals that are positive). Assume that only X is available rather than the actual sequence of test results.


(a) Derive the maximum likelihood estimator of p.


If n = 25 and x = 3, what is the estimate?

(b) Is the estimator of part (a) unbiased?

(c) If n = 25 and x = 3, what is the mle of the probability (1 - p)5 that none of the next five tests done on disease-free individuals are positive? (Round your answer to four decimal places.)

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Answer:

Part a

The maximum likelihood estimator of p, is the sample proportion,  [tex]\bar p=\frac{x}{n}[/tex]

And the proportion of the estimate at n=25 and x = 3  is 0.12.

Part b

The maximum likelihood estimate from part (a) is an unbiased estimator.

Part c

The maximum likelihood estimate of the probability is 0.5277.

Step-by-step explanation:

a)

The likelihood function is, [tex]L(p;x) =\frac{n!}{x!(n-x)!}p*(1- p)^{n-x}[/tex]

Applying log on both sides:

[tex]\log L(p;x) =\log\frac{n!}{x!(n-x)!}+x\log p+(n-x)log(1-p)[/tex]

Find the differentiation of log L with respect to P and equate to zero to find the MLE for .

[tex]\frac{d}{dp}\log L(p,x)=\frac{x}{p}-\frac{n-x}{1-p}\\\\=\frac{x}{p}-\frac{(n-x)}{1-p}\\\\=x(1-p)-(n-x)p=0\\=x-xp-np+xp=0\\=x-np=0\\=x=np\\=\bar p=\frac{x}{n}[/tex]

Calculate the estimate of the function if n=25 and x = 3.

[tex]\bar p=\frac{x}{n}\\\\=\frac{3}{25}\\\\= 0.12 [/tex]

Hint for next step

The MLE for the population proportion is the sample proportion and there is a 0.12 proportion of estimate at n = 25 and x = 3  

b)

Consider  

[tex]E(\bar p)\\\\E(\bar p)=E(\frac{X}{n})\\\\=\frac{1}{n}E(X)\\\\=\frac{1}{n}(np)(since\, E(X) =np)\\\\=p[/tex]

Since [tex]E(\bar p)=p [/tex], the maximum likelihood estimate from part (a) is an unbiased estimator.

Hint for next step

Based on the part (a) calculation, since it is proved that , the maximum likelihood estimate is unbiased estimator for population proportion

(c)

Calculate the maximum likelihood estimate of the probability [tex](1-p)^5[/tex].

By the Invariance property of MLE, to find the MLE of [tex](1-P)^5[/tex], the MLE of p can be used.

That is, the MLE of [tex](1-p)^5[/tex] is [tex](1-\bar p)^5[/tex]

Therefore,

MLE of the probability,

[tex](1- p)^5 = (1-\bar P)^5\\\\=(1-\frac{3}{25})^5\\\\=(1-0.12)^5\\=(0.88)^5\\=0.5277[/tex]

Using proportion concepts, it is found that:

a) The maximum likelihood estimator of p is 0.12.

b) Unbiased, as the individuals are known to not have the disease, thus there is not a change of individuals with the disease counting as positive.

c) 0.5277 = 52.77% probability that none of the next five tests done on disease-free individuals are positive.

A proportion is the number of desired outcomes divided by the number of total outcomes.

Item a:

The maximum likelihood estimator is the sample proportion, which is given by:

[tex]p = \frac{x}{n}[/tex]

Then:

[tex]p = \frac{3}{25} = 0.12[/tex]

The maximum likelihood estimator of p is 0.12.

Item b:

Unbiased, as the individuals are known to not have the disease, thus there is not a change of individuals with the disease counting as positive.

Item c:

Since the tests are independent, this probability is:

[tex]P = (1 - p)^5[/tex]

Then

[tex]P = (0.88)^5 = 0.5277[/tex]

0.5277 = 52.77% probability that none of the next five tests done on disease-free individuals are positive.

A similar problem is given at https://brainly.com/question/24342347

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