Question

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 = 20 and x = 2, what is the estimate? p = (b) Is the estimator of part (a) unbiased? Yes No (c) If n = 20 and x = 2, what is the mle of the probability (1-P) that none of the next five tests done on disease-free individuals are positive? (Round your answer to four decimal places.)

Answer 1

Answer:-

Given That:-

A diagnostic test for a certain disease is applied to n individuals known to not have the disease . let x = is the number among the n test result that are positive Assume that only X is available rather than actual sequence of test  results .

X is the number among the n test result that are positive

P = the probability that a disease free individual test result positive

(a) The maximum likelihood estimate of P

SO, Pmf of X is

S(;P) = 0; p*(1 – p)--

$ln[f(x;p)]=ln\binom{n}{x} + xlnp+(n-x)ln(l-p)$

Taking $\frac{d}{dp}\begin{Bmatrix} ln[f(x;p)] \end{Bmatrix}$ and equating

To zero , we get ,

0= d-I (x - u) +0

$\frac{x}{p}=\frac{(n-x)}{(1-p)}$ (or)

(or)

  

  $p=\frac{x}{n}$

(a)Thus the estimate of pis

$p= \frac{x}{n}\Rightarrow \frac{2}{20}\Rightarrow 0.1$

estimate of $p=0.1$

(b) Given $E\begin{bmatrix} \frac{x}{n} \end{bmatrix}=\frac{1}{n}E(x)$

  $=\frac{1}{n}np$

$=p$

yes , the estimate of part (a) is unbiased

(c) Given data

  

Probability $=(1-p)^{5}$

P(none of the next five helmets examined are flawed }

$=(1-p)^{5}$

The estimate of Required probability is

$=(1-p)^{5}$

$=(1-\frac{2}{20})^{5}$

$=(\frac{18}{20})^{5}$

$=0.5904$