Question

When a chip fabrication facility is operating normally, the lifetime of a microchip operated at temperature T, measured in degrees Celsius, is given by an exponential (????) random variable X with expected value E[X]=1/????=(200/T)2 years. Occasionally, the chip fabrication plant has contamination problems and the chips tend to fail much more rapidly. To test for contamination problems, each day m chips are subjected to a one-day test at T=100∘C. Based on ????, the number of chips that fail in one day, design a significance test for the null hypothesis H0: the plant is operating normally.
a) Suppose the rejection set of the test is R = {N > 0}. Find the significance level of the test as a function of m, the number of chips tested.
(b) How many chips must be tested so that the significance level is α = 0.01.
(c) If we raise the temperature of the test, does the number of chips we need to test increase or decrease?

Answer 1

Answer:

a. 1 - e^{m/365}

b. 916.60 \approx 917.

c. - 365 T^2 \log(1 - \alpha)/10000 which is a deceasing function of T.

Step-by-step explanation:

a)

\alpha(m) = P_{H_0}(N>0) = 1 - P_{H_0}(N=0) \\ = 1 - p^m.

where p = P_{H_0}(\mbox{The chip survives for 1 day}) \\ = P_{H_0}(X> 1 \mbox{day}) = P_{H_0}(X> 1/365 \ \mbox{year}) = e^{-\frac{1}{365}} .

Since if T = 100 the lifetime follows exponential with mean = 1 year.

Therefore \alpha(m) = 1 - e^{m/365}

b) At T = 25 the lifetime is exponential with mean = 16 years. Therefore \lambda = \frac{1}{16} and so

\alpha(m) = 1 - e^{\frac{m}{50000}} = 0.01 \Rightarrow m = 916.60 \approx 917.

c) If we raise the temperature during the test the number of chips we need to test (for the same level of significance) will decrease since for a fixed value of \alpha

m = -365 T^2 \log(1 - \alpha)/10000 which is a deceasing function of T.

Answer 2

This exercise is necessary to use the exponential function to perform the calculations that remain:

A)$1-e^{m/365}$

B)$917$

C)$\frac{- 365 T^2 \log(1 - \alpha)}{10000}$

In the first part of the exercise it is necessary to find the significance level of the tested chip.

A)Thus, we have that the values ​​are:

$\alpha(m) = P_{H_0}(N>0) \\= 1 - P_{H_0}(N=0) \\ = 1 - p^m.$

where,

$p = P_{H_0}(\mbox{The chip survives for 1 day}) \\ = P_{H_0}(X> 1 \mbox{day}) \\= P_{H_0}(X> 1/365 \ \mbox{year})\\ = e^{-\frac{1}{365}} .$

Since if T = 100 the lifetime follows exponential with mean = 1 year. Therefore:

$\alpha(m) = 1 - e^{m/365}$

B) At T = 25 the lifetime is exponential with mean = 16 years. Therefore:

$\lambda = \frac{1}{16}$

and so:

$\alpha(m) = 1 - e^{\frac{m}{50000}} = 0.01 \\\Rightarrow m = 916.60 \approx 917.$

C) If we raise the temperature during the test the number of chips we need to test (for the same level of significance) will decrease since for a fixed value of $\alpha$:

$M= \frac{- 365 T^2 \log(1 - \alpha)}{10000}$

which is a deceasing function of T.

Learn more: brainly.com/question/18817620