Answered: Consider an exponential random variable…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

. Consider an exponential random variable X with pdf f(x | θ) = θ −1 e −x/θ, x > 0. A single observation of such a variable is used to test H0 : θ = 2 against H1 : θ = 5. The null hypothesis is rejected if the observed value is greater than 4. (a) What is the probability of committing a Type I error? (b) What is the probability of committing a Type II error? (c) What is the power of the test? (d) Find a test (i.e. determine a test statistic and critical region) of these hypotheses that has significance level 0.05.

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Expert Solution

Step 1

(a). Compute the probability of committing a Type I error:

It is given that, a random variable X is exponentially distributed with the probability density function as given below:

$f (x | θ) = \{\begin{cases} \frac{1}{θ} e^{- \frac{x}{θ}}; x > 0 a n d θ > 0 \\ 0; o t h e r w i s e \end{cases}$

The null and alternative hypotheses are given below:

Null hypothesis H₀:

H₀ : θ = 2

Alternative hypothesis H₁:

H₁ : θ = 5

It is given that, the null hypothesis is rejected if the observed value is greater than 4.

The rejection region is R = {x : x > 4}.

The probability of committing a Type I error is obtained as 0.1353 from the calculation given below:

$\begin{array}{rcl} P (T y p e I e r r o r) & = & P (R e j e c t i n g H_{0} w h e n H_{0} i s t r u e) \\ = & P_{H_{0}} (X \in R e j e c t i o n r e g i o n) \\ = & P_{θ = 2} (X > 4) \\ = & \int_{4}^{\infty} \frac{1}{2} e^{- \frac{x}{2}} d x \\ = & e^{- \frac{4}{2}} \\ = & 0.1353 \end{array}$

Thus, the probability of committing a Type I error is 0.1353 .

Step 2

(b). Compute the probability of committing a Type II error:

The probability of committing a Type II error is obtained as 0.5507 from the calculation given below:

$\begin{array}{rcl} P (T y p e I I e r r o r) & = & P (F a i l i n g t o R e j e c t H_{0} w h e n H_{0} i s f a l s e) \\ = & P_{H_{1}} (D o n o t r e j e c t H_{0}) \\ = & P_{θ = 5} (X \leq 4) \\ = & \int_{0}^{4} \frac{1}{5} e^{- \frac{x}{5}} d x \\ = & 1 - e^{- \frac{4}{5}} \\ = & 1 - 0.4493 \\ = & 0.5507 \end{array}$

Thus, the probability of committing a Type I error is 0.5507 .

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