Question

Suppose that a single observation X is taken from the normal distribution with unknown mean μ and known variance is 1. Suppose that it is known that the value of μ must be −5, 0, or 5, and it is desired to test the following hypotheses at the level of significance 0.05:

H₀: μ = 0, H₁: μ = −5 or μ = 5.

Suppose also that the test procedure to be used specifies rejecting H0 when |X| > c, where the constant c is chosen so that Pr(|X| >c|μ)=0.05.

a. Find the value of C, and show that if X = 2, then will be rejected.

b. Show that if X = 2, then the value of the likelihood function at μ = 0 is 12.2 times as large as its value at μ = 5 and is 5.9 × 10⁹ times as large as its value at μ = −5.

Short answer

(a) c=1.96

(b) Proved by obtaining the likelihood of various

Step-by-step solution

a) value of c 

(a) Under. Thus

P(|X|≥c) =2(1-ɸ(c)) =0.05

C =ɸ^-1(0.975)

= 1.96

And, since we reject the null if, hence at X=2 the null is rejected.

b) likelihood function

(b) The likelihood function is given by

∆ (x| µ ) = 1/√ 2π) exp(- 1/2 (x - µ)²)

Thus, we have,

μ	0	5	-5
˄(2|μ)	0.05399	0.0443	9.13472× 10-12

Using these values, we indeed obtain that if X=2, then the value of the likelihood function at μ =0 is 12.2 times as large as its value at μ = 5 and is 5.9× 10⁹ times as large as its value at μ = 5.