Hypothesis Testing for Normal Mean μ w/σ=1

[lildrea88]

Suppose X is normal mean $$\mu$$ unknown and standard deviation σ = 1.
(a) You are to show that the uniformly most powerful test for testing test of size α
based on a sample of size n for testing H0 :$$\mu$$  ≤ $$\mu$$0 verses H1$$\mu$$ :  ≥ $$\mu$$0
is of the form R = {$$\Sigma$$x_j ≥ c}
where c = z$$\sqrt{n}$$ + n$$\mu$$₀

really not sure where to start..i currrently have no examples to work off and my textbook doesn't help..i do know that i am suppose to use neyman pearson, but i just don't know where to apply it
help please!

 

[lippyka]

Answer: The uniformly most powerful test for testing the hypothesis H0 : \mu ≤ \mu0 versus H1 : \mu ≥ \mu0, based on a sample of size n from a normal distribution with mean \mu and standard deviation σ = 1, is given by R = \{\Sigma x_j ≥ c\}, where c = z\sqrt{n} + n\mu_0. This follows from the Neyman-Pearson Lemma, which states that the most powerful test for two simple hypotheses is a likelihood ratio test. In this case, the likelihood ratio test is given by \frac{L(H_1)}{L(H_0)}=\frac{f(x|H_1)}{f(x|H_0)}=\frac{f(\bar{x}|\mu \ge \mu_0)}{f(\bar{x}|\mu \le \mu_0)}=\frac{\prod_{i=1}^nf(x_i|\mu \ge \mu_0)}{\prod_{i=1}^nf(x_i|\mu \le \mu_0)}=\frac{\prod_{i=1}^ne^{-(x_i-\mu_0)^2/2}}{\prod_{i=1}^ne^{-(x_i-\mu)^2/2}}=e^{\Sigma (x_i-\mu_0)^2/2 - \Sigma (x_i-\mu)^2/2}. The test statistic is then given by R = \{\Sigma x_j ≥ c\}, where c = z\sqrt{n} + n\mu_0. This is the uniformly most powerful test for testing the hypothesis H0 : \mu ≤ \mu0 versus H1 : \mu ≥ \mu0.