Question

10 Problem 1 Let X is amount of coffeein a "100g coffeecan), EXu. With a sample of size 25 you test null of coffee can"), . Let X ~N(μ, σ2) and variance is known: σ- s amount of coffee in a against μ > 100 at 5% significance level (a) Find the decision rule for that test. (b) Suppose, actually100. What will be the probability to reject null of alternative μ > 100 in that case? Plot that probability as a function of μ 100 in favor of the

Answer 1

a)

decision rule

P(Z > z) = 0.05

z= 1.645

hence

Xbar > mean + z *sd/sqrt(n)

Xbar > 100 + 1.645* 1 /sqrt(25)

Xbar > 100.329

b)

P(Xbar > 100.329 | $\mu$ = $\mu_0$)

=P(Z > (100.329 -$\mu_0$)/(1/sqrt(25))

=P(Z > 5 *(100.329 - $\mu_0$))

mu_0	p
100	0.049985
99.95	0.029046
99.9	0.015976
99.85	0.00831
99.8	0.004085
99.75	0.001896
99.7	0.00083
99.65	0.000343
99.6	0.000134
99.55	4.91E-05
99.5	1.7E-05

code in R

powerZtest = function(alpha = 0.05, sigma, n, delta){
zcr = qnorm(p = 1-alpha, mean = 0, sd = 1)
s = sigma/sqrt(n)
power = 1 - pnorm(q = zcr, mean = (delta/s), sd = 1)
return(power)
}