Hypothesis test and finding type 2 error probability

[toothpaste666]

Homework Statement

From extensive records it is known that the duration of treating a disease by a standard therapy has a mean of 15 days. It is claimed that a new therapy can reduce the treatment time. To test this claim, the new therapy is to be tried on 70 patients and their times to recovery are to be recorded.
(a) Formulate the hypotheses and determine the rejection region of the test with a level of significance α = 0.05.
(b) If x = 14.6 and s = 3.0, what does the test conclude?
(c) Repeat (a) and (b) using the p-value.
(d) Using σ = 3.0, what is the Type II error probability β(μ′ ) of the test for the alternative μ′ =14?

The Attempt at a Solution

a) H0: μ= μ0 = 15
H1: μ < μ0 = 15
since the alternative hypotheses is μ1 < μ0 we reject H0 if Z < -zα = -z.05 = -1.645

b) X = 14.6, s = 3

Z = (X - μ)/(s/sqrt(n)) = (14.6 - 15)/(3/sqrt(70)) = -1.12 > -1.645 so we cannot reject the null hypothesis .

c) the P value is P(Z < -1.12) =.1314
I am not sure how they want me to use it in a) and b) though

d) μ1 = 14 μ0 = 15
μ1< μ0 and it is a one sided test
so we use
Z < -zα + sqrt(n)((μ0 - μ1)/σ) = -1.96 + sqrt(70)((15-14)/3) = .829
Z < .829
γ = P(Z < .829) = .7967
β = 1 - γ = 1 - .7967 = .2033

I am not sure if I am understanding part d) correctly either

 

[RUber]

For part d), you used the 2-sided Z score for alpha = .05. It should be the same as the one used in parts a and b, right? The calculation steps look correct to me.

Also, for using the p-value, you simply would say to reject the null hypothesis if p < alpha.
You could write, H0: p( mu ≥ mu_0 ) ≥ .05, Ha: p(mu < mu_0) >.95 ...though I have never seen a null hypothesis written in terms of the probability.

 

[toothpaste666]

oops yes you are right. thank you