1a) The strength of concrete depends to some extent on the method used for drying it. Two different drying methods were tested independently on specimens. The strength using each of the methods follow a normal distribution with mean μ_x and μ_y respectively and the same variance. The results are:
method 1: n1=7, x bar=3250, s1=210
method 2: n2=10, y bar=3240, s2=190
Do the methods appear (use alpha=0.05) to produce concrete with different mean strength?
1b) Suppose σ_x=210, σ_y=190 (n1, n2, x bar, y bar same as part a). Find the probability of deciding that the methods are not different when the true difference in means is 2.
Hypothesis testing for pairs of means
The Attempt at a Solution
I am OK with part a). Here the test is H_o: μ_x=μ_y v.s. H_a: μ_x≠μ_y. I computed a p-value of >0.2, so the p-value is greater than alpha(which is 0.05), and so we fail to reject H_o and the answer is "no".
Now I have some troubles with part b)...
Here we have:
Ho: μ_x = μ_y
Ha: μ_x - μ_y = 2
I think in part b) we have to find P(type II error) = P(fail to reject H_o | μ_x-μ_y=2)
How can we translate "fail to reject H_o" into a mathematical statement for which we can compute the probability? How can we find the "rejection region" in this case??
Thanks a lot!