# Thread: Hypothesis testing for pairs of means

1. ## Hypothesis testing for pairs of means

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 ?

For this one, the test is H_o: μ_x=μ_y v.s. H_a: μ_xμ_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 am stuck with part b...
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.

I am having trouble understanding what the question is asking for. Is it asking for P(type II error)? If so, how can I find it in this case?

Any help is greatly appreciated!

P.S. By the way, this topic is also discussed in SOS mathematics cyberboard forum.

2. They want you to compute $\beta$ under the SIMPLE hypothesis of $H_a:\mu_1-\mu_2=2$

3. Now I am a bit confused...

(i) P(fail to reject H_o | H_o is false) = β

(ii) P(fail to reject H_o | μ_x-μ_y=2)

Are the two probabilities above equal? In (i), we are given (conditional on) that H_o is false, so μ_x≠μ_y. In (ii), we are given that μ_x-μ_y=2.

But μ_x≠μ_y and μ_x-μ_y=2 aren't equivalent (μ_x≠μ_y does not imply that μ_x-μ_y=2, it could be that μ_x-μ_y=3, μ_x-μ_y=4, etc.), so I think (i) and (ii) are not equal. And it looks like the question is asking for (ii), right?

Thanks!

4. Originally Posted by kingwinner
Now I am a bit confused...

(i) P(fail to reject H_o | H_o is false) = β

(ii) P(fail to reject H_o | μ_x-μ_y=2)

Are the two probabilities above equal? In (i), we are given (conditional on) that H_o is false, so μ_x≠μ_y. In (ii), we are given that μ_x-μ_y=2.

But μ_x≠μ_y and μ_x-μ_y=2 aren't equivalent (μ_x≠μ_y does not imply that μ_x-μ_y=2, it could be that μ_x-μ_y=3, μ_x-μ_y=4, etc.), so I think (i) and (ii) are not equal. And it looks like the question is asking for (ii), right?

Thanks!

There's a different value for beta for every single possibility under the alternative hypothesis.
There's $\beta (\mu_1-\mu_2 =1)$, $\beta (\mu_1-\mu_2 =2)$....

You are asked to find $\beta (\mu_1-\mu_2 =2)$.

5. In hypothesis testing, if H_o is false, does this imply that H_a is true?
e.g. H_o: μ_x=μ_y is false => H_a: μ_x-μ_y=2 is true ?

6. But let's rephrase my question in this way: from a purely mathematical point of view, μ_x≠μ_y does not GUARANTEE that μ_x-μ_y=2, no arguments about that, right? So is there something special about hypothesis testing?

7. one set is contained in the other
μ_x-μ_y=2 GUARANTEEs μ_x≠μ_y

8. Originally Posted by matheagle
one set is contained in the other
μ_x-μ_y=2 GUARANTEEs μ_x≠μ_y
Yes, but for P(fail to reject H_o | H_o is false) and P(fail to reject H_o | μ_x-μ_y=2) to be equal, the condition must be perfectly equivalent, i.e. must have H_o is false if and only if μ_x-μ_y=2

μ_x-μ_y=2 guarantees μ_x≠μ_y, but the converse is false, hence not if and only if...

9. As I wrote yesterday. There is a different beta for every different possible outcome under the alternative hypothesis. And one minus beta is the power FUNCTION. There are a lot of beta's. YOU were asked to obtain just one, when the difference of the two populations means were equal to 2.

10. I've been seeing two versions of the definition for β...

Definition 1: β = P(fail to reject H_o | H_o is false)

Definition 2: β = P(fail to reject H_o | H_a is true)

Are these two definitions consistent?

So in hypothesis testing, saying H_o is false is EXACTLY THE SAME thing as saying that H_a is true, am I right?

11. yes, we end up deciding on one or the other
But most statisticians either say we accept $H_a$
or we fail to accept $H_a$.
Some won't say we accept $H_0$, because this really is like a proof by contradiction.
We assume $H_0$ to be correct, only to hope that it isn't and thus deciding on $H_a$.

12. So we have:
Ho: μ_x = μ_y
Ha: μ_x - μ_y = 2
Need to find: P(type II error) = P(fail to reject H_o | μ_x-μ_y=2)

Now I am having some trouble with the red part. How can I find the "rejection region" in this case??

Thanks!