µ0 since rejecting the null hypothesis in favor of the alternative would mean that the sample mean would have to be “low”

A Note About “Composite” Null Hypotheses (Optional Reading)

In this course/book, we are only considering hypothesis tests where the null hypothesis has the form

H0 :µ=µ0

as opposed to other possibilities like

H0 :µ≤µ0 and H0 :µ≥µ0.

This is not that big of a restriction since a test of (for example) H0:µ=µ0 versus Ha:µ < µ0

would be to look at the sample meanxand rejectH0if thexis “low”, suggesting that the alternative hypothesis test is true. You would do the same thing to test

H0 :µ≥µ0 versus H_a:µ < µ0

since rejecting the null hypothesis in favor of the alternative would mean that the sample mean would have to be “low”.

However, there is a slight added complication when considering the latter test. It involves the level of significance α.

Let’s first consider the type of test we can do now:

H⁰:µ=µ⁰ versus Ha:µ < µ⁰

(We could also consider any of the other two alternative hypotheses, but let’s just consider this one.)

The test would be to look atxand rejectH0 in favor ofH_aifxis “low”, sayx < cfor some number c.

In this case,

α = P(Type I error)

= P(reject H⁰ when it’s true)

= P(X < c when µ=µ0)

Now if you know that the true meanµisµ⁰, then you know that the mean for X isµ⁰, so you can proceed by standardizing X and “turning it into aZ”, and looking in a table.

On the other hand, if we are trying to test

H0:µ≥µ0 versus H_a:µ < µ0,

we would still want to rejectH0 in favor ofH_a if the sample meanx is “low”, say x < c, but now α = P(Type I error)

= P(reject H0 when it’s true)

= P(X < c when µ≥µ⁰)

The problem here is that we can’t compute this probability since now µ can be a lot of things–

actually, it can be any number greater than µ0. So, how can we standardize X to “turn it into a Z”?!

The answer is that for a “composite” null hypothesis such as the one above, we have to slightly alter our definition ofα. In this case

α= max P(Type I error)

where the maximum is taken over all possible values included in the null hypothesis.

How do we do this?

This is actually not a difficult thing to do (for a hypothesis test of means based on a normal or at distribution). The answer is going to be

α = maxµ≥µ0 P(Type I error)

= P(Type I error when µ=µ0)

This can be seen:

1. By a picture! A Type I error is when you reject the null hypothesis when it is really true.

You are going to reject ifxis less than some numberc. Fix a numbercon a number line and consider two different bell curves that x could have come from– one with mean µ¹ and the other with meanµ2 > µ1.

Regardless of where I putcin this picture, the probability thatx is less thanc(so we reject) is higher for theµ1 curve than for theµ2 curve. So, the Type I error in theµ1 case is greater than the Type I error for the µ2 case. In other words, the higher the µ, the smaller the probability of a Type I error. So, to get the maximum Type I error, we take the smallest possibleµ included in the null hypothesisH0 :µ≥µ0.

So we can conclude:

α = maxµ≥µ0 P(Type I error)

= P(Type I error when µ=µ0).

2. The other way to reach this same conclusion that a largerµ results in a smaller probability of a Type I error is to work outα for a µ1 andµ2 where, againµ2 > µ1. Then we have

P(Type I error when µ=µ¹) which is

P(reject H⁰ when µ=µ¹) =P(X < c when µ=µ¹) which when standardized “to aZ” becomes

P(Z < c−µ1

σ/√

n) = Φ

c−µ1

σ/√ n

Do the same thing withµ2 and we end up with

P(Type I error when µ=µ2) as

Φ

c−µ2

σ/√ n

Nowµ2 > µ1 implies that

c−µ2

σ/√

n < c−µ1

σ/√ n and since Φ(·) is an increasing function, we have that

Φ

c−µ2

σ/√ n

<Φ

c−µ1

σ/√ n

So, again, the larger probability of a Type I error comes from the smallerµand therefore the largest or maximum probability of a Type I error comes from the smallest possible µ in the null hypothesis, namely µ0!