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### 45-733 PROBABILITY AND STATISTICS I Topic #8A

29 February 2000

#### Hypothesis Testing

1. The Classical Theory of Two Simple Hypotheses

```We have a large shipment of devices delivered to our manufacturing
plant. Suppose we know with certainty that either the proportion of
defective devices is either .01 or .001.  We take a random sample and
Ù         Ù
compute p.  Given p, how do we decide between .01 and .001?
```
In the classical theory of two simple hypotheses we denote these two possibilities as:

H0: p = p0
H1: p = p1

Where H0: is known as the Null Hypothesis and H1: is known as the Alternative Hypothesis.

Given a decision, there are four possibilities:

Accept H0: and H0: is True.
Accept H0: and H1: is True.
Reject H0: and H0: is True.
Reject H0: and H1: is True.

We can represent these possibilites by a two by two table:
```
True State of World

H0     H1
--------------
Accept H0  |1 - a |  b  |
|      |     |
Decision       |------|-----|
|      |     |
Reject H0  |  a   |1 - b|
|      |     |
--------------
```

Where:
a = TYPE I Error = P[Reject H0: | H0: is True] and
b = TYPE II Error = P[Accept H0: | H1: is True]
2. Hypothesis Test Between Two Means for the Normal Distribution (s2 is known)

H0: m = m0
H1: m = m1 > m0

A reasonable decision rule for this problem is:
```   _
If Xn > m0 + c then Reject H0:
_
If Xn < m0 + c then Do Not Reject H0:
```
Where c is some constant. In most circumstances m0 < m0 + c < m1.
The a and b errors are:
```      _
a = P[Xn > m0 + c | m = m0]
_
b = P[Xn < m0 + c | m = m1]
```

3. Example: Suppose n = 25, s2 = 400, and a = .05, and we have the hypothesis test:

H0: m = 100
H1: m = 110
```      _
a = P[Xn > 100 + c ] = .05 =
_
P[(Xn - 100)/20/5 > (100 + c - 100)/4 ] = P[Z > c/4]
Now, since P[Z > 1.645] = .05, c/4 = 1.645, and c = 6.58
_
b = P[Xn < 106.58 | m = 110] =
_
P[(Xn - 110)/4 < (106.58 - 110)/4 ] =

P[Z < -.855] = F(-.855) = 1 - F(.855) = .1967
```
4. The only way to simultaneously reduce a and b is to increase the sample size. With fixed sample size, n, reducing a causes b to increase and vice versa. There are many situations in which it is desirable to make a or b as small as possible even at the cost of greatly increasing the other error. A good example of this is disease testing:
```
True State of World

Has   Not
Disease Have
Disease
--------------
Patient Has Disease |1 - a |  b  |
|      |     |
Doctor's Decision  |------|-----|
|      |     |
Patient Does Not Have   |  a   |1 - b|
Disease  |      |     |
--------------
```

Clearly, telling a patient that he/she does not have a disease when they in fact have the disease -- the Type I error -- is much more costly than telling a patient that he/she has a disease when they in fact do not have the disease -- the Type II error. In the first instance, a sick person can go infect other people and cause great harm. In the second, the harm is to scare a healthy person. Clearly, in disease testing, minimizing the a probability makes sense.

5. Recall that the Hypothesis Test Between Two Means for the Normal Distribution where s2 is known is:

H0: m = mo
H1: m = m1 > mo

The decision rule for this problem is:

```   _
If Xn > mo + c then Reject H0:
_
If Xn < mo + c then Do Not Reject H0:
```
Which is equivalent to:
```    _
If (Xn - mo)/s/n1/2 > za then Reject H0:
_
If (Xn - mo)/s/n1/2 < za then do not Reject H0:
``` 