45-733 PROBABILITY AND STATISTICS I Topic #8A

29 February 2000

- 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

In the classical theory of two simple hypotheses we denote these two possibilities as:_{Ù}compute_{Ù}**p**. Given**p**, how do we decide between .01 and .001?

**H**_{0}: p = p_{0}

H_{1}: p = p_{1}

Where**H**is known as the Null Hypothesis and_{0}:**H**is known as the Alternative Hypothesis._{1}:

Given a decision, there are four possibilities:

Accept**H**and_{0}:**H**is True._{0}:

Accept**H**and_{0}:**H**is True._{1}:

Reject**H**and_{0}:**H**is True._{0}:

Reject**H**and_{0}:**H**is True._{1}:

We can represent these possibilites by a two by two table:

**True State of World H**_{0}H_{1}-------------- Accept H_{0}|1 - a | b | | | | Decision |------|-----| | | | Reject H_{0}| a |1 - b| | | | --------------

Where:

**a = TYPE I Error = P[Reject H**and_{0}: | H_{0}: is True]

**b = TYPE II Error = P[Accept H**_{0}: | H_{1}: is True] - Hypothesis Test Between Two Means for the Normal Distribution
(
**s**is known)^{2}

**H**_{0}: m = m_{0}

H_{1}: m = m_{1}> m_{0}

A reasonable decision rule for this problem is:_ If

Where**X**then Reject_{n}> m_{0}+ c**H**_ If_{0}:**X**then Do Not Reject_{n}< m_{0}+ c**H**_{0}:**c**is some constant. In most circumstances**m**._{0}< m_{0}+ c < m_{1}

The**a**and**b**errors are:_

**a = P[X**_{n}> m_{0}+ c | m = m_{0}] _ b = P[X_{n}< m_{0}+ c | m = m_{1}]

- Example: Suppose
**n**= 25,**s**= 400, and^{2}**a**= .05, and we have the hypothesis test:

**H**_{0}: m = 100

H_{1}: m = 110_

**a = P[X**Now, since_{n}> 100 + c ] = .05 = _ P[(X_{n}- 100)/20/5 > (100 + c - 100)/4 ] = P[Z > c/4]**P[Z > 1.645]**= .05,**c**/4 = 1.645, and**c**= 6.58 _**b = P[X**= .1967_{n}< 106.58 | m = 110] = _ P[(X_{n}- 110)/4 < (106.58 - 110)/4 ] = P[Z < -.855] = F(-.855) = 1 - F(.855) - 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.

- Recall that the Hypothesis Test Between Two Means for the Normal
Distribution where
**s**is known is:^{2}

**H**_{0}: m = m_{o}

H_{1}: m = m_{1}> m_{o}

The decision rule for this problem is:

_ If

Which is equivalent to:**X**then Reject_{n}> m_{o}+ c**H**_ If_{0}:**X**then Do Not Reject_{n}< m_{o}+ c**H**_{0}:_ If

**(X**then Reject_{n}- m_{o})/s/n^{1/2}> z_{a}**H**_ If_{0}:**(X**then do not Reject_{n}- m_{o})/s/n^{1/2}< z_{a}**H**_{0}: