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 = m_o
                H₁: m = m₁ > m_o
   
        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: