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: