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					                                                                        Process Quality Engineering


HOMEWORK PROBLEMS AND SOLUTIONS

Readings: Basic Statistics: Tools for Continuous Improvement, Kiemele and Schmidt,
             Sections 6.1 - 6.5


Exercises:


K&S Problems 6.1, 6.10, 6.14, Practice Exam Questions




                                              Contents
                            Problem                             Page
                            Exercise 6.1                           2
                            Exercise 6.10                          3
                            Exercise 6.14                          5
                            Practice Exam Questions                 9




                                                                     HW5.DOC, May 31, 1995, page 1
                                                                               Process Quality Engineering




Exercise 6.1, page 6-63

The trailing edge of a turbine blade is to be robotically machined to a target of 6 mm. A simple random
sample of 50 blades are inspected and found to have a mean of 5.98 mm with a standard deviation of
.047 mm. Test at the   .01 level whether the process is centered at the target value.


SOLUTION
One sample hypothesis test of the mean ().
Step 1
   State the hypotheses to be tested.
        H 0 :   6 mm
          H1 :   6 mm


Step 2
   Determine a planning value for  .
         = .01


Step 3
   Take a sample of size n  30 and compute y and s2.
        y  5.98 mm          s  0.047 mm      n  50

   Compute the standardized test statistic:
When n  30 the
                   
z  dist. is a good            y  0    5.98  6     0.02
                     0
                     z                                     3.01
  approx. to the                s       0.047        0.00665
     t  dist.                   n            50


Step 4
   Using a standardized normal distribution, compute the area in the tail beyond z 0 .
   Pz  z0  Pz  3.01 1  Pz  3.01 1  .9987  0.0013


Step 5
   Compute p as 2 times the area found in Step 4.
   p = 2(.0013) = 0.0026




                                                                           HW5.DOC, May 31, 1995, page 2
                                                                                   Process Quality Engineering


Step 6
   Given that p (= .0026) < (= 0.01), we reject H 0    0  with
         (1 - p)% = (1 - .0026)% = 99.74% confidence.
   We do not believe that the process is centered at the target value (6 mm).



Exercise 6.10, page 6-65

A medical researcher is testing a new drug to see if it slows the growth a malignant tumors in rats. The
control group of 40 rats is given none of the drug and the test group of 40 rats is given a dose every day
for two weeks. At the end of the test, the following statistics were calculated.


                            Control Rats         Test Rats
 Average Growth
    of Tumor 6.2 grams        4.1 grams

Standard Deviation            3.1 grams          2.8 grams


Conduct the appropriate hypothesis test to determine if the new drug appears to slow the mean rate of
tumor growth.


SOLUTION
Two Samples Hypothesis Test of the Mean

                             Control Rats(y1 )          Test Rats(y2 )
 Average Growth
    of Tumor 6.2 grams           4.1 grams

Standard Deviation               3.1 grams                 2.8 grams


Step 1
  State the hypothesis
       H 0  1   2
         H1  1   2


Step 2
   Select 
         = 0.01


                                                                             HW5.DOC, May 31, 1995, page 3
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Step 3
   Obtain n1, n2 , y1, y2, s2 , s2
                            1    2

   n1  n2  40       y1  6.2            y2  4.1        s1  3.1    s2  2.8


   t0 
             y1  y2
                             sp 
                                      n1  1s12  n 2  1s2
                                                              2

               1     1                       n1  n2  1
          sp      
              n1 n 2
                                       393.1  392.8
                                                   2            2

                              sp 
                                          40  40  2
                                       680.55
                                              2.953811
                                         78

                y1  y 2        6.2  4.1       2.1
    t0                                             3.18
                  1      1                2   0.66049
             sp             2.953811
                 n 1 n2                  40
   Or, if n1 , n2  30 use

                 y1  y 2             6.2  4.1                 2.1
          z0                                                        3.18
                 s1 s2
                   2
                                    3.12       2.82       0.66049
                      2                     
                 n1 n 2              40           40


Step 4
   If n1 , n2  30 use z-distribution to estimate area beyond z 0 .
   Pz  z 0   Pz  3.18  1  Pz  3.18  1 .9993  0.0007


Step 5
   Find p (one-tailed test)
        p  Pz  z 0   0.0007


Step 6 Conclusion
   Since p  .0007    0.01, we reject H 0 1   2  with 1  p%  1  .0007%  99.93%
   confidence.
    1   2 or the new drug appears to slow the mean rate of tumor growth.




                                                                                 HW5.DOC, May 31, 1995, page 4
                                                                              Process Quality Engineering




Exercise 6.14, page 6-67

A marketing company wants to determine if a new product will sell better on the east coast or west
coast. Fifty individuals on each coast are randomly selected to test the new product. The following
table shows the results of the market test:


                             Estimated Market Share with New Product

        East coast           18% (i.e., 9 of 50 strongly in favor)
        West coast           26% (i.e., 13 of 50 strongly in favor


a) The company will first market the product in the area which shows the most potential for high
   sales. Are the market shares of the two coasts significantly different?
b) The company feels that the new market share must be at least 15% if the product is to be
   successful. Test each sample to see if they exceed this level.


SOLUTION
a) Two sample test of hypothesis for proportions
              n1  n 2  30
   Step 1
        State hypothesis
          H 0 : 1  2
            H1 :  1   2


   Step 2
       Determine 
          = .05

   Step 3
                                    x1        x2 
                                                 
        Compute proportions p1  n , p2  n 
                                       1         2
                                            9
         p1 = 9 of 50 strongly in favor =      = 0.18 (East coast)
                                           50
                                          13
         p2 = 13 of 50 strongly in favor      = 0.26 (West coast)
                                          50




                                                                          HW5.DOC, May 31, 1995, page 5
                                                                                   Process Quality Engineering

                             p1  p 2                                0.18  0.26
        z0                                             
               x1  x 2  x1  x 2  1 1           9  13  9  13  2 
                        
                             1                                   1
               n 1  n2  n 1  n2 n1 n 2        50  50  50  50 50 


                 0.08
                              0.966
             0.22.78.04

   Step 4
       Use z-distribution to determine area beyond z 0 .
                Pz  z0  Pz  0.966 1  Pz  0.966
       By interpolating:
                    
                       0.96      .8315                      x  .8315  d
               0.006
                                     d
                                       
        0.01          
                       0.966        x  0.0025
                                                               d     0.006
               0.97                                                  
                                .8340 
                                                              0.0025   0.01

                     0.006 
        d  0.0025               0.0015
                      0.01 

        x  0.8315  0.0015  0.8330  Pz  0.966
            Pz  0.966  1 0.8330  0.167


   Step 5
       Compute p (two-tailed test)
        p  2Pz  0.966  20.1670  0.3340


   Step 6
       Conclusion:
        Since p  0.3340    0.05, we fail to reject H 0 1  2  ,with
                1 p%  1 0.3340% 66.6% confidence.
                                   
        There is no significant difference in market share between either coast.

b) Perform one sample test of hypothesis for proportions on each sample from each coast.
   Step 1
       State hypothesis:
            East        West
          1  0.15  2  0.15
          1  0.15  2  0.15

                                                                                HW5.DOC, May 31, 1995, page 6
                                                                           Process Quality Engineering




Step 2
    Determine 
           = 0.05
Step 3
    Compute proportion p1 , p2 , then compute

                     p i  0
            z0 
                    0  0 
                        1
                         n

        East                     West
        9  0.18
    p1  50                  p2  13 50  0.26

           0.18  0.15              0.26  0.15
    z0                      z0 
            0.150.85               0.150.85
                50                       50
            0.03                     0.11
                               
           0.0505                   0.0505
        0.5940                  2.18


Step 4
    Use z-distribution to compute area beyond z0 (upper tailed test).

             East                            West
    Pz  z 0   Pz  0.59         Pz  z0   Pz  2.18
               1  Pz  0.59                    1 Pz  2.18
               1  0.7224                         1 0.9854
               0.2776                             0.0146


Step 5
    Compute p (one tailed test)
       East         West
     p  0.2776     p  0.0146




                                                                        HW5.DOC, May 31, 1995, page 7
                                                                              Process Quality Engineering


  Step 6
       Conclusion:


                       East                                 West
       Since p (= 0.2776) >  (= 0.05), we               Since p (= 0.0146) <  (=0.05), we
       failed to reject H0 ( 1  0.15) with             reject H0 (  2  0.15) with
       (1 - p)% = 72.24% confidence.                     (1 - p)% = 98.54% confidence.
        East coast sample does not                       West coast sample does exceed
           exceed 15% (market share).                        15% (market share).

Practice Exam Questions

1. You have been asked to determine whether your company’s “lightning”batteries have longer
   life than your competition’s “durable”batteries. Call the types “L” and “D”.

   a) Describe the null and alternative hypotheses that you would use to test the relation between
      L and D .

   b) Describe IN PLAIN ENGLISH what you would tell your boss if the hypothesis was
      rejected.

   c) Describe IN PLAIN ENGLISH what you would tell your boss if the hypothesis was
      accepted.


2. Suppose that you are testing the hypothesis

               H 0 :   1000
                    vs.
               H1 :   1000


   and your set of 16 data values produces X = 989 and s = 12.


   a) What is the p-value for the test?


   b) What is your conclusion, statistically speaking?


   c) What is your conclusion, IN PLAIN ENGLISH?




                                                                          HW5.DOC, May 31, 1995, page 8
                                                                                 Process Quality Engineering


3. You are conducting a two-sample test of

                 H0 : A  B
                       vs.
                 H 1 :  A  B
     You have 11 observations of population A with sample mean 12.5 and s2  4 .
                                                                         A

     You have 21 observations of population B with sample mean 12.0 and s2  3 .
                                                                         B



     a) What is the test statistic that you would use? Compute it and the resulting p-value.


     b) What do you conclude?


4.   You are investigating whether your new "Bountiful" paper towels are stronger with a new
     manufacturing process, compared with the existing process. You want to decide whether to
     shut down the factory and switch to the new processing machines. You have decided to use a
     hypothesis test for mean strength to help you decide. You feel comfortable with a = .05. You
     have collected test data and found the strength (in pounds) for 11 New towels to average
                          2
     XNew = 13.9, with s New = 4 , and for 21 Existing towels you had XExi st ing = 12.0, and
     s2 st ing = 3.
      Exi



     a) Set up the appropriate hypotheses:

                H0 :
                vs
                H1 :

     b) Compute the appropriate test statistic (hint: sP = 10 30 )


     c) Do you accept/reject H0 ?


     d) Explain the result IN PLAIN ENGLISH




                                                                             HW5.DOC, May 31, 1995, page 9

				
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