CASE PROBLEM | Homework Helpers

CASE PROBLEM

EC 580, Instructor: Asiye Aydilek
statistical  procedures  that  can  be  used to  control  their  manufacturing  processes.  In  one
particular application,  a client ave Quality Associates a sample of 800 observations taken
during  a  time  in  which  that  client’s  process  was  operating  satisfactorily.  The  sample
standard deviation for these data was 0.21 and hence, with so much data, the population
standard  deviation  was  assumed  to  be  0.21.  Quality  Associates  then  suggested  that
random  samples  of  size  30  be  taken  periodically  to  monitor  the  process  on  an  ongoing
basis. By analyzing the new samples, the client could quickly learn whether the process
was operating satisfactorily. When the process was not operating satisfactorily, corrective
action could be taken to eliminate the problem.
The  design  specification  indicated  the  mean  for  the  process  should  be  12.  The
hypothesis test suggested by Quality Associates follows
H0
: µ=12
H1
: µ 12
Corrective action will be taken any time H0
is rejected.
The data set  Quality’ contains data from four samples, each of size 30, collected at hourly
intervals during the first day of operation of the new statistical control procedure.
1-  Conduct a hypothesis test for each sample at the 0.01 level of significance and determine
what action, if any, should be taken. Provide the test statistic and p-value for each test.

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Step 1: Enter the data range  (ex. A2:A31) into the  =count cell formula in cell F4
Write Sample Size to the cell E4
Step 2: Enter the data range  (ex. A2:A31) into the  =average cell formula in cell F5
Write Sample Mean to the cell E5
Step 3: Enter the population standard deviation 0.21 into cell F6
Write population std.deviation to the cell E6
Step 4: Enter the hypothesized value for the population mean 12 into cell F8.
Write Hypothesized value to the cell E8.
Step 5: Enter the formula of the standard error =F6/sqrt(F4) into cell F10
Write Standard Error to the cell E10
Step 6: Enter the formula of the test statistic =(F5-F8)/F10 into cell F11
Write Test Statistic to the cell E11
Step 7: Enter the formula of p-value of lower tail =normsdist(F11) into cell F13
Write p-value (Lower Tail) to the cell E13
Step 8: Enter the formula of p-value of upper tail = 1- F13 into cell F14
Write p-value (Upper Tail) to the cell E14
Step 9: Enter the formula of p-value of two tail = 2*(min(F13,F14)) into cell F15
Write p-value (Two Tail) to the cell E15

Repeat  the  steps  above  for  each  sample.  The  formula  for  the second  sample  should  be
written to G cells(replace F’s with G) and for the third sample to H cells and the fourth
sample    to  I  cells. Also    make  the  adjustments  for  the  data  range.  For  instance  if  the
second  sample  data  range  is  B2:B31,  the  formulas  in  G  cells  should  be  using  B2:B31
range.

Since our test is two-tailed test, use p-value (two tail) in cells (F15,G15,H15,I15) to make
the rejection decision for each sample.
If p-value>0.01 then the null hypothesis cannot be rejected.
If  p-value<0.01  then  the  null  hypothesis  is  rejected. In  such  a  case,  corrective  action  is
necessary.

2-  Compute  the  standard  deviation  for  each  of  the  four  samples.  Does  the  assumption  of
0.21 for the population standard deviation appear reasonable?
(Hint: Look at the range of the sample standard deviations for all four samples. If 0.21 is
in that range then the assumption of 0.21 is good)

3-  Discuss the implications of changing the level of significance to larger value. What error
(Type 1 or Type 2) could increase if the level of significance is increased?

(Hint: Decide whether you will reject the null hypothesis more or less with a larger Î±
value.
Decide  whether Rejecting  more often  means quicker or  slower corrective  action  when
the process is out of control.
Decide  we  have  higher  of  which  error  when there will  be  a  higher  error  probability  of
stopping  the  process  and  attempting  corrective  action  when  the  process  is  operating
satisfactorily.)

Sample 1    Sample 2    Sample 3    Sample 4
11.55    11.62    11.91    12.02
11.62    11.69    11.36    12.02
11.52    11.59    11.75    12.05
11.75    11.82    11.95    12.18
11.90    11.97    12.14    12.11
11.64    11.71    11.72    12.07
11.80    11.87    11.61    12.05
12.03    12.10    11.85    11.64
11.94    12.01    12.16    12.39
11.92    11.99    11.91    11.65
12.13    12.20    12.12    12.11
12.09    12.16    11.61    11.90
11.93    12.00    12.21    12.22
12.21    12.28    11.56    11.88
12.32    12.39    11.95    12.03
11.93    12.00    12.01    12.35
11.85    11.92    12.06    12.09
11.76    11.83    11.76    11.77
12.16    12.23    11.82    12.20
11.77    11.84    12.12    11.79
12.00    12.07    11.60    12.30
12.04    12.11    11.95    12.27
11.98    12.05    11.96    12.29
12.30    12.37    12.22    12.47
12.18    12.25    11.75    12.03
11.97    12.04    11.96    12.17
12.17    12.24    11.95    11.94
11.85    11.92    11.89    11.97
12.30    12.37    11.88    12.23
12.15    12.22    11.93    12.25

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