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                                         9-3 TESTS ON THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE UNKNOWN  305


                                   approximately. Thus, the probability of rejecting H :    0.82 if the true mean exceeds this
                                                                            0
                                   by 0.02 is approximately 1 
   1 
 0.10   0.90, and we conclude that a sample size of
                                   n   15 is adequate to provide the desired sensitivity.
                                   Minitab will also perform power and sample size computations for the one-sample t-test.
                                   Below are several calculations based on the golf club testing problem:


                                       Power and Sample Size

                                       1-Sample t Test
                                       Testing mean   null (versus   null)
                                       Calculating power for mean   null   difference
                                       Alpha   0.05 Sigma   0.02456
                                                   Sample
                                       Difference   Size    Power
                                         0.02        15     0.9117
                                       Power and Sample Size
                                       1-Sample t Test
                                       Testing mean   null (versus   null)
                                       Calculating power for mean   null   difference
                                       Alpha   0.05 Sigma   0.02456
                                                   Sample
                                       Difference   Size    Power
                                         0.01        15     0.4425
                                       Power and Sample Size
                                       1-Sample t Test
                                       Testing mean   null (versus   null)
                                       Calculating power for mean   null   difference
                                       Alpha   0.05 Sigma   0.02456
                                                   Sample   Target   Actual
                                       Difference   Size    Power    Power
                                         0.01        39     0.8000   0.8029


                                   In the first portion of the computer output, Minitab reproduces the solution to Example 9-7, veri-
                                   fying that a sample size of n   15 is adequate to give power of at least 0.8 if the mean coefficient
                                   of restitution exceeds 0.82 by at least 0.02. In the middle section of the output, we asked Minitab
                                   to compute the power if the difference in   and     0.82  we wanted to detect was 0.01. Notice
                                                                         0
                                   that with n   15, the power drops considerably to 0.4425. The final portion of the output is the
                                   sample size required to give a power of at least 0.8 if the difference between   and   of interest
                                                                                                     0
                                   is actually 0.01. A much larger n is required to detect this smaller difference.
                 9-3.4 Likelihood Ratio Approach to Development of Test Procedures
                        (CD Only)

                 EXERCISES FOR SECTION 9-3
                 9-30.  An article in the ASCE Journal of Energy Engineer-  building material. Five samples of the material were tested in
                 ing (1999, Vol. 125, pp. 59–75) describes a study of the ther-  a structure, and the average interior temperature (°C) reported
                 mal inertia properties of autoclaved aerated concrete used as a  was as follows: 23.01, 22.22, 22.04, 22.62, and 22.59.
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