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                                     TABLE 21.3
                                     Factors for Two-Sided 95% Statistical Intervals for a Standard
                                     Deviation of a Normal Distribution
                                                             Simultaneous Prediction Intervals
                                          Confidence Intervals  to Contain All m == == n Future Observations
                                      n     k 1      k 2         k 1            k 2
                                      4    0.57     3.73        0.25           3.93
                                      5    0.60     2.87        0.32           3.10
                                      6    0.62     2.45        0.37           2.67
                                      7    0.64     2.20        0.41           2.41
                                      8    0.66     2.04        0.45           2.23
                                     10    0.69     1.83        0.50           2.01
                                     15    0.73     1.58        0.58           1.73
                                     20    0.76     1.46        0.63           1.59
                                     40    0.82     1.28        0.73           1.38
                                     60    0.85     1.22        0.77           1.29
                                     ∞     1.00     1.00        1.00           1.00

                                     Source: Hahn, G. J. and W. Q. Meeker (1991). Statistical Intervals: A Guide
                                     for Practitioners, New York, John Wiley.

                       Example 21.3

                           A random sample of n = 5 observations yields the values y =  28.4  µg/L and σ = 1.18 µg/L.

                           1.  Using the factors in Table 21.3, find a two-sided confidence interval for the standard deviation
                             σ of the population. For n = 5, k 1  = 0.6 and k 2  = 2.87. The 95% confidence interval is:

                                                 [0.60(1.18), 2.87(1.18)] = [0.7, 3.4]

                             We are 95% confident that the interval 0.7 to 3.4 µg/L contains the unknown standard deviation
                             s of the population of concentration readings.
                           2.  Construct a two-sided 95% prediction interval to contain the standard deviation of five addi-
                             tional concentrations randomly sampled from the same population. For n = m = 5, k 1  = 0.32,
                             k 2  = 3.10, and the prediction interval is:

                                                  [0.32(1.18), 3.1(1.18)] = [0.4, 3.7]

                             We are 95% confident that the standard deviation of the five additional concentration readings
                             will be in the interval 0.4 to 3.7 µg/L.

                       Notice how wide the intervals are compared with confidence intervals and tolerance intervals for the
                       mean.




                       Case Study: Spare Parts Inventory
                       Village water supply projects in Africa have installed thousands of small pumps that use bearings from
                       a company that will soon discontinue the manufacture of bearings. The company has agreed to create
                       an inventory of bearings that will meet, with 95% confidence, the demand for replacement bearings for
                       at least 8 years. The number of replacement bearings required in each of the past 6 years were:

                                                 282, 380, 318, 298, 368, and 348

                       © 2002 By CRC Press LLC
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