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For the given data, y = 332.3 and s = 39.3. Assume that the number of units sold per year follows a
normal distribution with a mean and standard deviation that are constant from year to year, and will
continue to be so, and that the number of units sold in one year is independent of the number sold in
any other year.
Under the stated assumptions, y = 332.3 provides a prediction for the average yearly demand. The
demand for replacement bearings over 8 years is thus 8(332.3) = 2658. However, because of statistical
variability in both the past and future yearly demands, the actual total would be expected to differ from
this prediction. A one-sided upper 95% prediction bound (y U ) for the mean of the yearly sales for the
next m = 8 years is:


1

1
(
y U = y + t n−1, α s ---- + 1  1/2 = 332.3 + 2.015 39.3) --- + 1  1/2 = 375.1
---
---
6
n

8 
m
Thus, an upper 95% prediction bound for the total 8-year demand is 8(375.1) = 3001 bearings.
We are 95% confident that the total demand for the next 8 years will not exceed 3001 bearings. At
the same time, if the manufacturer actually built 3001 bearings, we would predict that the inventory would
most likely last for 3001/332.3 ≅ 9 years. A one-sided lower prediction bound for the total 8-year
demand is only 2316 bearings.
Case Study: Groundwater Monitoring
A hazardous waste landfill operator is required to take quarterly groundwater samples from m = 25
monitoring wells and analyze each sample for n = 20 constituents. The total number of quarterly
comparisons of measurements with published regulatory limits is nm = 25(20) = 500. There is a virtual
certainty that some comparisons will exceed the limits even if all wells truly are in compliance for
all constituents. The regulations make no provision for the “chance” failures, but substantial savings
would be possible if they would allow for a small (i.e., 1%) chance failure rate. This could be accom-
plished using a two-stage monitoring plan based on tolerance intervals for screening and prediction
intervals for resampling verification (Gibbons, 1994; ASTM, 1998).
The one-sided tolerance interval is of the form y + ks, where s, the standard deviation for each
constituent, has been estimated from an available sample of n b background measurements. Values of k
are tabulated in Gibbons (1994, Table 4.2).
Suppose that we want a tolerance interval that has 95% confidence (α = 0.05) of including 99% of
all values in the interval (coverage p = 0.99). This is n b = 20, 1 − α = 0.95, and p = 0.99, k = 3.295 and:

y + 3.295s

For the failure rate of (1 − 0.99) = 1%, we expect that k = 0.01(500) = 5 comparisons might exceed
the published standards. If there are more than the expected five exceedances, the offending wells should
be resampled, but only for those constituents that failed.
The resampling data should be compared to a 95% prediction interval for the expected number of
exceedances and not the number that happens to be observed. The one-sided prediction interval is:

1
y + ks 1 + -----
n b
If n b is reasonably large (i.e., n b ≥ 40), the quantity under the square root is approximately 1.0 and can be
ignored. Assuming this to be true, this case study uses k = 2.43, which is from Gibbons (1994, Table 1.2)
for n b = 40, 1 − α = 0.95, and p = 0.99. Thus, the prediction interval is y + 2.43s.
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