## 85%/90% 2-sided binomial tolerance intervals for a future
## lot of 2500 when a sample of 230 were drawn from a lot of
## 1000. All methods but Jeffreys' method are compared
## below.
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "LS")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "WS")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "AC")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "CP")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "AS")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "LO")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "PR")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "PO")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "CL")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "CC")
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 2, method = "CWS")
## Using Jeffreys' method to construct the 85%/90% 1-sided
## binomial tolerance limits. The first calculation assumes
## a prior on the proportion of defects which places greater
## density on values near 0. The second calculation assumes
## a prior on the proportion of defects which places greater
## density on values near 1.
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 1, method = "JF", a1 = 2, a2 = 10)
bintol.int(x = 230, n = 1000, m = 2500, alpha = 0.15, P = 0.90,
side = 1, method = "JF", a1 = 5, a2 = 1)
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