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294     Chapter 8/Statistical intervals for a single sample

                      Approximate
               One-Sided Coni dence   The approximate 100(1 – α)% lower and upper conidence bounds are

               Bounds on a Binomial                         ˆ p)                           ˆ p)
                        Proportion               ˆ p z−  ˆ p(1 −                ˆ p z+  ˆ p(1 −
                                                    α          ≤  p  and    p ≤    α                (8-26)
                                                          n                              n
                                      respectively.


                                   A Different Cofi dence interval on the Binomial Proportion
                                   There is a different way to construct a CI on a binomial proportion than the traditional approach
                                   in Equation 8-23. Starting with Equation 8-22 and replacing the inequalities with an equality
                                   and solving the resulting quadratic equation for p results in
                                                                 2        p − ˆ p)  2
                                                                           ( ˆ 1
                                                             p +  z α  /2  ±  z /α 2  +  z α  /2 2
                                                             ˆ
                                                          p =    2 n         n     4 n
                                                                      1 +  z α 2  /2  n
                                   This implies that a two-sided CI on a proportion p is as follows:

                                                             2        p − ˆ p)  2
                                                                       ( ˆ 1
                                                         p +  z α  /2  +  z /2  +  z α  /2
                                                         ˆ
                                                                  α
                                                   UCL =     2 n         n     4 n 2
                                                                  1 +  z α 2  /2  n
                                                             2        p − ˆ p)  2                          (8-27)
                                                                       ( ˆ 1
                                                         p +  z α  /2  −  z α/2  +  z α /2
                                                         ˆ
                                                              n
                                                   LCL =     2n         n      4 n 2
                                                                     2
                                                                 1 +  z α /2  n
                                   The article by Agresti and Coull in The American Statistician (“Approximate Better Than
                                   ‘Exact’ for Interval Estimation of a Binomial Proportion,” 1998, pp. 119–126) reports that the
                                   actual conidence level for the CI in Equation 8-27 is closer to the “advertised” or nominal level

                                   for almost all values of α and p than for the traditional CI in Equation 8-23. The authors also
                                   report that this new interval can be used with nearly all sample sizes. So the requirements that
                                   np ˆ ≥ 5 or 10 or n(1− ˆ p) •  5 or 10 are not too important. If the sample size is large, the quantity
                                                              ˆ
                                   2
                                                                      2
                                                                2
                                   z α/2 /(2 n) will be small relative to  p , z α/ /(4 a ) will be small relative to  ˆ(p 1−  p  2  2  n
                                                                                                ˆ) / n, and z α/ /
                                                                  2
                                   will be small, so as a result the Agresti-Coull CI in Equation 8-27 will reduce to the traditional
                                   CI given in Equation 8-23.
               Example 8-10    The Agresti-Coull CI on a Proportion  Reconsider the crankshaft bearing data introduced in
                               Example 8-8. In that example we reported that  ˆ p = 0 .12 and n = 85. The traditional 95% CI was
                                                       .
                                                                 .
                                                      0 0509 ≤ p ≤  0 2243
               To construct the new Agresti-Coull CI, we use Equation 8-27:
                                   2        p − ˆ p)  2          . 1 96 2  0 12 0 88)  1 96 2
                                                                                       .
                                                                                .
                                                                            .
                                                                               (
                                             ( ˆ 1
                               p +  z α  /2  +  z /α 2  +  z α  /2  . 0 12 +  + 1 96  +  2
                                                                       .
                               ˆ
                                                                                       (
                        UCL =     2 n         n     4 n 2  =     ( 2 885)     85      4 85 )  =  0 2024
                                                                                              .
                                                                      +
                                                                            2
                                                                         .
                                       1 +  z α 2  /2  n             1 ( 1 96 / 85)
                                   2         ( p ˆ 1 − ) p ˆ  2  . 1 96 2  0 . ( . )   . 1 96 2
                                                                               0
                                                                                  8
                                                                                 8
                                                                              2
                                                                             1
                               ˆ p +  z α /2  − z α/2  +  z α /2 2  . 0 12 +  −1.96  +   2
                                +
                                                                       .
                                                                 ( )
                         LCL  =   2 n         n     4 n  =      2 85          85       ( 4 85  )  =  . 0 0654
                                           2
                                                                        1
                                       1  + z α /2  n                1+  ( .96 2  /85 )
               The two CIs would agree more closely if the sample size were larger.
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