Page 326 - Applied statistics and probability for engineers
P. 326

304     Chapter 8/Statistical intervals for a single sample



                 of the interval is L = σ( z +  z α )  n . Prove that the length   8-117.  Students in the industrial statistics lab at ASU
                                       2
                                   α 1
                 of the interval L is minimized when α  = α  = α / 2. Hint:   calculate conidence intervals on μ. Suppose that all
                                             1
                                                 2
                                            −1
                                   1
                                                 )
                                                    α
                 Remember that Φ( )z a  = − α, so Φ ( 1 − α = z ,  and the  these CIs are independent of each other. Consider the
                 relationship between the derivative of a function y = f(x) and   next one thousand 95% conidence intervals that will
                 the inverse x =  f ( ) is (d dy f  −1 ( ) = 1  ⎡ ⎣ (d dy ) ( )⎤.  be calculated. How many of these CIs do you expect to
                             −1
                               y
                                                       x
                                           y
                                                      f
                                                         ⎦
                                       )
                 8-115.  It is possible to construct a nonparametric toler-  capture the true value of μ? What is the probability that
                                                                between 930 and 970 of these intervals contain the true
                 ance interval that is based on the extreme values in a ran-
                                                                value of μ?
                 dom sample of size n from any continuous population. If
                 p  is the minimum proportion of the population contained  8-118.  The conidence interval for a population propor-
                                                                tion depends on the central limit theorem. A common rule
                 between the smallest and largest sample observations with
                                                                of thumb is that to use the normal approximation for the
                 conidence 1 – α, it can be shown that
                                                                sampling distribution for ˆ p, you should have at least 10
                                          n
                              np n −1  −( n − ) 1  p = α        “successes” and 10 “failures.” However, Agresti and Coull
                                                                developed a method that can be used for smaller samples
                 and n is approximately                         and increases the accuracy of all conidence intervals for
                                            2
                             n =  1 +  ⎛ 1 +  p⎞ ⎛ X α,4 ⎞ ⎟    proportions. The idea is simply to add 4 “pseudo obser-
                                   ⎜
                                        ⎟ ⎜
                                                                vations” to the data set—2 successes and 2 failures. That
                                2  ⎝1 −  p⎠ ⎝  4  ⎠
                                                                is, if you have X successes from of n trials, use  ɶ p =  X + 2
                 (a)  To be 95% conident that at least 90% of the popula-                              n + 4
                    tion will be included between the extreme values of the   instead of the usual  ˆ p =  X   in the formulas for the coni-
                                                                                   n
                    sample, what sample size will be required?
                                                                dence interval.
                 (b) A random sample of 10 transistors gave the following
                                                                  A quality control engineer is inspecting defects on
                    measurements on saturation current (in milliamps): 10.25,
                                                                a newly designed printed circuit board. She inspects 50
                    10.41, 10.30, 10.26, 10.19, 10.37, 10.29, 10.34, 10.23,
                                                                boards and inds no defects. The usual estimate would be
                    10.38. Find the limits that contain a proportion p of the
                                                                ˆ p = 0, but she does not believe that there will ever be a no-
                    saturation current measurements at 95% conidence.
                                                                defects situation for this product. Use this Agresi-Coull
                    What is the proportion p contained by these limits?
                                                                estimate to come up with a 95% conidence interval for the
                 8-116.  Suppose that X , X , … , X  is a random sample   true proportion of defects. [Hint: The lower bound may be
                                            n
                                   1
                                      2
                 from a continuous probability distribution with median ɶ μ
                                                                negative, which is not feasible, so simply use the upper limit
                 (a) Show that
                                                                and report that she is 95% conident that the true defect pro-
                                                    n −1        portion is no higher than this number.]
                       p{min( ) < μ < max( X i } ) = − ⎛ ⎞ 1    8-119.  A biology student inds that of 35 males with Dros-
                                 ~
                                              1
                                                 ⎜ ⎟
                             X i
                                                 ⎝ ⎠ 2
                                                                ophila melanogaster, 2 have Adh genotypes with a male
                    [Hint:  The complement of the event [min( )X < ɶ μ   mating advantage.
                                                        i
                 <  max( )X i ]  is [max ( ) ≤ ɶ μ ]∪ [min ( ) ≤ ɶ μ ]  but max  (a)  Using the standard methods, ind a 95% conidence inter-
                                  X i
                                              X i
                 ( ) ≤ ɶ μ if and only if X i ≤ ɶ μ for all i.]    val for the true proportion of males with this genotype.
                  X i
                 (b) Write down a 100(1 – α)% conidence interval for the   (b)  Compare the conidence interval you found in part (a)
                    median  ɶ μ where    n  −1                     with the Agresti-Coull conidence interval by adding 4
                                   α =  ⎛ ⎞ 1                      pseudo-observations to the data discussed in the previ-
                                      ⎜ ⎟
                                      ⎝ ⎠ 2                        ous exercise.
                Important Terms and Concepts
               Chi-square              Conidence interval on the   Error in estimation  Two-sided
                  distribution            variance of a normal   Large-sample             conidence
               Conidence coeficient       distribution            conidence interval      interval
               Conidence interval      Conidence intervals on   One-sided conidence bounds  t distribution
               Conidence interval         the mean of a normal   Prediction interval
                  for a population        distribution         Tolerance interval
                  proportion           Conidence level
   321   322   323   324   325   326   327   328   329   330   331