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Unit 29: Estimation of Parameters: Criteria for Estimates



                                                                                                  Notes
                   Example 8: A random sample of 500 pineapples was taken from a large consignment and
            65 were found to be bad. Estimate the proportion of bad pineapples in the consignment and
            obtain the standard error of the estimator. Deduce that the percentage of bad pineapples in the
            consignment almost certainly lies between 8.5 and 17.5.
            Solution.
            Let p be the proportion of bad pine apples in the large consignment. Its estimate based on the
                                                ´
                        65                   0.13 0.87
                                       ( ) ˆ
            sample is  ˆ p =  =  0.13  with  . .S E p =  =  0.015
                       500                     500
            Thus, the 99.73% confidence limits of  p are  0.13 ± 3 × 0.015,  i.e., 0.085 and 0.175. Hence, the
            proportion of bad pineapples in the given consignment almost certainly lies between 8.5% and
            17.5%.

            Remarks: The width of a confidence interval can be controlled in two ways:
            (i)  By adjusting the sample size: More is the sample size the narrower will be the interval.
            (ii)  By adjusting the level of confidence: Lower the level of confidence the narrower will be
                 the interval.

            29.3.1 Determination of an Approximate Sample Size for a Given
                   Degree of Accuracy

            Let us assume that we want to find the size of a sample to be taken from the population such that
            the difference between sample mean and the population mean would not exceed a given value,
            say Î, with a given level of confidence. In other words, we want to find n such that

                        P ( X -     ) Î =  0.95 (say)        .... (1)


                                                                               
            Assuming that the sampling distribution of  X  is normal with mean m and  . .S E  =  , we can
                                                                            X
                                                                                n
            write
             æ       X        ö          æ  X      ö
                       -
                                             -
            P -  1.96      1.96 =  0.95  or  P    1.96 =  0.95
             ç                 ÷          ç          ÷
             è        / n     ø          è   / n   ø
                æ            ö
                   -
            or  P X    1.96 ×  ÷  =  0.95                    .... (2)
                ç
                è            n ø
            Comparing (1) and (2), we get
                                       æ  1.96 ö  2  3.84 2
                       Î=  1.96×    or  n = ç  ÷  =
                                n       è  Î  ø   Î 2

            Remarks:
            1.   The sample size required with a maximum error of estimation, Î and with a given level of
                                2
                               z  2
                 confidence is  n =  , where z is the value of standard normal variate for a given level of
                                Î 2
                              2
                 confidence and   is the variance of population.



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