How to find a confidence interval of a binomial distribution using a simulated random sample? ...

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How to find a confidence interval of a binomial distribution using a simulated random sample?



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$begingroup$


I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
So I have 1000 values from the distribution.



How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).



It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
    So I have 1000 values from the distribution.



    How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).



    It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
      So I have 1000 values from the distribution.



      How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).



      It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?










      share|cite|improve this question











      $endgroup$




      I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
      So I have 1000 values from the distribution.



      How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).



      It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?







      statistics statistical-inference binomial-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 15:56







      Pumpkinpeach

















      asked Mar 17 at 17:46









      PumpkinpeachPumpkinpeach

      578




      578






















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          0












          $begingroup$

          By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.



          In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:



          set.seed(2019);  x = rbinom(1000, 52, 44/52)


          Then we see that (a centrally located) 95% of the values $x/52$ lie
          within the interval $(0.73, 0.94).$



          set.seed(318);  x = rbinom(1000, 52, 44/52)
          quantile(x/52, c(.025, .975))
          2.5% 97.5%
          0.7307692 0.9423077


          This is close to the same interval we get with
          the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
          amounts to $(0.748, 0.944).$



          p.est = 44/52;  pm=c(-1,1) 
          p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
          [1] 0.7480870 0.9442207




          Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
          interval in R, using the
          quantile function (inverse CDF) of the appropriate
          binomial distribution.



          qbinom(c(.025,.975), 52, 44/52)/52
          [1] 0.7500000 0.9423077


          (2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.



          n = 56; p.est = 46/n;  pm=c(-1,1) 
          p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
          [1] 0.7173299 0.9255273





          share|cite|improve this answer











          $endgroup$














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            0












            $begingroup$

            By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.



            In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:



            set.seed(2019);  x = rbinom(1000, 52, 44/52)


            Then we see that (a centrally located) 95% of the values $x/52$ lie
            within the interval $(0.73, 0.94).$



            set.seed(318);  x = rbinom(1000, 52, 44/52)
            quantile(x/52, c(.025, .975))
            2.5% 97.5%
            0.7307692 0.9423077


            This is close to the same interval we get with
            the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
            amounts to $(0.748, 0.944).$



            p.est = 44/52;  pm=c(-1,1) 
            p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
            [1] 0.7480870 0.9442207




            Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
            interval in R, using the
            quantile function (inverse CDF) of the appropriate
            binomial distribution.



            qbinom(c(.025,.975), 52, 44/52)/52
            [1] 0.7500000 0.9423077


            (2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.



            n = 56; p.est = 46/n;  pm=c(-1,1) 
            p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
            [1] 0.7173299 0.9255273





            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.



              In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:



              set.seed(2019);  x = rbinom(1000, 52, 44/52)


              Then we see that (a centrally located) 95% of the values $x/52$ lie
              within the interval $(0.73, 0.94).$



              set.seed(318);  x = rbinom(1000, 52, 44/52)
              quantile(x/52, c(.025, .975))
              2.5% 97.5%
              0.7307692 0.9423077


              This is close to the same interval we get with
              the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
              amounts to $(0.748, 0.944).$



              p.est = 44/52;  pm=c(-1,1) 
              p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
              [1] 0.7480870 0.9442207




              Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
              interval in R, using the
              quantile function (inverse CDF) of the appropriate
              binomial distribution.



              qbinom(c(.025,.975), 52, 44/52)/52
              [1] 0.7500000 0.9423077


              (2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.



              n = 56; p.est = 46/n;  pm=c(-1,1) 
              p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
              [1] 0.7173299 0.9255273





              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.



                In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:



                set.seed(2019);  x = rbinom(1000, 52, 44/52)


                Then we see that (a centrally located) 95% of the values $x/52$ lie
                within the interval $(0.73, 0.94).$



                set.seed(318);  x = rbinom(1000, 52, 44/52)
                quantile(x/52, c(.025, .975))
                2.5% 97.5%
                0.7307692 0.9423077


                This is close to the same interval we get with
                the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
                amounts to $(0.748, 0.944).$



                p.est = 44/52;  pm=c(-1,1) 
                p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
                [1] 0.7480870 0.9442207




                Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
                interval in R, using the
                quantile function (inverse CDF) of the appropriate
                binomial distribution.



                qbinom(c(.025,.975), 52, 44/52)/52
                [1] 0.7500000 0.9423077


                (2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.



                n = 56; p.est = 46/n;  pm=c(-1,1) 
                p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
                [1] 0.7173299 0.9255273





                share|cite|improve this answer











                $endgroup$



                By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.



                In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:



                set.seed(2019);  x = rbinom(1000, 52, 44/52)


                Then we see that (a centrally located) 95% of the values $x/52$ lie
                within the interval $(0.73, 0.94).$



                set.seed(318);  x = rbinom(1000, 52, 44/52)
                quantile(x/52, c(.025, .975))
                2.5% 97.5%
                0.7307692 0.9423077


                This is close to the same interval we get with
                the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
                amounts to $(0.748, 0.944).$



                p.est = 44/52;  pm=c(-1,1) 
                p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
                [1] 0.7480870 0.9442207




                Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
                interval in R, using the
                quantile function (inverse CDF) of the appropriate
                binomial distribution.



                qbinom(c(.025,.975), 52, 44/52)/52
                [1] 0.7500000 0.9423077


                (2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.



                n = 56; p.est = 46/n;  pm=c(-1,1) 
                p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
                [1] 0.7173299 0.9255273






                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 18 at 23:11

























                answered Mar 18 at 22:37









                BruceETBruceET

                36.2k71540




                36.2k71540






























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