# Chapter9 - CHAPTER9 INFERENCESON PROPORTIONS...

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CHAPTER 9 INFERENCES ON  PROPORTIONS

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This chapter will   deal with inferences on proportions and  comparisons of two proportions. We  will  see  how  to  employ  the  standard  normal  distribution  to  construct  confidence  intervals  on  ‘p’  and  test hypotheses concerning its value for larger samples.
9.1 Estimating proportions: Estimation  of  proportion  should  be  done  in  the  following  situation: We have a population of interest, a particular trait is being  studied,  and  each  member  of  the  population  can  be  classified as either having the trait or not. For  this,  we  draw  a  random  sample  of  size  ‘n’  from  the  population,  and  associated  with  it  is  a  collection  of  n  independent random variables X 1,  X 2 , …..X n  where = . ' ' 0 . ' ' 1 trait the have not does sample the of member th i the if trait the has sample the of member th i the if X i

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In general,                   gives the number of objects in the  sample  with  the  trait  and  the  statistic  X/n  gives  the  proportion of the sample with the trait.  Then  the  sample  proportion  which  is  a  logical  point  estimator for p is given by = = n i i X X 1 size sample trait the with sample in number n X p = =
Confidence interval on ‘p’: For this distribution of      must be determined. This is nothing but the sample mean. Therefore, by Central Limit theorem,     is approximately  normally distributed with same mean as X i ’s and variance  equal to Var X i /n  Since X i  is 1 when the object being sampled has the trait, P[ X i =1]=p and P[ X i =0]=1-p p p

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x i 1 0 f(x i ) p 1-p Also, it is easy to see that E[X i ]=1(p)+0(1-p)=p E[X i 2 ]=1 2 (p)+0 2 (1-p)=p Var X i =E[X i 2 ]-(E[X i ]) 2 =p-p 2 =p(1-p). Therefore by CLT,      is approximately  normally distributed   with mean p and variance  p(1-p)/n. p
The random variable  follows a standard normal distribution. For a 100(1- α )% confidence interval Isolating p in the middle of the inequality,  n p p p p / ) 1 ( / - - α α α - = - - - 1 ] / ) 1 ( / ) ( [ 2 / 2 / z n p p p p z P α α α - = - - - 1 ] / ) 1 ( / ) 1 ( [ 2 / 2 / n p p z p p n p p z p P

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Based on the above expression, we get the confidence  intervals for p The above expression employs ‘p’ (which we don’t know).  So replace it by  The confidence intervals become n p p z p / ) 1 ( 2 / - ± α p n p p z p / ) 1 ( 2 / - ± α
Sample size for estimating ‘p’: We can be 100(1- α29 % sure that       and p differ by at most  d , where d is given by  d=   Sample size for estimating p, prior estimate available Another method for estimating n is based on the fact that                   can never be greater than 0.25. Hence replace this  term by  ¼  in the above formula.

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