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ch9-12 - Chapter 9 Inferences on Proportions This chapter...

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Chapter 9 Inferences on Proportions
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This  chapter  will   deal  with  inferences  on proportions and hypothesis tests on them. 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.
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9.1 Estimating proportions: Estimation  of  proportion  should  be  done  in  the following situation: We  have  population  of  interest,  particular  trait  is  being  studied,  and  each  member  of  the  population  can  be  classified  as either having the trait or not.
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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 = . trait the have not does sample the of member th i' ' the if 0 . trait the has sample the of member th i' ' the if 1 i X
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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 with the sample in number = = n X p
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Confidence interval on ‘p’: For this distribution of        must be  determined. Assume sample size n is large. 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 ˆ p ˆ
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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 ˆ
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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 ( / ) ˆ ( - - α α α - = - - - 1 ] / ) 1 ( / ) ˆ ( [ 2 / 2 / z n p p p p z P α α α - = - + - - 1 ] / ) 1 ( ˆ / ) 1 ( ˆ [ 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 ( - ± α/2 p n p p z p / ) 1 ( 2 / - ± α
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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 p n p p z / ) 1 ( 2 / - α 2 ) 1 ( 2 2 / d p p z n - = α p ˆ
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When prior estimate for p is not available, 
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