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  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.
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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 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 - = - - - 1 ] / ) 1 ( / ) ˆ ( [ 2 / 2 / z n p p p p z P - = - + - - 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 - = p ˆ
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This note was uploaded on 05/14/2010 for the course MATHEMATIC mathe taught by Professor Xyz during the Spring '10 term at Birla Institute of Technology & Science.

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

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