Question

please show excel formulas and calculations. thank you!!!!!!

Problem 3.13. The federal government claimed that one particular state had overpaid 21% of the Medicare recipients in the state. The director of the Department of Health of that state and Social Services planned to check this claim by selecting a random sample of 175 recipients of Medicare checks in the state and to determine the number of overpaid cases in the sample. Assuming the federal government's claim is correct; a, what is the probability that less than 17% of the people in the sample would be found to have been overpaid? what the probability that more than 24% of the people in the sample would be found to have been overpaid? b. what would be the percentage of people overpaid in the sample to be within 2.5% of the population proportion? c,

Answer 1

a)

population proportion ,p=   0.21      
n=   175      
          
std error , SE = √( p(1-p)/n ) =    0.0308      
          
sample proportion , p̂ =   0.17      
Z=( p̂ - p )/SE=   -1.299      
P ( p̂ <    0.17   ) =P(Z<( p̂ - p )/SE) =  
          
=P(Z <    -1.299   ) =    0.0969 [ excel formula =NORMSDIST(-1.299)

b)

population proportion ,p=   0.21      
n=   175      
          
std error , SE = √( p(1-p)/n ) =    0.0308      
          
sample proportion , p̂ =   0.24      
Z=( p̂ - p )/SE=   0.974      
P ( p̂ >    0.24   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   0.974   ) =    0.1649( excel formula =1 - NORMSDIST(0.974)

c)

within 2.5 % of population prportion

it means

we need to compute

P( 21%-2.5%< X < 21%+2.5%) = P(18.5%<X<23.5%)

                       population proportion ,p=   0.21          
                       n=   175          
                                      
                       std error , SE = √( p(1-p)/n ) =    0.0308          
                                      
                       we need to compute probability for               
                       0.185   < p̂ <   0.235      
                                      
                       Z1 =( p̂1 - p )/SE=   -0.812          
                       Z2 =( p̂2 - p )/SE=   0.812     

   
P(   0.185   < p̂ <   0.235   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -0.812   < Z <   0.812   )
                                      
= P ( Z <   0.812   ) - P (    -0.812   ) =    0.7916   -   0.208   =   0.5832