Question

In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008.

Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? (Round to four decimal places) Answer

Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? (Round to four decimal places)

Answer 1

a)

population proportion ,p=   0.75  
n=   450  
      
std error , SE = √( p(1-p)/n ) =    0.0204  
      
we need to compute probability for       
P(0.75-0.04 < p̂ <   0.75+0.04 )=P(0.71   < p̂ <   0.79)

Z1 =( p̂1 - p )/SE=   -1.960
Z2 =( p̂2 - p )/SE=   1.960
P(   0.71   < p̂ <   0.79   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.960   < Z <   1.960   )
                                      
= P ( Z <   1.960   ) - P (    -1.960   ) =    0.9750   -   0.025   =   0.9500  

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population proportion ,p=   0.75
n=   200
  
std error , SE = √( p(1-p)/n ) =    0.0306

we need to compute probability for         

P(0.75-0.04   < p̂ <    0.75+0.04 )=P(0.71    < p̂ <    0.79)

Z1 =( p̂1 - p )/SE=   -1.306
Z2 =( p̂2 - p )/SE=   1.306

P(   0.71   < p̂ <   0.79   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.306   < Z <   1.306   )
                                      
= P ( Z <   1.306   ) - P (    -1.306   ) =    0.9043   -   0.096   =   0.8086