1n 2008 the Better Business Bureau settled 75% of co plaints they received USA T

Question

1n 2008 the Better Business Bureau settled 75% of co plaints they received USA Today March 2 200 Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year in ol ing ne 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. Use z-table. a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of p. Can assume to be normally distributed because np> 5 and n1-p) p-020 to 2 decimals) standard error of the proporto p(to 4 decimals) s b. Based upon a sample of 450 complaints, what is the probabiity that the sample proportion will be within.04 of the population proportion (to 4s)? Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of p. c. Can assume to be normally distributed because np 5 and n(1-p)5 (o 2 decimals) standard error of the proportion o p) (to 4 dedimals) d. 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 (to 4 decimals)? e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in part (b)? in precision by increasing the sample size from 200 to 450. It is always The probability of the sample proportion being within 04 of the population mean was reduced with the smaller sample. So there is a gain onvenient using a larger sample, so long as the extra cost of using the larger sample sizeis not too great

Explanation / Answer

Answer

a) p=0.75

For N= 450

Standard error of the proportion= sqrt {p(1-p)/n }= sqrt(0.75*0.25/450) = 0.2041

b) For sample proportion to be within 0.04 of population proportion

=0.04/0.2041 = 0.196

Therefore, Probability of proportion within 0.04of population proportion = P(z<0.1960)-P(z<-0.1960)

= 0.5753-0.4247

=0.1506

c) For a sample of 200 complaints

p=0.75

N=200

Standard error of the proportion= sqrt {p(1-p)/n }= sqrt(0.75*0.25/200) = 0.0306

d)Based upon sample of 200 complaints

=0.04/0.0306=1.3072

Therefore, Probability of proportion within 0.04 of population proportion = P(z<1.3072)-P(z<-1.3072)

=0.8089

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