In 2008 the Better Business Bureau settled 75% of complaints they received (USA

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. Use z-table a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of p. Select 75 (to 2 decimals) standard error of the proportion ( p ) (to 4 decimals) b. Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion (to 4 decimals)? c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of p Select (to 2 decimals) standard error of the proportion ( p ) (to 4 decimals) 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)? The probability of the sample proportion being within 04 of the population mean was Selectwith the smaller sample. So there is a Selectin precision by Select the sample size from 200 to 450. It is always convenient using a larger sample, so long as the extra cost of using the larger sample size Select

Explanation / Answer

a)
Normal distribution
p = 0.75

SE for proportion = sqrt ( p * (1 - p) / n)
SE = sqrt(0.75 * 0.25 / 450) = 0.0204

b)
P(0.71 < pcap < 0.79)
= P(-0.04/0.0204 < z < 0.04/0.0204)
= P(-1.9596 < z < 1.9596)
= 0.95 (0.949956479)

c)
Normal distribution
p = 0.75

SE for proportion = sqrt ( p * (1 - p) / n)
SE = sqrt(0.75 * 0.25 / 200) = 0.0306

d)
P(0.71 < pcap < 0.79)
= P(-0.04/0.0306 < z < 0.04/0.0306)
= P(-1.3064 < z < 1.3064)
= 0.8086

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