sample of new car dealer complaints to estimate the proportion of complaints the
ID: 3260164 • Letter: S
Question
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 0.71, the same as the overall proportion of complaints settled in 2008.
Round your answers to four decimal places.
a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of p.
normal
not normalItem
b. Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion?
c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of p. In determining your answer, use the standard error found in part a. and the probability found using the tables in the text.
- Select your answer -
normal
not normal
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?
e. As measured by the increase in probability, how much more precise is the result, when you use a sample of 450 instead of a sample of 200?
The distribution is (select one)normal
not normalItem
Mean = Stand. Dev of p bar =Explanation / Answer
a)mean = 0.71
sd = sqrt(0.71*0.39/450) = 0.024805
Z = (p^ - 0.71)/sqrt(0.71*0.39/450)
= (p^ - 0.71)/0.024805
b) P( |p^ -p| <0.04)
P(-0.04 < p^ -p < 0.04)
=P(-0.04/0.024805 < Z< 0.04 /0.024805)
=P(1.61257 <Z< 1.61257)
=0.8932
c) if n= 200
mean = 0.71
sd = sqrt(0.71*0.39/200) = 0.0372088
Z= (p^ -0.71)/0.0372088
d)
P( |p^ -p| <0.04)
P(-0.04 < p^ -p < 0.04)
=P(-0.04/0.0372088< Z< 0.04 /0.0372088)
=P(-1.07501 <Z< 1.07501)
=0.7176
e) 0.8932 - 0.7176 = 0.1756
hence the result is 17.56 % more precise
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