Respuesta :
Given that 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.
Part a) :
Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).
Population proportion = 0.75
sample size. n = 450
[tex]Standard\ error=\sqrt{\frac{p(1-p)}{n}} \\ \\ =\sqrt{\frac{(0.75)(0.25)}{450}} = 0.0204 [/tex]
Therefore, the sampling distribution has a proportion of 0.75 and a standard error of 0.0204.
Part 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)?
The probability that the sample proportion will be within 0.04 of the population proportion is given by:
[tex]P(p\ \textless \ 0.75\pm0.04)=2P\left(z\ \textless \ \frac{0.04}{0.0204} \right)-1 \\ \\ 2P(z\ \textless \ 1.961)-1=2(0.97505)-1=1.9501-1 \\ \\ =\bold{0.9501}[/tex]
Therefore, the probability that the sample proportion will be within .04 of the population proportion is 0.9501.
Part C:
Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).
Population proportion = 0.75
sample size. n = 200
[tex]Standard\ error=\sqrt{\frac{p(1-p)}{n}} \\ \\ =\sqrt{\frac{(0.75)(0.25)}{200}} = 0.0306 [/tex]
Therefore, the sampling distribution has a proportion of 0.75 and a standard error of 0.0306.
Part 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)?
The probability that the sample proportion will be within 0.04 of the population proportion is given by:
[tex]P(p\ \textless \ 0.75\pm0.04)=2P\left(z\ \textless \ \frac{0.04}{0.0306} \right)-1 \\ \\ 2P(z\ \textless \ 1.306)-1=2(0.90429)-1=1.80858-1 \\ \\ =\bold{0.8086}[/tex]
Therefore, the probability that the sample proportion will be within .04 of the population proportion is 0.8086.
Part a) :
Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).
Population proportion = 0.75
sample size. n = 450
[tex]Standard\ error=\sqrt{\frac{p(1-p)}{n}} \\ \\ =\sqrt{\frac{(0.75)(0.25)}{450}} = 0.0204 [/tex]
Therefore, the sampling distribution has a proportion of 0.75 and a standard error of 0.0204.
Part 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)?
The probability that the sample proportion will be within 0.04 of the population proportion is given by:
[tex]P(p\ \textless \ 0.75\pm0.04)=2P\left(z\ \textless \ \frac{0.04}{0.0204} \right)-1 \\ \\ 2P(z\ \textless \ 1.961)-1=2(0.97505)-1=1.9501-1 \\ \\ =\bold{0.9501}[/tex]
Therefore, the probability that the sample proportion will be within .04 of the population proportion is 0.9501.
Part C:
Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).
Population proportion = 0.75
sample size. n = 200
[tex]Standard\ error=\sqrt{\frac{p(1-p)}{n}} \\ \\ =\sqrt{\frac{(0.75)(0.25)}{200}} = 0.0306 [/tex]
Therefore, the sampling distribution has a proportion of 0.75 and a standard error of 0.0306.
Part 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)?
The probability that the sample proportion will be within 0.04 of the population proportion is given by:
[tex]P(p\ \textless \ 0.75\pm0.04)=2P\left(z\ \textless \ \frac{0.04}{0.0306} \right)-1 \\ \\ 2P(z\ \textless \ 1.306)-1=2(0.90429)-1=1.80858-1 \\ \\ =\bold{0.8086}[/tex]
Therefore, the probability that the sample proportion will be within .04 of the population proportion is 0.8086.
