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
a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of .
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 .
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 do you gain in precision by taking the larger sample in part (b)?

Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 [tex]\mu_{\hat p}= p[/tex]

The standard deviation of this sampling distribution of sample proportion is:

 [tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]

(a)

The sample selected is of size n = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

[tex]\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204[/tex]

So, the sampling distribution of sample proportion is [tex]\hat p\sim N(0.75,0.0204^{2})[/tex].

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

[tex]P(p-0.04<\hat p<p+0.04)=P(\frac{-0.04}{0.0204}<\frac{\hat p-p}{\sigma_{\hat p}}<\frac{0.04}{0.0204})[/tex]

                                          [tex]=P(-1.96<Z<196)\\=P(Z<1.96)-P(Z<-1.96)\\=0.975-0.025\\=0.95[/tex]

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size n = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

[tex]\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306[/tex]

So, the sampling distribution of sample proportion is [tex]\hat p\sim N(0.75,0.0306^{2})[/tex].

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

[tex]P(p-0.04<\hat p<p+0.04)=P(\frac{-0.04}{0.0306}<\frac{\hat p-p}{\sigma_{\hat p}}<\frac{0.04}{0.0306})[/tex]

                                          [tex]=P(-1.31<Z<196)\\=P(Z<1.31)-P(Z<-1.96)\\=0.9049-0.0951\\=0.8098\\\approx 0.81[/tex]

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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