An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 96% confidence interval for the proportion who are not satisfied.(a) Past studies suggest that this proportion will be about 0.19. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.02.(You will need a critical value accurate to at least 4 decimal places.)Sample size:(b) Using the sample size above, when the sample is actually contacted, 21% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?MoE:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 96% of confidence, our significance level would be given by [tex]\alpha=1-0.96=0.04[/tex] and [tex]\alpha/2 =0.02[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.0538, z_{1-\alpha/2}=2.0538[/tex]

The margin of error for the proportion interval is given by this formula:

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)

And on this case we have that [tex]ME =\pm 0.02[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.19(1-0.19)}{(\frac{0.02}{2.0538})^2}=1622.912[/tex]

And rounded up we have that n=1623

Part b

The margin of error is given by:

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

So then if w replace the value of n obtained from part a we got:

[tex] ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208[/tex]