An environmental agency worries that many cars may be violating clean air emissions standards. The agency hopes to check a sample of vehicles in order to estimate that percentage with a margin of error of 55​% and 9090​% confidence. To gauge the size of the​ problem, the agency first picks 7070 cars and finds 1414 with faulty emissions systems. How many should be sampled for a full​ investigation?

Respuesta :

Answer:

[tex]n=\frac{0.2(1-0.2)}{(\frac{0.05}{1.64})^2}=172.13[/tex]  

And rounded up we have that n=173

Step-by-step explanation:

Previous concepts

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".  

[tex]p[/tex] represent the real population proportion of interest

[tex]\hat p[/tex] represent the estimated proportion for the sample

n is the sample size required (variable of interest)

[tex]z[/tex] represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

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

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 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.10[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:  

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/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 0.05[/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)

We can assume that the estimates proportion is [tex]\hat p=\frac{14}{70}=0.2[/tex]. And replacing into equation (b) the values from part a we got:  

[tex]n=\frac{0.2(1-0.2)}{(\frac{0.05}{1.64})^2}=172.13[/tex]  

And rounded up we have that n=173