Question

52. In exploring possible sites for a convenience store in a large neighborhood, the retail chain wants to know the proportion of ratepayers in favor of the proposal. If the estimate is required to be within 0.1 of the true proportion, would a random sample of size n=100 from the council records be sufficient for a 95% confidence interval of this precision? A. There is not enough information to answer this question B. No, because n=100 , the sample size, is too small C. No, because the length of the confidence interval would be greater than 0.1. D. Yes.

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$  

Then the minimum sample size in order to satisfy the condition of 0.1 for the margin of error is 97 and since the sample used is n =100 we can conclude that is sufficient and the best answer would be:

D. Yes.

Step-by-step explanation:

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. We know that we require a 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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

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

We want a margin of error of $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

Since we don't have prior info for the population proportion we can use as estimator the value of 0.5. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$  

Then the minimum sample size in order to satisfy the condition of 0.1 for the margin of error is 97 and since the sample used is n =100 we can conclude that is sufficient and the best answer would be:

D. Yes.