Question

In exploring possible sites for a convenience store in a large neighbourhood 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? If not, what should be the minimum sample size?

Answer 1

Answer:

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

And rounded up we have that n=97

Based on this with a sample of 100 we have a size large enough to satisfy the condition that the margin of error would be 0.1

Step-by-step explanation:

Notation and definitions

$X$ number of ratepayers in favor of the proposal

$n$ random sample taken  (variable of interest)

$\hat p$ estimated proportion of ratepayers in favor of the proposal

$p$ true population proportion of ratepayers in favor of the proposal

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

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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 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 formula:  

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

And on this case we have that $ME =\pm 0.1$ 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)  

We can use as estimate $\hat p =0.5$ since we don't have prior info for this. And replacing into equation (b) the values from part a we got:

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

And rounded up we have that n=97

Based on this with a sample of 100 we have a size large enough to satisfy the condition that the margin of error would be 0.1