Question

You wish to compute the 99% confidence interval for the population proportion. How large a sample should you draw to ensure that the sample proportion does not deviate from the population proportion by more than 0.13?

Answer 1

Answer:

We need a sample of at least 99

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error of the interval is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

How large a sample should you draw to ensure that the sample proportion does not deviate from the population proportion by more than 0.13?

We don't know the proportion, so we use $\pi = 0.5$. So we need a sample of size at least n, in which n is found when $M = 0.13$. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.13 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.13\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.13}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.13})^{2}$

$n = 98.08$

Rouding up

We need a sample of at least 99

Answer 2

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47$  

And rounded up we have that n=99

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".  

Solution to the problem

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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 $ME =\pm 0.13$ 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 assume the value for $\hat p =0.5$ since we don't have previous info. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47$  

And rounded up we have that n=99