Question

A private opinion poll is conducted for a politician to determine what proportion of the population favors adding more national parks. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.64})^2}=268.96$  

And rounded up we have that n=269

Step-by-step explanation:

Previous cocepts

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 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

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 $ME =\pm 0.05$ 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 a prior estimation for the true proportion we can use $\hat p =0.5$

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

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.64})^2}=268.96$  

And rounded up we have that n=269

Answer 2

Answer:

We need a sample size of at least 271

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 is:

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

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%

We need a sample size of at least n, in which n is found when $M = 0.05$

We do not know the true proportion, so we use $\pi = 0.5$, which is when we are going to need to use the largest sample size.

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

$0.05 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.05\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.05}$

$(\sqrt{n})^{2} = (\frac{1.645*0.5}{0.05})^{2}$

$n = 270.6$

Rounding up

We need a sample size of at least 271