Question

A researcher wishes to​ estimate, with 99​% ​confidence, the population proportion of adults who think Congress is doing a good or excellent job. Her estimate must be accurate within 5​% of the true proportion. ​(a) No preliminary estimate is available. Find the minimum sample size needed. ​(b) Find the minimum sample size​ needed, using a prior study that found that 40​% of the respondents said they think Congress is doing a good or excellent job. ​(c) Compare the results from parts​ (a) and​ (b).

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.58})^2}=665.64$  

And rounded up we have that n=666

b) $n=\frac{0.4(1-0.4)}{(\frac{0.05}{2.58})^2}=639.01$  

And rounded up we have that n=640

c) As we can se the change is just about 666-640 = 26 people, so the estimation of $\hat p =0.4$ reduce the number of sample size compared for the case when we don't have prior estimation of the population proportion

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".  

The population proportion have the following distribution  

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

Part a

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:  

$z_{\alpha/2}=\pm 2.58$  

$ME =0.05$ represent the margin of error desired

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)  

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

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.58})^2}=665.64$  

And rounded up we have that n=666

Part b

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{2.58})^2}=639.01$  

And rounded up we have that n=640

Part c

As we can se the change is just about 666-640 = 26 people, so the estimation of $\hat p =0.4$ reduce the number of sample size compared for the case when we don't have prior estimation of the population proportion