Question

A researcher wishes to​ estimate, with 95​% ​confidence, the population proportion of adults who think the president of their country can control the price of gasoline. 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 their president can control the price of gasoline. ​
(c) Compare the results from parts​ (a) and​ (b).

Answer 1

Answer:

a) n=385

b) n=369

c) We have a difference of 16 between the results obtained for prior estimation of the population proportion and without prior estimation.

Step-by-step explanation:

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 :

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

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)  

Part a

First we need to find the critical value, on this case the confidence is 95%, and the significance level would be $\alpha=1-0.95=0.05$, and $\alpha/2 =0.025$, and we can find the cirtical value on the normal standard distribution that accumulates 0.025 of the area on each tail on this case $z_{\alpha/2}=1.96$

Since we don't have prior estimate for the proportion of interest we can assume that $\hat p=0.05$.

And replacing into equation (b) the values given we have:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$  

And rounded up we have that n=385

Part b

On this case the critical value not changes since is the same confidence, but we have a prior estimate for the proportion $\aht p=0.40$ . And replacing into equation (b) the values given we have:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$  

And rounded up we have that n=369

Part c

We have a difference of 16 between the results obtained for prior estimation of the population proportion and without prior estimation.