Question

You wish to estimate, with 95% confidence, the population proportion of U.S. adults who think that the president can do a lot about the price of gasoline. Your estimate must be accurate within 4% of the population 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 48% of U.S. adults think the president can do a lot about the price of gasoline. (Source: CBS News/New York Times Poll) (c) Compare the results from parts (a) and (b).

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

And rounded up we have that n=601

b) $n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

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

$n=\frac{0.48(1-0.48)}{(\frac{0.04}{1.96})^2}=599.290$  

And rounded up we have that n=600

c) If we analyze the results we see that the difference is just one 1 individual for the answer of part b compared to the part a. So then we don't have significant differences in the sample sizes estimated.

Step-by-step explanation:

1) Notation and definitions

$X$ number of U.S. adults who think that the president can do a lot about the price of gasoline

$n$ random sample taken  (variable of interest)

$\hat p$ estimated proportion of U.S. adults who think that the president can do a lot about the price of gasoline

$p$ true population proportion of U.S. adults who think that the president can do a lot about the price of gasoline

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

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

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 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)  

Since we don't have prior estimate for the proportion we can use 0.5. And on this case we have that $ME =\pm 0.04$ 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)  

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

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

And rounded up we have that n=601

Part b

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

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

$n=\frac{0.48(1-0.48)}{(\frac{0.04}{1.96})^2}=599.290$  

And rounded up we have that n=600

Part c

If we analyze the results we see that the difference is just one 1 individual for the answer of part b compared to the part a. So then we don't have significant differences in the sample sizes estimated.