A researcher wishes to estimate, with 99% 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 3% 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 26% of the respondents said they think their president can control the price of gasoline. c) Compare the results from parts (a) and (b).

c) For this case we can see that if we have a prior estimate the minimum sample size required for the margin of error desired would be less as we can see in part b we reduce the sample size compared to the part a

Step-by-step explanation:

Part a

The critical value for a confidence level of 99% is for this case [tex]z_{\alpha/2} =2.58[/tex]

The margin of error for the proportion interval is given by this formula:

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)

And on this case we have that [tex]ME =\pm 0.03[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

Since we don't know any estimate for the true proportion we can use [tex]\hat p =0.5[/tex] as a godd estimator. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{2.58})^2}=1849[/tex]

And rounded up we have that n=1849

Part b

For this case we have a prior estimate [tex]\hat p =0.26[/tex] and replacing we got:

[tex]n=\frac{0.26(1-0.26)}{(\frac{0.03}{2.58})^2}=1422.99[/tex]

And rounded up we have that n=1423

Part c

For this case we can see that if we have a prior estimate the minimum sample size required for the margin of error desired would be less as we can see in part b we reduce the sample size compared to the part a