the local independent party wants to poll registered voters to see the proportion who would consider voting for an independent candidate. how many voters they need to survey to get a 97% confidence level that the true population proportion is within a 0.03 margin of error. how many people do they need to survey to achieve the desired margin of error?

Answer: 1308Step-by-step explanation:Given : Level of confidence = 0.97Significance level : [tex]\alpha=1-0.97=0.03[/tex]Critical value : [tex]z_{\alpha/2}=2.17[/tex]Margin of error : [tex]E=0.03[/tex]If prior proportion of population is unknown , then the formula to find the population proportion is given by :-[tex]n=0.25(\dfrac{z_{\alpha/2}}{E})^2[/tex] [tex]\Rightarrow n=0.25(\dfrac{2.17}{0.03})^2=1308.02777778\approx1308[/tex]Hence, the minimum sample size needed =1308