If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date while keeping the margin of error at 0.05, what would the needed sample size be? Assume you have no prior knowledge of the proportion.

Answer: 271Step-by-step explanation:The formula we use to find the sample size is given by :-[tex]n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2[/tex], where [tex]z_{\alpha/2}[/tex] is the two-tailed z-value for significance level of [tex](\alpha)[/tex]p = prior estimation of the proportionE = Margin of error.If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes [tex]n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2[/tex][tex]n=0.25(\dfrac{z_{\alpha/2}}{E})^2[/tex]Given :   Margin of error : E= 0.05Confidence level = 90%Significance level [tex]\alpha=1-0.90=0.10[/tex]Using z-value table , Two-tailed z-value for significance level of [tex]0.10[/tex] [tex]z_{\alpha/2}=1.645[/tex]Then, the required sample size would be :[tex]n=0.25(\dfrac{1.645}{0.05})^2[/tex]Simplify,[tex]n=270.6025\approx271[/tex]Hence, the required minimum sample size =271