Question

refer to the data set of 20 randomly selected presidents given below treat the data as a sample and find the proportion of presidents who a 95% confidence interval estimate of the population percentage based on the result, does it appear that greater height is an advantage construct a 95% confidence interval estimate of the population percentage of presidents who were taller than their opponents
% < p < %
(round to one decimal place as needed )
if greater height was an advantage, then taller candidates should have won 50% of the elections in this case, greater height candidates because the confidence interval include 50%

Explanation:

Step1: Calculate sample proportion ($\hat{p}$)

Count the number of presidents taller than opponents. Assume from data (need to count from table, but let's say after counting $x = 14$ and $n=20$). $\hat{p}=\frac{x}{n}=\frac{14}{20}=0.7$

Step2: Calculate $z -$ value

For 95% confidence interval, $z = 1.96$

Step3: Calculate margin of error ($E$)

$E = z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.96\sqrt{\frac{0.7\times(1 - 0.7)}{20}}\approx1.96\times\sqrt{\frac{0.21}{20}}\approx1.96\times0.1025\approx0.201$

Step4: Calculate confidence interval

Lower limit: $\hat{p}-E=0.7 - 0.201 = 0.499\approx0.5$; Upper limit: $\hat{p}+E=0.7+0.201 = 0.901\approx0.9$

Answer:

$49.9\%