Nikhil asked 120 randomly chosen moviegoers to watch a clip of an upcoming movie and choose the best title. The title "Everythin
g” was chosen by 36 of the moviegoers as the preferred title. To the nearest percent, with a confidence level of 90% (z*-score 1.645), what is the confidence interval for the proportion of moviegoers who preferred the title "Everything”?
The mean proportion is p = 36/120 = 0.3. The standard deviation of the proportion will be sqrt(p*(1-p)/n) = sqrt(0.3*0.7/120) = 0.0418. We multiply this by the z-score of 1.645 to get a deviation of 0.0688. Therefore, the confidence interval is (0.3 - 0.0688, 0.3 + 0.0688), which is (0.2312, 0.3688).
