Best Answer:
ANSWER: Margin of Sampling Error is d. 4% which is one-half of the 95% Resulting Confidence Interval for 'true mean'= [0.63, 0.71]

Why???

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

n: Number of samples = 500

k: Number of Successes = 335

p: Sample Proportion [335/500] = 0.67

significant digits: 2

Confidence Level = 95

"Look-up" Table 'z-critical value' = 1.960

This 'z-critical value' is a Look-up from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretive procedure to Look-Up the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 95%, there is a LEFT 'area' OUTSIDE. And due to symmetry there is a RIGHT 'area' OUTSIDE. Using a Look-up from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. For STANDARDIZED VARIABLE z = 1.96, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1 - 95/100) = 0.03 by a Look-up in the Table for Standard Normal Distribution.

95% Resulting Confidence Interval for 'true mean': p ± ('z critical value') * SQRT[p * (1 - p)/n]

= 0.67 ± 1.96 * SQRT[0.67 * (1 - 0.67)/500] = [0.63, 0.71]

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