In a sample of 110 bags, 12% are overweight. Assuming that the overweight bags occur randomly, find a 99% confidence interval for the proportion of all bags that are overweight.

Please explain and tell me what to enter on a calculator. I use a casio graphics.

## Answers (1)

ANSWER: 99% Resulting Confidence Interval for 'true mean'= [0.0402, 0.1998]

Why???

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

n: Number of samples = 110

p: Sample Proportion = 0.12

Confidence Level = 99

"Look-up" Table 'z-critical value' = 2.576

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) = 99, 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 = 2.5758, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1 - 99/100) = 0.005 by a Look-up in the Table for Standard Normal Distribution.

Or alternative; use Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical.

99% Resulting Confidence Interval for 'true mean': p +/- ('z critical value') * SQRT[p * (1 - p)/n]

= 0.12 +/- 2.5758 * SQRT[0.12 * (1 - 0.12)/110] = [0.0402, 0.1998]