What is the margin of error for the 95% confidence interval of (0.50, 0.66) of coworkers who went on a vacation last year away from home for at least a week

Margin of Error for the given CI is 0.08

Step-by-step explanation:

Given the data in the question;

it is a symmetric class interval hence, sample proportion p = (0.50 + 0.66) / 2

p = 1.16 / 2

sample proportion p = 0.58

Margin of Error E will be;

E = 1/2 × length of Confidence interval

Margin of Error E = 1/2 × ( 0.66 - 0.5)

Margin of Error E = 1/2 × 0.16

Margin of Error E = 0.08

Therefore; Margin of Error for the given CI is 0.08

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