Use the following information to answer the question. In a recent poll of 1100 randomly selected home delivery truck drivers,26% said they had encountered an aggressive dog on the job at least once.
-What is the margin of error, using a 95% confidence level, for estimating the true populationproportion of home delivery truck drivers who have encountered an aggressive dog on the job atleast once? (Round to the nearest thousandth)
A) 0.013
B) 0.053
C) 0.026
D) 0.004

Question

Quizplus · Accepted Answer

The margin of error (ME) for a proportion is calculated using the formula: ME = Z * sqrt[(p(1-p))/n], where Z is the Z-score for a 95% confidence level (approximately 1.96), p is the sample proportion (0.26), and n is the sample size (1100). Calculating this gives approximately 0.026.

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