Question

The National Fire Protection Association reported that house fires in the United States result in $6.8 billion in losses annually (National Fire Protection Association, September 2015). Suppose a random sample of 50 house fires in San Antonio yields a mean loss of $18,503. If the population standard deviation is s 5 $3300, what is the margin of error for a 95% confidence interval

Answer 1

914.7

Explanation:

The Formula for Margin of Error =

z × standard deviation/√n

From the above question,

z score for 95% confidence interval = 1.96

Standard deviation = $3300

n = random sample = 50

Margin of Error = 1.96 × 3300/√50

= 1.96 × 466.69047558

= 914.71333214

Margin of Error ≈ 914.7