Question

Based on sample results, a 90% confidence interval for the mean servings of fruit per day consumed by grade school children is (0.21, 2.45). What is the margin of error?

Answer 1

Answer: 1.12

Step-by-step explanation:

The confidence interval for the mean $\mu$ and margin of error E is given by :-

$(\mu-E,\ \mu+E)$                                             --------------(1)

The given confidence interval : (0.21, 2.45)              -------------(2)

From (1) and (2), we  have

$\mu-E=0.21-------(3)\\\\\mu+E=2.45--------(4)$

Subtract equation (3) from (4), we get

$2E=2.45-0.21\\\\\Rightarrow\ 2E=2.24\\\\\Rightarrow\ E=\dfrac{2.24}{2}=1.12$

Hence, the margin of error is 1.12 .