Answer: 1.12
Step-by-step explanation:
The confidence interval for the mean [tex]\mu[/tex] and margin of error E is given by :-
[tex](\mu-E,\ \mu+E)[/tex] --------------(1)
The given confidence interval : (0.21, 2.45) -------------(2)
From (1) and (2), we have
[tex]\mu-E=0.21-------(3)\\\\\mu+E=2.45--------(4)[/tex]
Subtract equation (3) from (4), we get
[tex]2E=2.45-0.21\\\\\Rightarrow\ 2E=2.24\\\\\Rightarrow\ E=\dfrac{2.24}{2}=1.12[/tex]
Hence, the margin of error is 1.12 .
The margin of error is 1.12.
Given
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).
The margin of error is a statistic expressing an amount of random sampling error in a survey’s results.
The confidence interval for the mean [tex]\rm \mu[/tex] and margin of error E is given by;
[tex]\rm Margin \ of \ error = \mu -E, \ and , \ \mu+E[/tex]
Then,
The mean servings of fruit per day consumed by grade school children are (0.21, 2.45).
[tex]\rm \mu-E=0.21\\\\\mu+E=2.45[/tex]
Subtracting both the equations;
[tex]\rm \mu-E-(\mu+E)=0.21-2.45\\\\\mu-E-\mu-E=-2.24\\\\-2E=-2.24\\\\E=\dfrac{-2.24}{-2}\\\\E=1.12[/tex]
Hence, the margin of error is 1.12.
To know more about the margin of error click the link given below.
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