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A survey was given to a random sample of the residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. The survey reported a confidence interval that between
57.5% and
60.5% of the residents supports the plan. What is the margin of error on the survey? Do not write
+- on the margin of error.
Answer:_______

A survey was given to a random sample of the residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. The survey reported a confidence interval that between $57.5 \%$ and $60.5 \%$ of the residents supports the plan. What is the margin of error on the survey? Do not write $\pm$ on the margin of error. $\newline$ Answer:________

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Interpret confidence intervals for population means

Full solution

Q. A survey was given to a random sample of the residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. The survey reported a confidence interval that between

57.5 \%

and

60.5 \%

of the residents supports the plan. What is the margin of error on the survey? Do not write

\pm

on the margin of error.

\newline

Answer:________

Calculate Confidence Interval Range: To find the margin of error, we need to calculate the range of the confidence interval and then divide it by $2$ . The confidence interval is from $57.5\%$ to $60.5\%$ .
Calculate Margin of Error: Calculate the range of the confidence interval. $\newline$ Range = Upper limit of confidence interval - Lower limit of confidence interval $\newline$ Range = $60.5\% - 57.5\%$ $\newline$ Range = $3\%$
Calculate Margin of Error: Calculate the range of the confidence interval. $\newline$ Range = Upper limit of confidence interval - Lower limit of confidence interval $\newline$ Range = $60.5\% - 57.5\%$ $\newline$ Range = $3\%$ Divide the range by $2$ to find the margin of error. $\newline$ Margin of error = Range / $2$ $\newline$ Margin of error = $3\% / 2$ $\newline$ Margin of error = $1.5\%$

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