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May 09, 11:58:07 PMA study was commissioned to find the mean weight of the residents incertain town. The study examined a random sample of 92 residentsand found the mean weight to be 181 pounds with a standard deviationof 40 pounds. At the 95% confidence level, find the margin of error forthe mean, rounding to the nearest tenth. (Do not write =).Submit AnswerAnswer:attempt 1 out of 2

Sagot :

Answer:

8.2

Explanation

Tyhe formula for calculating the margin of error is expressed as;

[tex]M\text{ = z}\times\sqrt[]{\frac{s^2}{n}}[/tex]

z is the z-score at 95% confidence interval =

s is the standard deviation = 40 pounds

n is the sample size = 92

Substitute

[tex]M\text{ = }1.96\times\sqrt[]{\frac{40^2}{92}}[/tex]

SOlve the resulting expression

[tex]\begin{gathered} M\text{ = 1.9}6\times\sqrt[]{\frac{1600}{92}} \\ M\text{ = 1.96 }\times\sqrt[]{17.391} \\ M\text{ = 1.96}\times4.17 \\ M\text{ = }8.174 \\ M\text{ }\approx8.2 \end{gathered}[/tex]

Hence the the margin of error for the mean, rounded to the nearest tenth is 8.2

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