Question

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

Answer 1

8.2

Step-by-step explanation

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

`M\text{ = z}*\sqrt[]{(s^2)/(n)}`

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

s is the standard deviation = 40 pounds

n is the sample size = 92

Substitute

`M\text{ = }1.96*\sqrt[]{(40^2)/(92)}`

SOlve the resulting expression

`\begin{gathered} M\text{ = 1.9}6*\sqrt[]{(1600)/(92)} \\ M\text{ = 1.96 }*\sqrt[]{17.391} \\ M\text{ = 1.96}*4.17 \\ M\text{ = }8.174 \\ M\text{ }\approx8.2 \end{gathered}`

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