Question

A study of a local high school tried to determine the mean amount of money that eachstudent had saved. The study surveyed a random sample of 74 students in the highschool and found a mean savings of 2600 dollars with a standard deviation of 1400dollars. At the 95% confidence level, find the margin of error for the mean, roundingto the nearest whole number. (Do not write +).

Answer 1

To calculate the margin of error, we use the formula:

`M_(\gamma)=z_(\gamma)\sqrt[]{(\sigma^2)/(n)}`

So, we got n, the number of students, sigma, the standard deviation, and gamma, the confidence level. z for 95% is 1.64, so:

`M_{95\text{ \%}}=1.64\sqrt[]{(1400^2)/(74)}=1.64\frac{1400}{\sqrt[]{74}}=266.9\approx267`

So, the margin of error goes from mean - 267 to mean + 267: