What if you used 11008 instead of 10591?

Question

Same question but with a 11008 as the value for Σ(Y-Y)2

Answer 1

I interpret the question as the variance of 8 Y values (Y value minus the mean of Y=314.375, where the value of Y is an amount in thousands of dollars (symbolised by $K), so the mean is $314,375). The variance is therefore:

(Y1-mean)^2+(Y2-mean)^2+...+(Y8-mean)^2=11,008, where Y1 to Y8 make up the statistical set. Divide this number by n, giving 11008/8, and take the square root to get the standard error (standard deviation): 37.094. This means that the standard error on the mean is +37.094 so:

277.281<Y<351.469, where Y is in $K, gives range of possible values for Y, where the mean sits in the exact middle of the range.