Margin of Error

Question

Find the sample size necessary for a 90% confidence level with maximal error of estimate E=0.43 for the mean price per 100 pounds of watermelon. (Enter a number and round up to the nearest whole number.) xbar =$6.88 per 100 pounds of watermelon. Assume that σ =$1.98 per 100 pounds.

_________= farming regions

I am getting stuck on basic algebra... zσ/√n is the equation I am using. Could you please show me how to solve this problem to get n singled out? Thank you.

Answer 1

e = z*s/sqrt(n) where e is the error margin and s is the standard deviation

sqrt(n)*e = z * s

sqrt(n) = z * s / e

n = (z * s/e)^2

So

n = (1.645 * 1.98/0.43)^2 = 58