ANSWER:

903

Good job! We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 5.86, therefore the margin of error (m) = 5.86/ 2 = 2.93. The sample size we need in order to obtain this is: 903:

n=(2×442.93)2$n={\left(\frac{2\times 44}{2.93}\right)}^2$

n=(882.93)2$n={\left(\frac{88}{2.93}\right)}^2$

n=902.05=(round up) 903