Question 866660
Use this <a href="http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg">table</a> to find the critical value to be z = 1.645 (look in the row that starts with {{{infinity}}}, look above the 90%)



So we know the following info



ME = 2 (given margin of error)
s = 30 (given standard deviation)
z = 1.645 (found using the table given above)



Margin of Error (ME)



ME = z*s/sqrt(n)



2 = 1.645*30/sqrt(n)



2 = 49.35/sqrt(n)



2*sqrt(n) = 49.35



sqrt(n) = 49.35/2



sqrt(n) = 24.675



n = (24.675)^2



n = 608.855625



n = 609  <font color="blue">round UP to the nearest whole number</font>



Why do we round up? Well if we round down to 608, then we will have a margin of error (ME) larger than 2. We round up to clear this hurdle (since larger n leads to smaller ME). This happens whenever you get a decimal result for these type of problems.



So you need a sample size of at least <font color="red">609</font> students to make sure the margin of error (ME) is 2 or less.