Question

A random sample of 120 students has a test score average with a standard deviation of 9.2. Find the margin of error if c= 0.98.

Answer 1

Solution :

Given that,


Population standard deviation =    = 9.2

Sample size = n =120

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z/2 * ( /n)

= 2.326* ( 9.2 /  120 )

E=1.9535