Question

A sports reporter wishes to establish the mean satisfaction level (on a scale from 0 to 10) to within a margin of error of 0.7. If it is known from previous studies that the standard deviation in the level of fan satisfaction is 3.4, what sample size will be needed to achieve the desired precision with 98% confidence?

Standard Normal Distribution Table

Round up to the next whole number

Answer 1

Solution :

standard deviation =    = 3.4

margin of error = E = 0.7

At 98% confidence level

= 1 - 98%

= 1 - 0.98 =0.02

/2 = 0.01

Z_/2 = Z_0.01 = 2.33

Z_/2 = Z_0.025 = 1.96  

sample size = n = [Z_/2* / E] ²

n = [2.33 * 3.4 / 0.7 ]²

n = 128.07

Sample size = n = 129