Question

determine the sample sieze erequired to estimate teh mean score on a standardizes test within 4 [pints of the trrue mean with 98% confidence

Answer 1

Answer: Minimum sample size = 110

Explanation:

Formula to find minimum sample size :

`n=((z* s)/(E))^2` , where z = critical z-value for confidence level, s= sample standard deviation , E = margin of error

Given: E= 4 , s= 18

Critical z-value for 98% confidence = 2.327

So,
`n=((2.327*18)/(4))^2`

`=((41.886)/(4))^2\\\\=(10.4715)^2=109.65231225\approx110`

Minimum sample size = 110