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

Step-by-step explanation:

Formula to find minimum sample size :

$n=(\dfrac{z\times s}{E})^2$ , 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,  $n=(\dfrac{2.327\times18}{4})^2$

$=(\dfrac{41.886}{4})^2\\\\=(10.4715)^2=109.65231225\approx110$

Minimum sample size = 110