Answer: It is true that required total sample size is less than the pilot sample already taken.
Step-by-step explanation:
Since we have given that
N = 30
Margin of error = E = 30
Standard deviation = 60
At 95% level of significance,
z = 1.96
So, the value of sample would be
[tex]n=(\dfrac{z.\sigma}{E})^2\\\\n=(\dfrac{1.96\times 60}{30})^2\\\\n=(1.96\times 2)^2\\\\n=3.92^2\\\\n=15.366[/tex]
So, it is true that required total sample size is less than the pilot sample already taken.