Answer :
Answer:
(1) The planning value for the population standard deviation:
[tex]=\frac{50,000-30,000}{4}[/tex]
= 5,000
(2) Given a = 0.05, Z(0.025) = 1.96 (from standard normal table)
[tex]n = (\frac{Z\times SD}{Error}) ^{2}[/tex]
(a)
[tex]n = (\frac{1.96\times 5,000}{500}) ^{2}[/tex]
n = 384.16
(b)
[tex]n = (\frac{1.96\times 5,000}{200}) ^{2}[/tex]
n = 2,401
(c)
[tex]n = (\frac{1.96\times 5,000}{100}) ^{2}[/tex]
n = 9,604
(3) No, because the sample size of the study is too larger.