Answer:
a) 231
The sample to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00
n = 231
Step-by-step explanation:
Explanation:-
Given data the population standard deviation is known σ =$15.50
Given the margin of error ±$2.00
we know that  95 percent confidence interval of margin of error is determined by Â
[tex]M.E = \frac{Z_{\alpha }S.D }{\sqrt{n} }[/tex]
cross multiplication √n we get ,
[tex]\sqrt{n} = \frac{Z_{\alpha }S.D }{M.E }[/tex]
squaring on both sides, we get
[tex](\sqrt{n} )^2 = (\frac{Z_{\alpha }S.D }{M.E })^2[/tex]
[tex]n = (\frac{Z_{\alpha }S.D }{M.E })^2[/tex]
the tabulated z- value = 1.96 at 95% of level of significance.
[tex]n = (\frac{1.96(15.50) }{2 })^2[/tex]
n = 230.7≅231
Conclusion:-
The sample to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00
n = 231
