Respuesta :
Answer:
a) The large sample size 'n' = 1320.59
b) If the standard deviation and the margin of error are both doubled also the sample size is not changed.
Step-by-step explanation:
Explanation:-
a)
Given data the standard deviation of the population
σ = 70.5
Given the margin error = 5 units
We know that the estimate of the population mean is defined by
that is margin error = [tex]\frac{z_{\alpha } S.D }{\sqrt{n} }[/tex]
[tex]M.E = \frac{z_{\alpha } S.D }{\sqrt{n} }[/tex]
cross multiplication , we get
[tex]M.E (\sqrt{n} ) = z_{\alpha } S.D[/tex]
[tex]\sqrt{n} = \frac{z_{\alpha } S.D }{M.E }[/tex]
[tex]\sqrt{n} = \frac{2.578 X 70.5}{5} }[/tex]
√n = 36.34
squaring on both sides , we get
n = 1320.59
b) The margin error of the mean
[tex]\sqrt{n} = \frac{z_{\alpha } S.D }{M.E }[/tex]
the standard deviation and the margin of error are both doubled
√n = zₓ2σ/2M.E
√n = 36.34
squaring on both sides , we get
n = 1320.59
If the standard deviation and the margin of error are both doubled also the sample size is not changed.
