Question

We wish to estimate the population mean of a variable that has standard deviation 70.5. We want to estimate it with an error no greater than 5 units with probability 0.99. How big a sample should we take from the population? What happens if the standard deviation and the margin of error are both doubled?

Answer 1

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 = $\frac{z_{\alpha } S.D }{\sqrt{n} }$

            $M.E = \frac{z_{\alpha } S.D }{\sqrt{n} }$

cross multiplication , we get

$M.E (\sqrt{n} ) = z_{\alpha } S.D$

$\sqrt{n} = \frac{z_{\alpha } S.D }{M.E }$

$\sqrt{n} = \frac{2.578 X 70.5}{5} }$

√n = 36.34

squaring on both sides , we get

n = 1320.59

b) The margin error of the mean

   $\sqrt{n} = \frac{z_{\alpha } S.D }{M.E }$

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.