Answer:
The sample size to obtain the desired margin of error is 160.
Step-by-step explanation:
The Margin of Error is given as
[tex]MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}[/tex]
Rearranging this equation in terms of n gives
[tex]n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2[/tex]
Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as
[tex]n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n[/tex]
As n is given as 40 so the new sample size is given as
[tex]n_2=4n\\n_2=4*40\\n_2=160[/tex]
So the sample size to obtain the desired margin of error is 160.
