Answer :
Answer:
[tex]n=(\frac{1.960(1)}{0.0001})^2 =384160000[/tex]
So the answer for this case would be n=384160000 rounded up to the nearest integer
Step-by-step explanation:
We know the following info:
[tex] ME = 0.0001[/tex] represent the margin of error desired
[tex] \sigma= 1[/tex] we assume that the population deviation is this value
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.0001 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
The critical value for 95% of confidence interval now can be founded using the normal distribution. If we use the normal standard distribution or excel we got: [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(1)}{0.0001})^2 =384160000[/tex]
So the answer for this case would be n=384160000 rounded up to the nearest integer