Respuesta :
Answer:
[tex]n=(\frac{2.58(0.15)}{0.025})^2 =239.63 \approx 240[/tex]
So the answer for this case would be n=240 rounded up to the nearest integer
Step-by-step explanation:
We know the following info:
[tex]\sigma =0.15[/tex] represent the population deviation
[tex] ME= 0.025[/tex] the margin of error desired
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.025 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} s}{ME})^2[/tex] (b)
The critical value for 99% of confidence interval now can be founded using the normal distribution. And we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.58(0.15)}{0.025})^2 =239.63 \approx 240[/tex]
So the answer for this case would be n=240 rounded up to the nearest integer
