Answer:
[tex]n=(\frac{1.960(840)}{150})^2 =120.47 \approx 121[/tex]
So the answer for this case would be n=12 rounded up to the next integer
Explanation:
[tex]\bar X=2500[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=840 represent the sample standard deviation
n represent the sample size
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 =150 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, the significance level if 5% and the critical value would be [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(840)}{150})^2 =120.47 \approx 121[/tex]
So the answer for this case would be n=12 rounded up to the next integer
