Answer:
[tex]n=(\frac{1.960(0.18)}{0.03})^2 =138.30[/tex]
So the answer for this case would be n=139 rounded up to the nearest integer
Step-by-step explanation:
For this case we have the following info given:
[tex]\bar X = 5.5[/tex] the sample mean
[tex]s = 0.18[/tex] the standard deviation
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 =0.03 and we are interested in order to find the value of n, if we solve n from equation (b) we got:
[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex] (b)
The critical value for 95% of confidence interval now can be founded using the normal distribution and if we look in the normal standard distirbution we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(0.18)}{0.03})^2 =138.30[/tex]
So the answer for this case would be n=139 rounded up to the nearest integer
