Answer:
[tex]n=(\frac{1.440(2.3)}{0.12})^2 =761.76 \approx 762[/tex]
So the answer for this case would be n=762 rounded up to the nearest integer
Step-by-step explanation:
Information given
[tex]\bar X = 17[/tex] represent the mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma= \sqrt{5.29}= 2.3[/tex] represent the standard deviation
Solution to the problem
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 =0.12 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 confidence level is 85%, the significance level would be [tex] \alpha=1-0.85 = 0.15[/tex] and [tex]\alpha/2 =0.075[/tex] the critical value for this case would be [tex]z_{\alpha/2}=1.440[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.440(2.3)}{0.12})^2 =761.76 \approx 762[/tex]
So the answer for this case would be n=762 rounded up to the nearest integer
