Respuesta :
Answer:
[tex]n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38[/tex]
So then the minimum sample to ensure the condition given is n= 38
Step-by-step explanation:
Notation
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=12[/tex] represent the population standard deviation
n represent the sample size
ME = 4 the margin of error desired
Solution to the problem
When we create a confidence interval for the mean 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 =4 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 96% of confidence interval now can be founded using the normal distribution. The significance is [tex]\alpha=1-0.96 =0.04[/tex]. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got [tex]z_{\alpha/2}=2.054[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38[/tex]
So then the minimum sample to ensure the condition given is n= 38
