Respuesta :
Answer:
[tex]n=(\frac{2.58(17)}{10})^2 =19.24 \approx 20[/tex]
So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma =17[/tex] represent the population standard deviation
n represent the sample size
Solution to the problem
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 =10 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 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.58(17)}{10})^2 =19.24 \approx 20[/tex]
So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence
