Respuesta :
Answer:
[tex]n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16[/tex]
So the answer for this case would be n=16 rounded up to the nearest integer
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=10[/tex] represent the sample standard deviation
n represent the sample size
ME=5 represent the margin of error
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (2)
And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (3)
The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (3) we got:
[tex]n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16[/tex]
So the answer for this case would be n=16 rounded up to the nearest integer
