Respuesta :
Answer:
[tex]n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573[/tex]
And if we use a sample level of n =2000 the margin of error would be higher as we can see here:
[tex] ME=2.33\frac{500}{\sqrt{2000}}=26.05[/tex]
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=500[/tex] represent the population standard deviation assumed
n represent the sample size (variable of interest)
Confidence =98% or 0.98
ME = 10 represent the margin of error desired
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 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got [tex]z_{\alpha/2}=2.33[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573[/tex]
So the answer for this case would be n=13573 rounded up to the nearest integer
And if we use a sample level of n =2000 the margin of error would be higher as we can see here:
[tex] ME=2.33\frac{500}{\sqrt{2000}}=26.05[/tex]
