Respuesta :
Answer:
[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex] (5)
Replacing into formula (5) we got:
[tex]n=(\frac{1.99(2)}{1})^2 =15.84 \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=20[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\hat \sigma = s= 2[/tex] represent the sample standard deviation
n represent the sample size
Solution to the problem
Since the Confidence is 0.9544 or 95.44%, the value of [tex]\alpha=0.0456[/tex] and [tex]\alpha/2 =0.0228[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.0228,0,1)".And we see that [tex]z_{\alpha/2}=1.999[/tex]
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (4)
And on this case we have that ME =1 and we are interested in order to find the value of n, if we solve n from equation (4) we got:
[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex] (5)
Replacing into formula (5) we got:
[tex]n=(\frac{1.99(2)}{1})^2 =15.84 \approx 16[/tex]
So the answer for this case would be n=16 rounded up to the nearest integer
