Respuesta :
Answer:
n=465 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=3.3[/tex] represent the population standard deviation
Confidence =0.95 or 95%
Me= 0.3, represent the margion of error
n represent the sample size (variable of interest)
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 =0.3 mm 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} \sigma}{ME})^2[/tex] (b)
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}=1.960[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(3.3)}{0.3})^2 =464.8336 \approx 465[/tex]
So the answer for this case would be n=465 rounded up to the nearest integer
