Respuesta :
Answer:
[tex]n=(\frac{1.960(2.3)}{0.5})^2 =81.28 \approx 82[/tex]
So the answer for this case would be n=82 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=6[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=2.3 represent the sample standard deviation
n represent the sample size
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 =0.5 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} s}{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}=1.960[/tex], replacing into formula (3) we got:
[tex]n=(\frac{1.960(2.3)}{0.5})^2 =81.28 \approx 82[/tex]
So the answer for this case would be n=82 rounded up to the nearest integer
Using the z-distribution, as we have the standard deviation for the population, it is found that a sample size of 82 should be used in the study.
What is the margin of error for a z-distribution confidence interval?
It is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- z is the critical value.
- [tex]\sigma[/tex] is the population standard deviation.
- n is the sample size.
In this problem, the parameters are:
[tex]z = 1.96, \sigma = 2.3, M = 0.5[/tex]
Hence:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96\frac{2.3}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 1.96(2.3)[/tex]
[tex]\sqrt{n} = \frac{1.96(2.3)}{0.5}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(2.3)}{0.5}\right)^2[/tex]
[tex]n = 81.2[/tex]
Rounding up, a sample size of 82 should be used in the study.
More can be learned about the z-distribution at https://brainly.com/question/25256953
