Respuesta :
Using the z-distribution, as we have the standard deviation for the population, it is found that 215 students must be sampled.
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- z is the critical value.
- n is the sample size.
- [tex]\sigma[/tex] is the standard deviation for the sample.
The margin of error is:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
The margin of error is of M = 0.25, with a population standard deviation of [tex]\sigma = 1.87[/tex], hence we solve for n to find the minimum sample size.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.25 = 1.96\frac{1.87}{\sqrt{n}}[/tex]
[tex]0.25\sqrt{n} = 1.96(1.87)[/tex]
[tex]\sqrt{n} = \frac{1.96(1.87)}{0.25}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(1.87)}{0.25}\right)^2[/tex]
[tex]n = 214.9[/tex]
Rounding up, 215 students must be sampled.
More can be learned about the z-distribution at https://brainly.com/question/25890103
