Respuesta :
Using the z-distribution, it is found that 217 sample measurements should be taken at each site.
The margin of error of a z-confidence interval 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.
The first step is finding the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.
In this problem, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], which means that it is z = 1.96.
Estimate of the standard deviation of 1.5, thus, [tex]\sigma = 1.5[/tex].
We want the sample for a margin of error of 0.2ºC, thus, we have to solve for n when [tex]M = 0.2[/tex].
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.2 = 1.96\frac{1.5}{\sqrt{n}}[/tex]
[tex]0.2\sqrt{n} = 1.96(1.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(1.5)}{0.2}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96(1.5)}{0.2})^2[/tex]
[tex]n = 216.1[/tex]
Rounding up:
217 sample measurements should be taken at each site.
A similar problem is given at https://brainly.com/question/17039768
