Respuesta :
Answer:
16
Step-by-step explanation:
Standard deviation is 2
Margin error for the problem is 1
Probability 95%, that means thet the siginficance level α is 1 – p
α = 1 – 0.95 = 0.05
margin of error (ME) can be defined as follows
ME = Z(α/2) * standard deviation/ √n
Where n is the sample size
Z(0.05/2) = Z(0.025)
Using a z table Z = 1.96
Now, replacing in the equation and find n
1 = 1.96 * 2/ √n
1 = 3.92 / √n
√n = 3.92
n = 3.92^2
n = 15.36 = 16
Using the z-distribution, it is found that 16 students should be sampled.
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 2, thus, [tex]\sigma = 2[/tex].
We want the sample for a margin of error of 1, thus, we have to solve for n when [tex]M = 1[/tex].
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.96\frac{2}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.96(2)[/tex]
[tex](\sqrt{n})^2 = [1.96(2)]^2[/tex]
[tex]n = 15.4[/tex]
Rounding up:
16 students should be sampled.
A similar problem is given at https://brainly.com/question/14936818
