Respuesta :
Answer: 56
Step-by-step explanation:
The formula to find the required sample size :
[tex]n=(\dfrac{\sigma\times z^*}{E})^2[/tex]
, where n= sample size.
[tex]\sigma[/tex] = Population standard deviation from previous studies.
As per given , we have
[tex]\sigma=5.8[/tex]
E= 2
The critical value of z for 99% confidence interval from z-table is 2.576.
Put theses values in the formula , we will get
[tex]n=(\dfrac{(5.8)(2.576)}{2})^2[/tex]
[tex]n=(7.4704)^2=55.80687616\approx56[/tex]
Therefore , the required sample size of mice = 56
Using the z-distribution, it is found that a sample of 56 is needed.
We have the standard deviation for the population, hence, the z-distribution is used.
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 = 0.99[/tex], thus, z with a p-value of [tex]\frac{1 + 0.99}{2} = 0.995[/tex], which means that it is z = 2.575.
As stated in the problem, the population standard deviation is of [tex]\sigma = 5.8[/tex].
The sample size needed is n for which M = 2, hence:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]2 = 2.575\frac{5.8}{\sqrt{n}}[/tex]
[tex]2\sqrt{n} = 2.575(5.8)[/tex]
[tex]\sqrt{n} = \frac{2.575(5.8)}{2}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575(5.8)}{2}\right)^2[/tex]
[tex]n = 55.8[/tex]
Rounding up, a sample of 56 is needed.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/25816292
