Respuesta :
Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.
What is a z-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 population.
The margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that the parameters are given as follows:
[tex]M = 3, z = 1.645, \sigma = 16[/tex].
Hence, solving for n, we find the sample size.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 1.645\frac{16}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 1.645 \times 16[/tex]
[tex]\sqrt{n} = \frac{1.645 \times 16}{3}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2[/tex]
[tex]n = 76.97[/tex]
Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.
More can be learned about the z-distribution at https://brainly.com/question/25890103
