Respuesta :
Using the z-distribution, as we have the standard deviation for the population, it is found that 8 stores must be analyzed.
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 an 82% confidence level, hence[tex]\alpha = 0.82[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.82}{2} = 0.91[/tex], so the critical value is z = 1.34.
The standard deviation for the population is of [tex]\sigma = 0.06[/tex], and we want a margin of error of M = 0.03, hence we solve for n to find the sample size needed.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.03 = 1.34\frac{0.06}{\sqrt{n}}[/tex]
[tex]0.03\sqrt{n} = 1.34 \times 0.06[/tex]
Simplifying by 0.03:
[tex]\sqrt{n} = 1.34 \times 2[/tex]
[tex](\sqrt{n})^2 = (1.34 \times 2)^2[/tex]
[tex]n = 7.2[/tex]
Rounding up, 8 stores must be analyzed.
More can be learned about the z-distribution at https://brainly.com/question/25890103
