Get comprehensive solutions to your questions with the help of IDNLearn.com's experts. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
Using the z-distribution, as we have the standard deviation for the population, the margin of error is given by:
b. 0.39.
What is a t-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 sample.
The margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
As for the other parameters, we have that [tex]\sigma = 1.8, n = 81[/tex].
Hence, the margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.96\frac{1.8}{\sqrt{81}}[/tex]
[tex]M = 0.39[/tex]
Hence option b is correct.
More can be learned about the z-distribution at https://brainly.com/question/25890103
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.
