Respuesta :
Answer:
A sample size of 3 is required.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.96}{2} = 0.02[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.02 = 0.98[/tex], so Z = 2.054.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Based on previous evidence, he believes that the standard deviation is approximately 3.6 hours.
This means that [tex]\sigma = 3.6[/tex]
He would like to be 96% confident that his estimate is within 5 hours of the true population mean. Use RStudio to determine how large of a sample size is required without rounding any interim calculations.
The sample size needed is of n, and n is found when M = 5. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]5 = 2.054\frac{3.6}{\sqrt{n}}[/tex]
[tex]5\sqrt{n} = 2.054*3.6[/tex]
[tex]\sqrt{n} = \frac{2.054*3.6}{5}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.054*3.6}{5})^2[/tex]
[tex]n = 2.19[/tex]
Rounding up:
A sample size of 3 is required.
