Respuesta :
Answer:
A sample of 385 is needed.
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 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
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, which is the square root of the variance, and n is the size of the sample.
Assume the variance is known to be 3.61.
This means that [tex]\sigma = \sqrt{3.61}[/tex]
How large of a sample would be required in order to estimate the mean number of reproductions per hour at the 95% confidence level with an error of at most 0.19 reproductions?
A sample of n is needed.
n is found when [tex]M = 0.19[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.19 = 1.96\frac{\sqrt{3.61}}{\sqrt{n}}[/tex]
[tex]0.19\sqrt{n} = 1.96\sqrt{3.61}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{3.61}}{0.19}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{3.61}}{0.19})^2[/tex]
[tex]n = 384.16[/tex]
Rounding up
A sample of 385 is needed.
