Respuesta :
Answer:
A sample size 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 and n is the size of the sample.
You feel that a reasonable estimate of the standard deviation is 10.0 hours.
This means that [tex]\sigma = 10[/tex]
What sample size is needed so that the expected margin of error of your estimate is not larger than one hour for 95% confidence?
A sample size of n is needed. n is found when M = 1. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.96\frac{10}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.96*10[/tex]
[tex](\sqrt{n})^2 = (1.96*10)^2[/tex]
[tex]n = 384.16[/tex]
Rounding up:
A sample size of 385 is needed.
