Respuesta :
Answer:
She has to test 166 students in order to come up with a valid estimation.
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{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find 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.
If she wants to be 99% confident that her sample mean is off by no more than 3 points, how many students she has to test in order to come up with a valid estimation?
She test have at least n students, in which n is found [tex]M = 3, \sigma = 15[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 2.575*\frac{15}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = \frac{2.575*15}{3}[/tex]
[tex]\sqrt{n} = 12.875[/tex]
[tex](\sqrt{n})^{2} = (12.875)^{2}[/tex]
[tex]n = 165.7[/tex]
Rouding up
She has to test 166 students in order to come up with a valid estimation.
