Answer:
A sample size of at least 328 students 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{1-0.9}{2} = 0.05[/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]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the width 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.
In this problem, we have that:
[tex]M = 10, \sigma = 110[/tex]
So:
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 1.645*\frac{110}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 180.95[/tex]
[tex]\sqrt{n} = 18.095[/tex]
[tex]n = 327.43[/tex]
A sample size of at least 328 students is required.
Answer:
A sample size of at least 328 students is required.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the width M as such
In which is the standard deviation of the population and n is the size of the sample.
In this problem, we have that:
So:
A sample size of at least 328 students is required.
