Answer:
The sample size necessary is 237.
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 Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.02 = 0.98[/tex], so Z = 2.056.
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.
Find the sample size necessary to estimate that mean with a 15-minute sampling error.
A sample size of n is necessary.
n is found when [tex]M = 15[/tex]
We have that [tex]\sigma = 112.2[/tex]
So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]15 = 2.056\frac{112.2}{\sqrt{n}}[/tex]
[tex]15\sqrt{n} = 2.056*112.2[/tex]
[tex]\sqrt{n} = \frac{2.056*112.2}{15}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.056*112.2}{15})^2[/tex]
[tex]n = 236.5[/tex]
Rounding up,
The sample size necessary is 237.
