Respuesta :
At least 832 teenagers must be interviewed.
It is given that the standard deviation is of 1.24 liters.
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2}[/tex] = 0.005
Now, we have to find z in the Ztable as such z has a p-value of [tex]1-\alpha[/tex].
So it is z with a p-value of [tex]1-0.005 = 0.995[/tex], so [tex]z= 2.575[/tex]
Now, find the margin of error M as such
[tex]M = Z * \frac{SD}{\sqrt{n} }[/tex]
In which SD is the standard deviation of the population and n is the size of the sample.
How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liters when constructing a 99% confidence interval?
At least n teenager must be interviewed.
n is found when M = 0.1.
We have that SD = 1.12
So
[tex]0.1 = 2.575*\frac{1.12}{\sqrt{N} }[/tex]
[tex]\sqrt{n} = \frac{2.575*1.12}{0.1}[/tex]
[tex]n = 831.7[/tex]
So rounding up to at least 832 teenagers must be interviewed.
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