Respuesta :
Answer:
At least 832 teenargers must be interviewed.
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 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.
How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval
At least n teenargers must be interviewed.
n is found when M = 0.1.
We have that [tex]\sigma = 1.12[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.1 = 2.575*\frac{1.12}{\sqrt{n}}[/tex]
[tex]0.1\sqrt{n} = 2.575*1.12[/tex]
[tex]\sqrt{n} = \frac{2.575*1.12}{0.1}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*1.12}{0.1})^{2}[/tex]
[tex]n = 831.7[/tex]
Rounding up
At least 832 teenargers must be interviewed.
