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Sagot :
Answer:
48 cans must be sampled.
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].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
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.
The machine is calibrated so that the population standard deviation is 0.04 ounces.
This means that [tex]\sigma = 0.04[/tex]
How many filled cans must be sampled so that we estimate the mean fill volume within 0.015 ounces with 99% confidence?
n cans must be sampled, and n is found when M = 0.015. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.015 = 2.575\frac{0.04}{\sqrt{n}}[/tex]
[tex]0.015\sqrt{n} = 2.575*0.04[/tex]
[tex]\sqrt{n} = \frac{2.575*0.04}{0.015}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*0.04}{0.015})^2[/tex]
[tex]n = 47.2[/tex]
Rounding up, 48 cans must be sampled.
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