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Sagot :
Answer:
A sample of 151 cans must be selected.
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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
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 standard deviation of fill volumes is 0.05 ounce.
This means that [tex]\sigma = 0.05[/tex]
How large a sample of cans must be selected to estimate the population mean fill volume with a 95% confidence interval that has a margin of error of 0.008 ounce?
We need a sample of n.
n is found when M = 0.008. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.008 = 1.96\frac{0.05}{\sqrt{n}}[/tex]
[tex]0.008\sqrt{n} = 1.96*0.05[/tex]
[tex]\sqrt{n} = \frac{1.96*0.05}{0.008}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.05}{0.008})^2[/tex]
[tex]n = 150.1[/tex]
Rounding up:
A sample of 151 cans must be selected.
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