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Sagot :
Answer:
A sample of 179 is needed.
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 - 0.85}{2} = 0.075[/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 - 0.075 = 0.925[/tex], so Z = 1.44.
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.
A previous study found that for an average family the variance is 1.69 gallon?
This means that [tex]\sigma = \sqrt{1.69} = 1.3[/tex]
If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?
A sample of n is needed, and n is found for M = 0.14. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.14 = 1.44\frac{1.3}{\sqrt{n}}[/tex]
[tex]0.14\sqrt{n} = 1.44*1.3[/tex]
[tex]\sqrt{n} = \frac{1.44*1.3}{0.14}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2[/tex]
[tex]n = 178.8[/tex]
Rounding up
A sample of 179 is needed.
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