Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
