Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Answer:
She needs a sample size of 25.
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 - 0.68}{2} = 0.16[/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 - 0.16 = 0.84[/tex], so Z = 0.995.
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 population SD is 2 grams.
This means that [tex]\sigma = 2[/tex]
What is the minimum sample size she needs to create a confidence interval that has a width of 0.4 grams?
She needs a sample size of n.
n is found when M = 0.4. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 0.995\frac{2}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = \frac{0.995*2}{0.4}[/tex]
[tex](\sqrt{n})^2 = (\frac{0.995*2}{0.4})^2[/tex]
[tex]n = 24.8[/tex]
Rounding up:
She needs a sample size of 25.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
