Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Answer:
The minimum sample size is 239.
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 Z-table as such z has a p-value 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.
Population standard deviation is equal to 1.5
This means that [tex]\sigma = 1.5[/tex]
Margin of error of 0.25
This means that [tex]M = 0.25[/tex]
What's the minimum size of the sample?
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.25 = 2.575\frac{1.5}{\sqrt{n}}[/tex]
[tex]0.25\sqrt{n} = 2.575*1.5[/tex]
[tex]\sqrt{n} = \frac{2.575*1.5}{0.25}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*1.5}{0.25})^2[/tex]
[tex]n = 238.7[/tex]
Rounding up:
The minimum sample size is 239.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
