Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
554 executives should be surveyed.
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 Z-table as such z has a p-value 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.
Standard deviation of 3 hours.
This means that [tex]\sigma = 3[/tex]
The 95% level of confidence is to be used. How many executives should be surveyed?
n executives should be surveyed, and n is found with [tex]M = 0.25[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.25 = 1.96\frac{3}{\sqrt{n}}[/tex]
[tex]0.25\sqrt{n} = 1.96*3[/tex]
[tex]\sqrt{n} = \frac{1.96*3}{0.25}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*3}{0.25})^2[/tex]
[tex]n = 553.2[/tex]
Rounding up:
554 executives should be surveyed.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.