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.
