Respuesta :
Answer:
The minimum sample size that we should consider is of 60 employees.
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 Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
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.
We want our 99 percent confidence interval to have a margin of error of no more than plus or minus 2 minutes. What is the smallest sample size that we should consider?
We need to find n for which [tex]M = 2, \sigma = 6[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]2 = 2.575*\frac{6}{\sqrt{n}}[/tex]
[tex]2\sqrt{n} = 2.575*6[/tex]
Simplifying by 2
[tex]\sqrt{n} = 2.575*3[/tex]
[tex](\sqrt{n})^{2} = (2.575*3)^{2}[/tex]
[tex]n = 59.67[/tex]
Rounding up
The minimum sample size that we should consider is of 60 employees.
Answer:
[tex]n=(\frac{2.58(6)}{2})^2 =59.907 \approx 60[/tex]
So we need a sample of at least 60 in order to satisfy the condition.
Step-by-step explanation:
Notation
[tex]\bar X[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=6[/tex] represent the population standard deviation assumed
n represent the sample size
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (2)
And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (3)
The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (3) we got:
[tex]n=(\frac{2.58(6)}{2})^2 =59.907 \approx 60[/tex]
So we need a sample of at least 60 in order to satisfy the condition.
