Answer:
You need to have 538 people in your study.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.94}{2} = 0.03](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.94%7D%7B2%7D%20%3D%200.03)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.88.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
You practiced on the people at work and found a standard deviation of about 0.37 seconds.
This means that ![\sigma = 0.37](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.37)
You want to get a 94% confidence interval that is only 0.06 in width.
This means that ![M = \frac{0.06}{2} = 0.03](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B0.06%7D%7B2%7D%20%3D%200.03)
How many people do you need to have in your study?
This is n for which M = 0.03. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.03 = 1.88\frac{0.37}{\sqrt{n}}](https://tex.z-dn.net/?f=0.03%20%3D%201.88%5Cfrac%7B0.37%7D%7B%5Csqrt%7Bn%7D%7D)
![0.03\sqrt{n} = 1.88*0.37](https://tex.z-dn.net/?f=0.03%5Csqrt%7Bn%7D%20%3D%201.88%2A0.37)
![\sqrt{n} = \frac{1.88*0.37}{0.03}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.88%2A0.37%7D%7B0.03%7D)
![(\sqrt{n})^2 = (\frac{1.88*0.37}{0.03})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B1.88%2A0.37%7D%7B0.03%7D%29%5E2)
![n = 537.6](https://tex.z-dn.net/?f=n%20%3D%20537.6)
Rounding up:
You need to have 538 people in your study.