Answer:
We would have to take a sample of 62 to achieve this result.
Step-by-step explanation:
Confidence level of 95%.
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
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 = 1.96.
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.
Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.
This means that ![\sigma = 2](https://tex.z-dn.net/?f=%5Csigma%20%3D%202)
If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?
We would need a sample of n.
n is found when
. 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)
![0.5 = 1.96\frac{2}{\sqrt{n}}](https://tex.z-dn.net/?f=0.5%20%3D%201.96%5Cfrac%7B2%7D%7B%5Csqrt%7Bn%7D%7D)
![0.5\sqrt{n} = 2*1.96](https://tex.z-dn.net/?f=0.5%5Csqrt%7Bn%7D%20%3D%202%2A1.96)
Dividing both sides by 0.5
![\sqrt{n} = 4*1.96](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%204%2A1.96)
![(\sqrt{n})^2 = (4*1.96)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%284%2A1.96%29%5E2)
![n = 61.5](https://tex.z-dn.net/?f=n%20%3D%2061.5)
Rounding up
We would have to take a sample of 62 to achieve this result.