Answer:
The minimum sample size required to create the specified confidence interval is 1804.
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{1-0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
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%2A%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.
Minimum sample size for a margin of error of 0.06:
This sample size is n.
n is found when ![M = 0.06, \sigma = 1.3](https://tex.z-dn.net/?f=M%20%3D%200.06%2C%20%5Csigma%20%3D%201.3)
So
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.06 = 1.96*\frac{1.3}{\sqrt{n}}](https://tex.z-dn.net/?f=0.06%20%3D%201.96%2A%5Cfrac%7B1.3%7D%7B%5Csqrt%7Bn%7D%7D)
![0.06\sqrt{n} = 1.96*1.3](https://tex.z-dn.net/?f=0.06%5Csqrt%7Bn%7D%20%3D%201.96%2A1.3)
![\sqrt{n} = \frac{1.96*1.3}{0.06}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A1.3%7D%7B0.06%7D)
![(\sqrt{n})^{2} = (\frac{1.96*1.3}{0.06})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%2A1.3%7D%7B0.06%7D%29%5E%7B2%7D)
![n = 1803.41](https://tex.z-dn.net/?f=n%20%3D%201803.41)
Rounding up to the next integer
The minimum sample size required to create the specified confidence interval is 1804.