Answer:
We need to sample at least 37 weeks of data.
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{1-0.98}{2} = 0.01](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.98%7D%7B2%7D%20%3D%200.01)
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 = 2.327](https://tex.z-dn.net/?f=z%20%3D%202.327)
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)
We want 98% confidence that the sample mean is within $500 of the population mean, and the population standard deviation is known to be $1300
This is at least n weeks, in which n is found when ![M = 500, \sigma = 1300](https://tex.z-dn.net/?f=M%20%3D%20500%2C%20%5Csigma%20%3D%201300)
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)
![500 = 2.327*\frac{1300}{\sqrt{n}}](https://tex.z-dn.net/?f=500%20%3D%202.327%2A%5Cfrac%7B1300%7D%7B%5Csqrt%7Bn%7D%7D)
![500\sqrt{n} = 2.327*1300](https://tex.z-dn.net/?f=500%5Csqrt%7Bn%7D%20%3D%202.327%2A1300)
![\sqrt{n} = \frac{2.327*1300}{500}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.327%2A1300%7D%7B500%7D)
![(\sqrt{n})^{2} = (\frac{2.327*1300}{500})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.327%2A1300%7D%7B500%7D%29%5E%7B2%7D)
![n = 36.6](https://tex.z-dn.net/?f=n%20%3D%2036.6)
Rounding up
We need to sample at least 37 weeks of data.
Answer:
79
Step-by-step explanation:
Let the original number be x.
x/2 = 39.5
x = 39.5 × 2
x = 79
the original cost was $111.60.
Answer:
3 + 8x
Step-by-step explanation: