Answer:
a) The 95% CI for the true average porosity is (4.51, 5.19).
b) The 98% CI for true average porosity is (4.11, 5.01)
c) A sample size of 15 is needed.
d) A sample size of 101 is needed.
Step-by-step explanation:
a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.
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.
![M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34](https://tex.z-dn.net/?f=M%20%3D%201.96%2A%5Cfrac%7B0.78%7D%7B%5Csqrt%7B20%7D%7D%20%3D%200.34)
The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51
The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19
The 95% CI for the true average porosity is (4.51, 5.19).
b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.
Following the same logic as a.
98% C.I., so ![z = 2.327](https://tex.z-dn.net/?f=z%20%3D%202.327)
![M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45](https://tex.z-dn.net/?f=M%20%3D%202.327%2A%5Cfrac%7B0.78%7D%7B%5Csqrt%7B16%7D%7D%20%3D%200.45)
4.56 - 0.45 = 4.11
4.56 + 0.45 = 5.01
The 98% CI for true average porosity is (4.11, 5.01)
c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?
A sample size of n is needed.
n is found when M = 0.4.
95% C.I., so Z = 1.96.
![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)
![0.4 = 1.96*\frac{0.78}{\sqrt{n}}](https://tex.z-dn.net/?f=0.4%20%3D%201.96%2A%5Cfrac%7B0.78%7D%7B%5Csqrt%7Bn%7D%7D)
![0.4\sqrt{n} = 1.96*0.78](https://tex.z-dn.net/?f=0.4%5Csqrt%7Bn%7D%20%3D%201.96%2A0.78)
![\sqrt{n} = \frac{1.96*0.78}{0.4}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A0.78%7D%7B0.4%7D)
![(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%2A0.78%7D%7B0.4%7D%29%5E%7B2%7D)
![n = 14.6](https://tex.z-dn.net/?f=n%20%3D%2014.6)
Rounding up
A sample size of 15 is needed.
d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?
99% C.I., so z = 2.575
n when M = 0.2.
![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)
![0.2 = 2.575*\frac{0.78}{\sqrt{n}}](https://tex.z-dn.net/?f=0.2%20%3D%202.575%2A%5Cfrac%7B0.78%7D%7B%5Csqrt%7Bn%7D%7D)
![0.2\sqrt{n} = 2.575*0.78](https://tex.z-dn.net/?f=0.2%5Csqrt%7Bn%7D%20%3D%202.575%2A0.78)
![\sqrt{n} = \frac{2.575*0.78}{0.2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A0.78%7D%7B0.2%7D)
![(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.575%2A0.78%7D%7B0.2%7D%29%5E%7B2%7D)
![n = 100.85](https://tex.z-dn.net/?f=n%20%3D%20100.85)
Rounding up
A sample size of 101 is needed.