Respuesta :
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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34[/tex]
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 [tex]z = 2.327[/tex]
[tex]M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45[/tex]
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.
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 1.96*\frac{0.78}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.96*0.78[/tex]
[tex]\sqrt{n} = \frac{1.96*0.78}{0.4}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}[/tex]
[tex]n = 14.6[/tex]
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.
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.2 = 2.575*\frac{0.78}{\sqrt{n}}[/tex]
[tex]0.2\sqrt{n} = 2.575*0.78[/tex]
[tex]\sqrt{n} = \frac{2.575*0.78}{0.2}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}[/tex]
[tex]n = 100.85[/tex]
Rounding up
A sample size of 101 is needed.
