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Answer:
a) The 95% CI for the true average porosity of a certain seam if the average porosity for 15 specimens from the seam was 4.85 is (4.46, 5.24).
b) The 98% CI for true average porosity of another seam based on 11 specimens with a sample average porosity of 4.56 is (4.01, 5.11).
c) A sample size of 12 is needed.
d) A sample size of 125 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 15 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].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
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{15}} = 0.39[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.39 = 4.46
The upper end of the interval is the sample mean added to M. So it is 4.85 + 0.39 = 5.24
The 95% CI for the true average porosity of a certain seam if the average porosity for 15 specimens from the seam was 4.85 is (4.46, 5.24).
(b) Compute a 98% CI for true average porosity of another seam based on 11 specimens with a sample average porosity of 4.56. (Round your answers to two decimal places.)
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 - 0.98}{2} = 0.01[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.01 = 0.99[/tex], so Z = 2.327.
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 = 2.327\frac{0.78}{\sqrt{11}} = 0.55[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 4.56 - 0.55 = 4.01
The upper end of the interval is the sample mean added to M. So it is 4.56 + 0.55 = 5.11
The 98% CI for true average porosity of another seam based on 11 specimens with a sample average porosity of 4.56 is (4.01, 5.11).
(c) How large a sample size is necessary if the width of the 95% interval is to be 0.46?
95% CI means that [tex]z = 1.96[/tex]
A sample of n is needed, and n is found when M = 0.46. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.46 = 1.96\frac{0.78}{\sqrt{n}}[/tex]
[tex]0.46\sqrt{n} = 1.96*0.78[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.78}{0.46})^2[/tex]
[tex]n = 11.05[/tex]
Rounding up
A sample size of 12 is needed.
(d) What sample size is necessary to estimate true average porosity to within 0.18 with 99% confidence?
99% CI means that [tex]z = 2.575[/tex]
A sample of n is needed, and n is found when M = 0.18. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.18 = 2.575\frac{0.78}{\sqrt{n}}[/tex]
[tex]0.18\sqrt{n} = 2.575*0.78[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*0.78}{0.18})^2[/tex]
[tex]n = 124.5[/tex]
Rounding up
A sample size of 125 is needed.
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