Respuesta :
Answer:
a) The 95% CI (Confidence interval)for the true average porosity of a certain seam is between 4.52 and 5.18
b) The 98% CI (Confidence interval)for the true average porosity of the seam is between 4.19 and 4.93.
c) A sample size of 14 is necessary.
d) A sample size of 94 is necessary
Step-by-step explanation:
a.Compute a 95% CI (Confidence interval)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{0.75}{\sqrt{20}} = 0.33[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.33 = 4.52
The upper end of the interval is the sample mean added to M. So it is 4.85 + 0.33 = 5.18
The 95% CI (Confidence interval)for the true average porosity of a certain seam is between 4.52 and 5.18
b.Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample averageporosity of 4.56.
98% C.I, so [tex]a = 2.327[/tex], following the logic explained in a.
[tex]M = 2.327*\frac{0.75}{\sqrt{16}} = 0.37[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 4.56 - 0.37 = 4.19
The upper end of the interval is the sample mean added to M. So it is 4.56 + 0.37 = 4.93
The 98% CI (Confidence interval)for the true average porosity of the seam is between 4.19 and 4.93.
c.How large a sample size is necessary if the width of the 95% interval is to be .40? d
95% C.I, so Z = 1.96.
The sample size is n when [tex]M = 0.4, \sigma = 0.75[/tex]
[tex]0.4 = 1.96*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.96*0.75[/tex]
[tex]\sqrt{n} = \frac{1.96*0.75}{0.4}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.75}{0.4})^{2}[/tex]
[tex]n = 13.5[/tex]
Rouding up
A sample size of 14 is necessary.
d.What sample size is necessary to estimate true average porosity to within .2with 99% confidence?
99% C.I, so Z = 2.575.
The sample size is n when [tex]M = 0.2, \sigma = 0.75[/tex]
[tex]0.2 = 2.575*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.2\sqrt{n} = 2.575*0.75[/tex]
[tex]\sqrt{n} = \frac{2.575*0.75}{0.2}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.75}{0.2})^{2}[/tex]
[tex]n = 93.24[/tex]
Rouding up
A sample size of 94 is necessary
