Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation .75. 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. b.Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample averageporosity of 4.56. c.How large a sample size is necessary if the width of the 95% interval is to be .40? d.What sample size is necessary to estimate true average porosity to within .2with 99% confidence?

Answer 1

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.75}{\sqrt{20}} = 0.33$

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 $a = 2.327$, following the logic explained in a.

$M = 2.327*\frac{0.75}{\sqrt{16}} = 0.37$

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 $M = 0.4, \sigma = 0.75$

$0.4 = 1.96*\frac{0.75}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.75$

$\sqrt{n} = \frac{1.96*0.75}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.75}{0.4})^{2}$

$n = 13.5$

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 $M = 0.2, \sigma = 0.75$

$0.2 = 2.575*\frac{0.75}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.75$

$\sqrt{n} = \frac{2.575*0.75}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.75}{0.2})^{2}$

$n = 93.24$

Rouding up

A sample size of 94 is necessary