Mathematics, 08.03.2021 20:00 mprjug6

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.78. (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. (Round your answers to two decimal places.)
(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.)
(c) How large a sample size is necessary if the width of the 95% interval is to be 0.46? (Round your answer up to the nearest whole number.)
(d) What sample size is necessary to estimate true average porosity to within 0.18 with 99% confidence? (Round your answer up to the nearest whole number.)

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Answers

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 $\alpha$ 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 $1 - \alpha$ .

That 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.

$M = 1.96\frac{0.78}{\sqrt{15}} = 0.39$

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

$\alpha = \frac{1 - 0.98}{2} = 0.01$

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

That is z with a pvalue of 1 - 0.01 = 0.99 , so Z = 2.327.

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.

$M = 2.327\frac{0.78}{\sqrt{11}} = 0.55$

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).

95% CI means that z = 1.96

A sample of n is needed, and n is found when M = 0.46. So

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

$0.46 = 1.96\frac{0.78}{\sqrt{n}}$

$0.46\sqrt{n} = 1.96*0.78$

$(\sqrt{n})^2 = (\frac{1.96*0.78}{0.46})^2$

n = 11.05

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 z = 2.575

A sample of n is needed, and n is found when M = 0.18. So

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

$0.18 = 2.575\frac{0.78}{\sqrt{n}}$

$0.18\sqrt{n} = 2.575*0.78$

$(\sqrt{n})^2 = (\frac{2.575*0.78}{0.18})^2$

n = 124.5

Rounding up

A sample size of 125 is needed.

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