Mathematics, 07.04.2020 19:53 cuzhernamesari

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?

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Answer from: danielaguardado63

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

Answer from: Quest

answer: the answer is c.

step-by-step explanation:

she’s dividing 1/3 of the flour into 2 jars.

hope this : )

Use the expression at the top right to answer the question (1/3 divided by 2)

Answer from: Quest

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)

Need 45 ! compare the rates at which each of the three students read: which student read at a fa

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