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

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

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 = (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*(\sigma)/(√(n))`

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

`M = 1.96*(0.75)/(√(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*(0.75)/(√(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*(0.75)/(√(n))`

`0.4√(n) = 1.96*0.75`

`√(n) = (1.96*0.75)/(0.4)`

`(√(n))^(2) = ((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*(0.75)/(√(n))`

`0.2√(n) = 2.575*0.75`

`√(n) = (2.575*0.75)/(0.2)`

`(√(n))^(2) = ((2.575*0.75)/(0.2))^(2)`

`n = 93.24`

Rouding up

A sample size of 94 is necessary