Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.77. (a) Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 21 specimens from the seam was 4.85. (Round your answers to two decimal places.)

Answer 1

The 95% CI for the true average porosity of a certain seam is (4.52, 5.18).

Explanation:

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`.

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(\sigma)/(√(n))`

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

`M = 1.96(0.77)/(√(21)) = 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 for the true average porosity of a certain seam is (4.52, 5.18).