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 25 specimens from the seam was 4.85. (Round your answers to two decimal places.)

Respuesta :

Answer:

The 95% CI for the true average porosity of a certain seam if the average porosity for 25 specimens from the seam was 4.85 is between 4.54 and 5.16.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

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

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96*\frac{0.78}{\sqrt{25}} = 0.31[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.31 = 4.54.

The upper end of the interval is the sample mean added to M. So it is 5.85 + 0.31 = 5.16.

The 95% CI for the true average porosity of a certain seam if the average porosity for 25 specimens from the seam was 4.85 is between 4.54 and 5.16.