Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75. (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.) , (b) compute a 98% ci for true average porosity of another seam based on 15 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

Answer 1

A) $4.85\pm0.29$
B) $4.56\pm0.45$

We first find the z-score associated with the level of confidence.  For 95% confidence:
Convert 95% to a decimal:  0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2 (this is the area in the tails):  0.05/2 = 0.025
Subtract from 1 (we don't want the tails, we want the area in the middle):  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we find the z-score for this is 1.96.  We now use the formula
$\overline{x} \pm z*(\frac{\sigma}{\sqrt{n}}) =4.85\pm 1.96*(\frac{0.75}{\sqrt{25}})=4.85\pm0.294\approx 4.85\pm 0.29$

For a 98% confidence level:
Convert 98% to a decimal:  0.98
Subtract from 1:  1 - 0.98 = 0.02
Divide by 2:  0.02/2 = 0.01
Subtract from 1:  1 - 0.01 = 0.99

Using the z-table we see the closest number to this is for the z-score 2.33:
$\overline{x} \pm z* (\frac{\sigma}{\sqrt{n}}) =4.56\pm 2.33*(\frac{0.75}{\sqrt{15}})=4.56\pm0.45$