Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed.

Compute a 90% CI for the true average porosity of a certain seam if the average porosity for 50 specimens from the seam was 4.85 with a sample standard deviation of .65.

Answer 1

Given that,

= 4.85

s =0.65

n = 50

Degrees of freedom = df = n - 1 =50 - 1 =49

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,49= 1.677 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=1.677 * (0.65 / 50) = 0.1542

The 90% confidence interval estimate of the population mean is,

- E < < + E

4.85- 0.1542< < 4.85+ 0.1542

4.6958 < < 5.0042

(4.6958, 5.0042 )