Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.74.

(a) Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 23 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 13 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 of the 95% interval is to be 0.41? (Round your answer up to the nearest whole number.)
____________specimens

(d) What sample size is necessary to estimate true average porosity to within 0.21 with 99% confidence? (Round your answer up to the nearest whole number.)
____________specimens

Answer 1

Solution :

Given that,

a) Z/2 = Z0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 0.74 /  23 )

= 0.30

At 95% confidence interval estimate of the population mean is,

  ± E

= 4.85   ± 0.30

= ( 4.55, 5.15 )

b) Z/2 = Z0.01 = 2.33

Margin of error = E = Z/2 * ( /n)

= 2.33 * ( 0.74 /  13 )

= 0.48

At 95% confidence interval estimate of the population mean is,

  ± E

= 4.56 ± 0.48

= ( 4.08, 5.04 )

c) margin of error = E = 0.41/ 2 = 0.205

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [ 1.96 * 0.74 / 0.205 ]2

n = 50.05

Sample size = n = 51 specimens.

d) margin of error = E = 0.21

Z/2 = Z0.005 = 2.576

sample size = n = [Z/2* / E] 2

n = [ 2.576 * 0.74 / 0.21 ]2

n = 82.39

Sample size = n = 83 specimens