Assume that the helium porosity (in percentage) of coal samples taken from any p
ID: 3395554 • Letter: A
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 16 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 of the 95% interval is to be 0.33? (Round your answer up to the nearest whole number.)
specimens
(d) What sample size is necessary to estimate true average porosity to within 0.23 with 99% confidence? (Round your answer up to the nearest whole number.)
specimens
Explanation / Answer
a)
n =16
SD=0.75
Mean =4.85
z for 95%CI = 1.96
Margin of Error = z*SD/sqrt(n)
= (1.96 * 0.75)/sqrt(16)
= 0.3675
Therefore,
CI: (4.85 - 0.3675 , 4.85+0.3675)
= (4.48 , 5.22) Answer
b)
n =15
SD=0.75
Mean =4.56
z for 98%CI = 2.33
Margin of Error = z*SD/sqrt(n)
= (2.33 * 0.75)/sqrt(15)
=0.4512
Therefore,
CI: (4.56 - 0.4512 , 4.56+0.4512)
= (4.11 , 5.01) Answer
c)
Width = 2 * Margin of Error
=> 0.33 = 2 * (1.96 * 0.75)/sqrt(n)
=> sqrt(n) = 2 * (1.96 * 0.75) / 0.33
=>n = 79.37
=> n = 79 Answer
d)
Width = 2 * Margin of Error
=> 0.23 = 2 * (2.58* 0.75)/sqrt(n)
=> sqrt(n) = 2 * (2.58* 0.75) / 0.23
=> n = 283 Answer
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