Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.76.(a) Compute a 99% CI for the true average porosity of a certain seam if the average porosity for 16 specimens from the seam was 4.85. (Give answers accurate to 2 decimal places.)

Answer 1

Answer: $(4.36, 5.34)$

Step-by-step explanation:

Formula to calculate the confidence interval for population mean is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = sample mean

n= sample size.

z*= Critical value

$\sigma$ = Population standard deviation.

As per given , we have

$\sigma=0.76$

$\overline{x}=4.85$

n= 16

From z-table , the critical value for 99% confidence = z*=2.576

Now , 99% confidence interval for true average porosity of a certain seam will be :

$4.85\pm (2.576)\dfrac{0.76}{\sqrt{16}}$

$4.85\pm (2.576)\dfrac{0.76}{4}$

$4.85\pm (2.576)(0.19)$

$4.85\pm (0.48944)$

$=(4.85-0.48944\ 4.85+0.48944) \\\\=(4.36056,\ 5.33944)\approx(4.36,\ 5.34)$

Hence, the required  99% CI for the true average porosity of a certain seam = $(4.36, 5.34)$