Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.71. (a) Compute a 95 % CI for the true average porosity of a certain seam if the average porosity for 17 specimens from the seam was 4.85. (Round your answers to tw decimal places.) 4.51 ] ) 5.19 (b) Compute a 98% CI for true average porosity of another seam based on 20 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 0.46? (Round your answer up to the nearest whole number.) specimens (d) What sample size is necessary to estimate true average porosity to within 0.2 with 99% confidence? (Round your answer up to the nearest whole number.) specimens

Answer 1

wice r=0.74 s krown, a 95% confidence interval for M would be Iu - (ā- Zor Pro , 8+ Zak = (4.85 - 70.025 04.85+ Zo.oas = (4.85-1.96 x 4.85+ 1.96x 07) = (4.512 ,5.187) ) since r=0.71 is known, a 98% Confidence interval for u would be Iu= (– Zayz , Zayed on) (4.56-70.0 0 , 4.56+ 20.01 ore ) = (4.5-3.33x 4.56 +2.33x options = C4.190.4.989), * the manger Ersione is er oogt = 0.23 = [1.96 x 0.71 ]? 0.23 = 36.60 , So rrounding up gives R=37 as the minimur number of Sample to produce a 95% CI widths of no more than 0.46.

margo Error es e- 0.1 = [2.58 x 0.71 12 = 335.54. , So sounding upto given D=336 as the menemum number of Sample to produce a 99% CI width of no more than o.a.