Question

The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed Complete parts (a) through (c) on the right 15.42 29.80 27.10 16.51 15.42 8.81 15.42 20.46 14.90 33.67 30.91 14.86 1750 15.35 9.72 19.80 14.86 8.09 15.42 18.30 (a) Find the sample mean The sample meanis L (Round to two decimal places as needed.) (b) Find the sample standard deviation The sample standard deviation is (Round to two decimal places as needed) (c) Construct a 95% confidence interval for the population mean Enter your answer in each of the answer boxes

Answer 1

Join O X: Concerpation of dissolve organic carbon. Ex Exi= 15,42+8.817 3o.glt - - - t 83,67+9.72+ 18.30 Exio=365.32 - .. + 33.612 9.727 18.302 {x; ²= 15.42²+ 8.81² 30.912 Exp2 = 7523-8386. y 365.32 - - 18. 266 n 20 $ = sample stelo depiction - Exiz-nx? - 752318386 – 20x18.2002 49.7844 = 6.692) 2001 366.718 S = 6.6921 Sa 0493.063., 401:59,3668974 , 242.35 250.705 + 287.066+ . SO 95% confidance intewal for d e since 6 is unknown, na 20 which is mat Small sample t test we can useel hen. ( I box tan 266 +60699 | 6012-09 $- trgo-os) 7 - ( 18.266 + 18766.6.6921 x 2.09) -----| - 2.og 12o from tas • (18.266 - 3.1275, 18.2 66 + 3.1275) - (15:1385 , 21.3935) lower limit= 15.1385 po n t Upper limit = 21'3935.

x: anual earning n=35 Exi= gg.ug7+ 80.537+78.066t. n..t 61.955+ 82.200 exi= 2553.877 x= {xi - 2553.877 = 72.7679 - sumple menn 35 (b) sample standard deviations Не е Exi²- gg.ugg²+ 80.537²+ . . . + 61.955 27 82.200² Exi² = 188591.3467 ? | Skien - 158591-3467 - 35x72.8679? n-l 35-1 595 = 92.4968. a sample Stal-deviation goro Confidance intervall : (X + 2) = (12.9499 + going cp 2004) 2 ( 72.7679 + 15.663448 x 1.90). = (72.7679- 42.1237, 727679+421237) -730.6442, 119.8916) lower limit = 30.3 6942 Upper limit= 114.8916