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. Find sample mean, standard deviation and construct a 99% confidence interval the population mean m.

5.20 29.80 27.10 16.51 14.00

8.81 15.42 20.46 14.90 33.67

30.91 14.86 7.40 15.35 9.72

19.80 14.86 8.09 14.00 18.30

Answer 1

Solution:

x	x2
5.2	27.04
29.8	888.04
27.1	734.41
16.51	272.5801
14	196
8.81	77.6161
15.42	237.7764
20.46	418.6116
14.9	222.01
33.67	1133.6689
30.91	955.4281
14.86	220.8196
7.4	54.76
15.35	235.6225
9.72	94.4784
19.8	392.04
14.86	220.8196
8.09	65.4481
14	196
18.3	334.89
x=339.16	x2=6978.0594

The sample mean is

Mean = (x / n) )

=5.2+29.8+27.1+16.51+14+8.81+15.42+20.46+14.9+33.67+30.91+14.86+7.4+15.35+9.72+19.8+14.86+8.09+14+18.3/20

=339.16/20

=16.958

The sample standard is S

  S =( x2 ) - (( x)2 / n ) n -1

=6978.0594-(339.16)220/19

=6978.0594-5751.4753/19

=1226.5841/19

=64.5571

=8.0347

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,19 =2.861

Margin of error = E = t/2,df * (s /n)

= 2.861 * (8.03 / 20)

= 5.14

Margin of error = 5.14

The 99% confidence interval estimate of the population mean is,

- E < < + E

16.96 - 5.14 < < 16.96 + 5.14

11.82 < < 22.10

(11.82, 22.10 )