TUGAS STATISTIKA KIMIA Exercise 2 1 – 8 “ Statistics of repeated repeated measurements measurements An gg ota ot a Kelo Ke lomp mpok ok 1. Rut Novalia R. Sianipar Siani par (J1B114601) (J1B114601) 2. Rina Apriani (J1B114602) (J1B114602) 3. Diky Subhanuddin (J1B114603) 4. Wenny Yuniandari (J1B108213)
1.
The reproducibility reproducibility of a method method for the determination determination of selenium in foods was investigated by taking nine samples from a single batch of brown rice and determining the selenium concentration in each. The f ollowing ollowing results w ere obtained: 0.07 0.07 0.08 0.07 0.07 0.08 0.08 0.09 0.08 μg g-1 (Moreno-Dominguez, (Moreno-Dominguez, T., Garcia-Moreno, C. and Marine-Font, A., 1983, Analyst, 108: 108: 505) Calculate the mean, standard deviation and relative standard deviation of these results.
Answer :
No Sample
Xi (μg/g)
(Xi - )
(Xi - )^2
1
0.07
0.077
-0.007
0.00005
2
0.07
0.077
-0.007
0.00005
3
0.08
0.077
0.003
0.00001
4
0.07
0.077
-0.007
0.00005
5
0.07
0.077
-0.007
0.00005
6
0.08
0.077
0.003
0.00001
7
0.08
0.077
0.003
0.00001
8
0.09
0.077
0.013
0.00017
9
0.08
0.077
0.003
0.00001
Total
0.69
0.0005
∑ = . = 0.077 (μg/g) ∑( ) = . = 0.007 Standard Standard d eviation eviation , s = ( ) Mean,
=
Relati Relati ve Standard deviation , RSD RSD = 100 . s /
=
× .⁄.
= 9. 09 %
2. The morphine levels levels (%) of seven batches of seized heroin heroin were determined,with determined,with the following results:15.1 21.2 18.5 25.3 19.2 16.0 17.8 Calculate the 95% and 99% confidence limits for these measurements Answer :
No Sample
Xi (μg/g)
1
15.10
19.01
-3.91
15.2881
2
21.20
19.01
2.19
4.7961
3
18.50
19.01
-0.51
0.2601
4
25.30
19.01
6.29
39.5641
5
19.20
19.01
0.19
0.0361
6
16.00
19.01
-3.01
9.0601
7
17.80
19.01
-1.21
1.4641
Total
(Xi - )
133.1
(Xi - )^2
70.4687
∑ = . = 19.01 % ∑( ) = . = 3.43 % Standard d eviation , s = ( ) Mean,
=
Conf idence 95 % =
± t n-1 . s /
√
= 19.01 ± 2.45 . 3.43 / Confi dence 99 % =
± t n-1 . s /
√
= 19.01 ± 7.71 . 3.43 /
√ = 19.01 ± 3.17 √ = 19.01 ± 4.81
3.
Ten replicate analyses of the concentration of mercury in a sample of commercial gas condensate gave the following results: 23.3 22.5 21.9 21.5 19.9 21.3 21.7 23.8 22.6 24.7 ng ml-1 (Shafawi, A., Ebdon, L., Foulkes, M., Stockwell, P. and Corns, W., 1999, Analyst, 124: 185) Calculate the mean, standard deviation, relative standard deviation and 99% confidence limits of the mean. Six replicate analyses on another sample gave the f ollowing values: 13.8 14.0 13.2 11.9 12.0 12.1 ng ml-1 Repeat the calculations for these values. Answer : First Data
Xi (μg/g)
(Xi - )
(Xi - )^2
1
23.30
22.32
0.98
0.9604
2
22.50
22.32
0.18
0.0324
3
21.90
22.32
-0.42
0.1764
4
21.50
22.32
-0.82
0.6724
5
19.90
22.32
-2.42
5.8564
6
21.30
22.32
-1.02
1.0404
7
21.70
22.32
-0.62
0.3844
8
23.80
22.32
1.48
2.1904
9
22.60
22.32
0.28
0.0784
10
24.70
22.32
2.38
5.6644
No Sample
Total
Mean,
=
223.2
∑ = . = 22.32 (ng/m l)
17.0560
∑ = = 1.377 ng/ml Relati ve Standard deviation , RSD = 100 . s / = × . ⁄. Standard d eviation , s =
(
)
(
.
)
= 6.17 % Confi dence 99 % =
± t n-1 . s /
√
= 22.32 ± 3.25 . 1.377 /
√ = 22.32 ± 1.41 ng /ml
Untuk d ata kedua :
No Sample
Xi (μg/g)
(Xi - )
(Xi - )^2
1
13.80
12.83
0.97
0.9409
2
14.00
12.83
1.17
1.3689
3
13.20
12.83
0.37
0.1369
4
11.90
12.83
-0.93
0.8649
5
12.00
12.83
-0.83
0.6889
6
12.10
12.83
-0.73
0.5329
Total
77
∑ = = 12.83 (ng/m l) ∑( ) = . = 0.95 ng/ml Standard d eviation , s = ( ) Mean,
4.5334
=
Relati ve Standard deviation , RSD = 100 . s /
=
× . ⁄.
= 7.42 % Confi dence 99 % =
± t n-1 . s /
√
= 12.83 ± 4.03 . 0.95 /
√ = 12.83 ± 1.57 ng /ml
4. The concentration of lead in the bloodstream was measured for a sa mple of 50 children from a large school near a busy main road. The sample mean was 10.12 ng ml -1 and the standard deviation was 0.64 ng ml-1. Calculate the 95% confidence interval for the mean lead concentration for all the children in the school. About how big should the sample have been to reduce the range of the confidence interval to 0.2 ng ml -1 (i.e. ±0.1 ng ml -1)? Answer :
Diketahui : n = 50 ,
= 10.12 ng/ml ,
s = 0.64 ng/ml
Ditanyakan : - Hitung selang kepercayaan 95 % untuk rataan kadar timbal (Pb) semua anak dari sekolah tersebut? , Berapakah jumlah sampling (n) untuk memperkecil selang kepercayaan (yaitu ± 0.1 ng/ml) Jawab : Conf idence 95 % =
± t n-1 . s /
√
= 10.12 ± 2.01 . 0.64 /
√ = 10.12 ± 0.18
Untuk memperkecil rentang selang kepercayaan, memerlukan data yang banyak (sampling) dan cukup aman bila dimisalkan bahwa t akan dengan dengan 1,96 jadi, bil a ± 0.1 ng/ml maka n adalah. 0.1 = 1.96 . s /
√
√ √ = . .⁄. = 12. 544 0.1 = 1.96 x 0.64 /
5.
n = 157
In an evaluation of a method for the determination of fluorene in seawater, a synthetic sample of seawater was spiked with 50 ng ml -1 of fluorene. Ten replicate determinations of the fluorene concentration in the sample had a mean of 49.5 ng ml -1 with a standard deviation of 1.5 ng ml -1. (Gonsalez, M.A. and Lopez, M.H., 1998, Analyst , 123: 2217) Calculate the 95% confidence limits of the mean. Is the spiked value of 50 ng ml -1 within the 95% confidence limits?
Answer :
Diketahui : n = 10 ,
= 49.5 mg/ml , s = 1.5 mg/ml
Ditanyakan : Jika didapatkan spike 50 ng/ml, apakah masuk kedalam tingkat kepercayaan 95 %. Confi dence 95 % =
± t n-1 . s /
√
= 49.5 ± 2.26 . 1.5 /
√ = 49.5 ± 1.07 ng/ml
Berdasarkan data tersebut, nilai 50 ng/ml masih termasuk dalam range tingk at kepercayaan 95 %. 6.
A 0.1 M solution of acid was used to titrate 10 ml of 0.1 M solution of alkali and the following volumes of acid were recorded: 9.88 10.18 10.23 10.39 10.21 ml Calculate the 95% confidence limits of the mean and use them to d ecide whether there is any evidence of systematic error. Answer :
No Sample
Xi (μg/g)
(Xi - )
(Xi - )^2
1
9.88
10.18
-0.3
0.0900
2
10.18
10.18
0
0.0000
3
10.23
10.18
0.05
0.0025
4
10.39
10.18
0.21
0.0441
5
10.21
10.18
0.03
0.0009
Total
50.89
∑ = . = 10.18 ng/ml Mean, = ∑( ) = . = 0.185 % Standard d eviation, s = ( ) Conf id ence 95 % =
± t n-1 . s /
√
= 10.18 ± 2.78 . 0.185 /
0.1375
√ = 10.18 ± 0.23
Berdasarkan data diatas, tid ak ada ditemukan bukti kesalahan sistematis.
7.
A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, using weighing by difference. The standard deviation of each weighing was 0.0001 g: what were the standard deviation and relative standard deviation of the weight of reagent used? The standard deviation of the volume of solvent used was 0.05 ml. Express this as a relative standard deviation. Hence calculate the relative standard deviation of the molarity of the solution. Repeat the calculation for a reagent of formula weight 392.
Answer:
Standard devi asi seti ap penimbang an adalah 0.0001 g Standard d eviasi untuk penimbangan adalah S=
(.) + (.) = 0.00014 g = 0.14 mg
Diketahui V larutan = 250 ml, Kon setrasi 0.05 M, Masa Relatif 40 Maka berat yang di timb ang adalah. Massa = Kons entrasi x mr x V larutan = 0.05 * 250 * 40 = 500 mg Relati ve Standard deviation , RSD = 100 . s /
=
× . ⁄ = 0.028
%
SD untuk volume solvent adalah 0.05 ml Relati ve Standard deviation, RSD = 100 . s /
=
× . ⁄ = 0.02
%
RSD untuk konsentrasi
( ) + ( ) = √ . + . = 0.034 %
RSD =
Perhitungan untuk pereaksi dengan massa relative (mr) 392, dengan asumsi V dan konsentrasi tetap. Maka banyaknya pereaksi ditimbang adalah Massa = Kons entrasi x mr x V larutan = 0.05 * 250 * 392 = 4900 mg Relative Standard deviation , RSD = 100 . s /
=
%
RSD untuk konsentrasi
( ) + ( ) = √ . + . = 0.020 %
RSD =
× . ⁄ = 0.0029
-10
8.
-10
The solubility product of barium sulphate is 1.3 x 10 , with a standard deviation of 0.1 x 10 Calculate the standard deviation of the calculated solubility of barium sulphate in w ater. Answer :
Ksp BaSO4 = 1.3 x 10 -10 , s BaSO4 = 0.1 x 10 -10
BaSO4 dalam air maka.. Ba 2+ (aq) + SO42Ksp = [Ba
2+
] . [SO42-]
1.3 . 10-10 = x . x
, dimana x = Ksp dalam air
X2 = 1.3 . 10-10 Ksp dalam air = 1.14 . 10 -5, Dengan rum us pad a (2.11.6)
= .. . . = .. .
s.d dalam air =
= 4.4 x 10-7 = 0.44 x 10-6 M
