TUGAS STATISTIKA KIMIA Exercise 2 1 8 Statistics of repeated measurements Anggota Kelompok

1. Rut Novalia R. Sianipar (J1B114601)

2. Rina Apriani (J1B114602)

3. Diky Subhanuddin (J1B114603)

4. Wenny Yuniandari (J1B108213)

1. The reproducibility of a method for the 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 following results were obtained:

0.07 0.07 0.08 0.07 0.07 0.08 0.08 0.09 0.08 g g-1

(Moreno-Dominguez, T., Garcia-Moreno, C. and Marine-Font, A., 1983, Analyst, 108: 505) Calculate the mean, standard deviation and relative standard deviation of these results.

Answer :

No Sample Xi (g/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

Mean, =

= .

= 0.077 (g/g)

Standard deviation, s = ( )() = . = 0.007 Relative Standard deviation, RSD = 100 . s / = . .

= 9. 09 %

2. The morphine levels (%) of seven batches of seized heroin were 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 :

Mean, =

= .

= 19.01 %

Standard deviation, s = ( )() = . = 3.43 % Confidence 95 % = tn-1 . s /

= 19.01 2.45 . 3.43 / = 19.01 3.17

Confidence 99 % = tn-1 . s /

= 19.01 7.71 . 3.43 / = 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 following 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

No Sample Xi (g/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 Total 223.2 17.0560

Mean, =

= .

= 22.32 (ng/ml)

No Sample Xi (g/g) (Xi - ) (Xi - )^2 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 133.1 70.4687

Standard deviation, s = ( )() = . = 1.377 ng/ml Relative Standard deviation, RSD = 100 . s / = . .

= 6.17 %

Confidence 99 % = tn-1 . s /

= 22.32 3.25 . 1.377 / = 22.32 1.41 ng/ml Untuk data kedua :

No Sample Xi (g/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

4.5334

Mean, =

=

= 12.83 (ng/ml)

Standard deviation, s = ( )() = . = 0.95 ng/ml Relative Standard deviation, RSD = 100 . s / = . .

= 7.42 %

Confidence 99 % = tn-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 sample 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 :

Confidence 95 % = tn-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, bila 0.1 ng/ml maka n adalah.

0.1 = 1.96 . s /

0.1 = 1.96 x 0.64 /

= .. . = 12. 544 n = 157

5. 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 %.

Confidence 95 % = tn-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 tingkat 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 decide whether there is any evidence of systematic error.

Answer :

No Sample Xi (g/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 0.1375

Mean, =

= .

= 10.18 ng/ml

Standard deviation, s = ( )() = . = 0.185 % Confidence 95 % = tn-1 . s /

= 10.18 2.78 . 0.185 / = 10.18 0.23 Berdasarkan data diatas, tidak 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 deviasi setiap penimbangan adalah 0.0001 g

Standard deviasi untuk penimbangan adalah

S = (.) + (.) = 0.00014 g = 0.14 mg Diketahui V larutan = 250 ml, Konsetrasi 0.05 M, Masa Relatif 40 Maka berat yang ditimbang adalah. Massa = Konsentrasi x mr x V larutan

= 0.05 * 250 * 40 = 500 mg

Relative Standard deviation, RSD = 100 . s / = . = 0.028 %

SD untuk volume solvent adalah 0.05 ml

Relative Standard deviation, RSD = 100 . s / = . = 0.02 %

RSD untuk konsentrasi

RSD = () + () = . + . = 0.034 %

Perhitungan untuk pereaksi dengan massa relative (mr) 392, dengan asumsi V

dan konsentrasi tetap. Maka banyaknya pereaksi ditimbang adalah Massa = Konsentrasi x mr x V larutan = 0.05 * 250 * 392 = 4900 mg

Relative Standard deviation, RSD = 100 . s / = . = 0.0029 % RSD untuk konsentrasi

RSD = () + () = . + . = 0.020 %

8. The solubility product of barium sulphate is 1.3 x 10-10 , with a standard deviation of 0.1 x 10-10 Calculate the standard deviation of the calculated solubility of barium sulphate in water.

Answer : Ksp BaSO4 = 1.3 x 10-10 , s BaSO4 = 0.1 x 10-10

BaSO4 dalam air maka.. Ba 2+ (aq) + SO42-

Ksp = [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 rumus pada (2.11.6)

=

s.d dalam air =

= .......

= 4.4 x 10-7 = 0.44 x 10-6 M