Penggunaan Sari Buah Jeruk Nipis (Citrus aurantifolia (Christm&Panzer) Swingle) sebagai Asam pada Pembuatan Granul Effervescent Vitamin C

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Lampiran 1. Gambar buah jeruk nipis

Gambar buah jeruk nipis

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Lampiran 3. Bagan alir proses penyarian dan pengeringan sari buah jeruk nipis

- dicuci bersih

- dipotong menjadi dua bagian

- diperas menggunakan alat pemeras jeruk

- dikeringkan menggunakan Freeze Dryer

- ditambahkan maltodekstrin sebanyak 5 bagian - dikeringkan di lemari pengering

10 kg Buah JerukNipis

Sari Buah Jeruk Nipis

Sari Kental

Ampas

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Lampiran 4. Gambar sari buah jeruk nipis dan serbuk sari buah jeruk nipis

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Lampiran 5. Data dan perhitungan kadar vitamin C dari serbuk sari buah jeruk nipis

Rumus Kesetaraan :

Kesetaraan = Va×W×%Kadar Vc×(Vt-Vb) Keterangan:

Va = Volume aliquot (ml) W = Berat vitamin C (mg) Vt = Volume titrasi (ml) Vb = Volume blanko (ml) Vc = Volume labu tentukur (ml) Rumus Kadar Vitamin C :

Kadar vitamin C = (Vt-Vb)×Kesetaraan×Vl Vp×Bs

Keterangan:

Vt : Volume titrasi (ml) Vb : Volume blanko (ml) Vl : Volume labu tentukur (ml) Vp : Volume pemipetan (ml) Bs : Berat sampel (g)

Data Kesetaraan : Volume

Volume titrasi (ml) Volume blanko

Kesetaraan = Va×W×%Kadar Vc×(Vt-Vb)

Kesetaraan =

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Lampiran 5. (Lanjutan)

Kesetaraan = 1,998 mg 10,9 ml Kesetaraan = 0,183mg�_ml

Data Kadar Vitamin C :

Kadar vitamin C = (24,5-0,1)×0,183 mg

ml � ×50 2×0,2

Kadar vitamin C = 223,26 0,4

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Lampiran 6. Bagan alir proses pembuatan granul effervescent

- digerus sebagian laktosa - digerus sebagian laktosa - ditambah orange pasta - ditambah orange pasta - digerus sampai homogen - digerus sampai homogen - ditambahkan natrium bikarbonat - ditambahkan serbuk sari dan sebagian maltodekstrin buah jeruk nipis dan

sebagian maltodekstrin

- dicampur hingga homogen - dicampur hingga homogen

- ditambahkan beberapa tetes - ditambahkan beberapa tetes alkohol 96% sampai di dapat alkohol 96% sampai di konsistensi yang mudah dikepal dapat konsistensi yang

mudah dikepal

- massa diayak dengan ayakan 16 mesh - massa diayak dengan ayakan 16 mesh - dikeringkan di lemari pengering - dikeringkan di lemari

pengering

- diayak dengan 20 mesh - diayak dengan 20 mesh

- ditambahkan vitamin C

- ditambahkan natrium metabisulfit - ditambahkan PEG 6000

- diaduk sampai homogen Bagian Asam Bagian Basa

Bagian Basa + Bagian Asam

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Lampiran 7. Perhitungan bahan granul effervescent

R/ Vitamin C 500mg

Natrium Metabisulfit 0,1% Natrium Bikarbonat X Sari Buah Jeruk Nipis Y

Maltodekstrin 2%

Sakarin 1%

Orange Pasta 2%

PEG 6000 3%

Laktosa q.s

Keterangan : X, Y perbandingan asam basa

• Formula 1

Natrium bikarbonat 100 mg + Sari Buah Jeruk Nipis 100 mg

• Formula 2

• Formula 3

• Formula 4

• Formula 5

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Lampiran 7. (Lanjutan) I. Rencana Kerja

1. Berat 1 sachet = 2000 mg

2. Metode = Granulasi Kering II. Perhitungan Bahan

1. Vitamin C = 500 mg x 100 sachet = 50 gram 2. Natrium meta bisulfit = 2 mg x 100 sachet = 0,2 gram 3. Asam-Basa

a. Formula 1

• Natrium bikarbonat = 100 mg x 100 sachet = 10 gram

• Sari Buah Jeruk Nipis = 100 mg x 100 sachet = 10 gram b. Formula 2

• Sari Buah Jeruk Nipis = 200 mg x 100 sachet = 20 gram c. Formula 3

• Sari Buah Jeruk Nipis = 300 mg x 100 sachet = 30 gram d. Formula 4

• Sari Buah Jeruk Nipis = 100 mg x 100 sachet = 10 gram e. Formula 5

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5. Sakarin = 20 mg x 100 sachet = 2 gram 6. PEG 6000 = 60 mg x 100 sachet = 6 gram

7. Orange Pasta = 40 mg x 100 sachet = 4 gram

8. Laktosa

• Formula 1 = 1138 mg x 100 sachet = 113,8 gram

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Lampiran 8. Data dan perhitungan uji sudut diam Rumus : Tg Ɵ = 2h/d

Keterangan : Ɵ = sudut diam

h = tinggi kerucut (cm) d = diameter (cm) Formula 1

Uji ke- Diameter (d) Tinggi (h)

1 14,3 3,8

2 14,2 3,7

3 14,8 4,0

Rata-rata 14,43 3,83

Tg Ɵ = 2h d Tg Ɵ = 2×3,83

14,43 Ɵ = arc tg 0,5308 Ɵ = 27,96° Formula 2

1 13,8 3,6

2 14,5 3,4

3 14,0 3,4

Rata-rata 14,1 3,47

Tg Ɵ = 2h d Tg Ɵ = 2×3,47

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Lampiran 8. (Lanjutan) Formula 3

1 13,2 3,9

2 13,7 4,1

3 13,1 3,8

Tg Ɵ = 2h d Tg Ɵ = 2×3,93

13,33 Ɵ = arc tg 0,5896 Ɵ = 30,52° Formula 4

1 13,9 4,0

2 14,5 3,9

3 14,1 3,9

Tg Ɵ = 2h d Tg Ɵ = 2×3,93

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1 14,6 4,1

2 13,9 3,9

3 14,3 3,9

Tg Ɵ = 2h d Tg Ɵ = 2×3,97

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Lampiran 9. Data dan perhitungan uji indeks tap Rumus :

Indeks tap = V1-V2

V₁ x 100% Keterangan : V1 = volume sebelum hentakan

V2 = volume setelah hentakan Formula 1

Uji ke- V1 V2 Indeks Tap (%)

1 24 20,75 13,54

2 23,75 21 11,58

3 23,75 20,5 13,68

Rata-rata 12,93

Uji ke-1 Uji ke-3 Indeks tap = 24-20,75

24 x 100% Indeks tap = Indeks tap = 23,75-21

23,75 x 100%

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Lampiran 9. (Lanjutan) Indeks tap = 23,75-18,75

23,75 x 100% Indeks tap =

Rata-rata 21,48

Uji ke-1 Uji ke-3 Indeks tap = 23,75-19,00

23,75 x 100% Indeks tap =

Indeks tap = 23,75-18,25

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Rata-rata 17,89

Uji ke-1 Uji ke-3 Indeks tap = 23,75-19,5

23,75 x 100% Indeks tap = Indeks tap = 23,5-19,5

23,5 x 100%

Rata-rata 14,32

Uji ke-1 Uji ke-3 Indeks tap = 24-19,75

24 x 100% Indeks tap =

Indeks tap = 23,75-21

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Lampiran 10. Data dan perhitungan uji kadar air Rumus :

Kadar air = berat awal - berat akhir

berat awal x 100%

berat awal x 100% Kadar air =

berat awal - berat akhir

berat awal x 100% Kadar air = 14,210-14,200

14,210 x 100% Kadar air =

14,232-14,232

14,232 x 100%

Kadar air = 0,07 % Kadar air = 0 %

Uji ke-2

14,275 x 100%

13,420 x 100% Kadar air =

13,228-13,228

13,228 x 100%

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Lampiran 10. (Lanjutan) Uji ke-3

13,105 x 100%

14,070 x 100% Kadar air =

14,153-14,153

14,153 x 100%

Kadar air = 0,14 % Kadar air = 0 %

Uji ke-2

13,970 x 100%

14,010 x 100% Kadar air =

14,015-14,000

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Lampiran 10. (Lanjutan) Uji ke-3

13,971 x 100% Kadar air = 17,02 %

Formula 5

Uji ke- Berat awal (g) Berat akhir (g) Kadar Air (%)

1 21,280 21,250 0,14

2 21,280 21,260 0,09

3 21,177 21,160 0,08

Rata-rata 0,11

Uji ke-1 Uji ke-3

21,280 x 100% Kadar air =

21,177-21,160

21,177 x 100% Kadar air = 0,14 % Kadar air = 0,08 %

Uji ke-2

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Lampiran 11. Data dan perhitungan uji kadar vitamin C dari granul effervescent Rumus Kesetaraan :

Kesetaraan = Va×W×%Kadar Vc×(Vt-Vb) Keterangan:

Va = Volume aliquot (ml) W = Berat vitamin C (mg) Vt = Volume titrasi (ml) Vb = Volume blanko (ml) Vc = Volume labu tentukur (ml) Rumus Kadar Vitamin C :

Keterangan:

Vt : Volume titrasi (ml) Data Kesetaraan :

Volume

Kesetaraan =

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Kesetaraan = 1,998 mg 15 ml Kesetaraan = 0,1332mg�_ml Data Kadar Vitamin C :

Kadar vitamin C secara praktek :

Kadar vitamin C praktek = (Vt-Vb)×Kesetaraan×Vl Vp×Bs

Kadar vitamin C praktek = (19,73-0,2)×0,1332 mg

ml � ×50 2×0,2065

Kadar vitamin C praktek = 130,0698 0,413 Kadar vitamin C praktek = 314,94mg�_g Kadar vitamin C praktek = 0,315mg�_mg Kadar vitamin C secara teoritis :

Kadar vitamin C teoritis = 2,72668 mg 8,26 mg Kadar vitamin C teoritis = 0,3301 mg�_mg Kadar vitamin C :

Kadar vitamin C = Kadar vitamin C praktek

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Kesetaraan =

2×50×99,9�₁₀₀ 50×(15,2-0,2) Kesetaraan = 1,998 mg

15 ml Kesetaraan = 0,1332mg�_ml Data Kadar Vitamin C :

ml � ×50 2×0,206

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Kadar vitamin C secara teoritis : Kadar vitamin C teoritis = 2,7 mg

8,24 mg Kadar vitamin C teoritis = 0,3277 mg�_mg Kadar vitamin C :

Kadar vitamin C teoritis× 100% Kadar vitamin C = 0,3212

mg Data Kesetaraan :

Volume

Kesetaraan =

2×50×99,9�₁₀₀ 50×(11-0,1) Kesetaraan = 1,998 mg

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ml � ×50 2×0,201

Kadar vitamin C praktek = 133,8645 0,402

Kadar vitamin C praktek = 332,9963mg�_g Kadar vitamin C praktek = 0,333mg�_mg Kadar vitamin C secara teoritis :

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Kesetaraan =

2×50×99,9�₁₀₀ 50×(15,2-0,2) Kesetaraan = 1,998 mg

ml � ×50 2×0,2065

Kadar vitamin C praktek = 313,3264mg�_g mg

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Kadar vitamin C secara teoritis : Kadar vitamin C teoritis = 2,7166 mg

8,26 mg Kadar vitamin C teoritis = 0,3289 mg�_mg Kadar vitamin C :

Kadar vitamin C teoritis× 100% Kadar vitamin C = 0,3133 Data Kesetaraan :

Volume

Kesetaraan =

2×50×99,9�₁₀₀ 50×(15,2-0,2) Kesetaraan = 1,998 mg

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ml � ×50 2×0,2065

Kadar vitamin C praktek = 315,5840mg�_g Kadar vitamin C praktek = 0,3156mg�_mg Kadar vitamin C secara teoritis :

Kadar vitamin C teoritis× 100% Kadar vitamin C = 0,3156

mg mg �

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Lampiran 12. Gambar granul effervescent dan kemasan granul effervescent

Gambar sachet tanpa etiket Gambar granul effervescent F1

Gambar granul effervescent F2 Gambar granul effervescent F3

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Gambar sachet dengan etiket (depan)

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Lampiran 13. Kuesioner uji kesukaan granul effervescent Nama :

Usia :

Pekerjaan : Petunjuk :

1. Anda akan menerima 5 (lima) sampel serbuk effervescent vitamin C 2. Larutkan serbuk tersebut ke dalam air putih yang telah tersedia 3. Amati dispersa yang terjadi, setelah selesai kemudian dicoba

4. Sebelum mencoba, netralkan mulut anda dengan meminum air putih yang telah tersedia

5. Setelah mencoba formula 1, netralkan kembali mulut anda dengan air putih untuk mencoba formula 2.

6. Setelah mencoba formula 2 netralkan kembali mulut anda dengan air putih untuk mencoba formula 3. Begitu seterusnya hingga formula 5

7. Berikan penilaian pada kolom di bawah ini

Formula 1 Formula 2 Formula 3 Formula 4 Formula 5 Dispersa

Rasa Warna Aroma

Medan, Mei 2015

Volunteer

( ) Keterangan :

1. Sangat Tidak Suka 2. Tidak Suka

3. Netral 4. Suka

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Lampiran 18. Rumus perhitungan nilai kesukaan granul effervescent

Untuk menghitung nilai kesukaan rerata dari setiap panelis digunakan rumus sebagai berikut :

�(ỹ−(1,96 ×�/√�))≤ � ≤(ỹ+ (1,96 ×�/√�))≅ 95%

ỹ=∑ ��

� �=1

�

�2 ₌ ∑ (�� −ỹ) 2

� �=1

�

�=��2

Keterangan :

n : banyak panelis

S2 : keragaman nilai kesukaan

1,96 : koefisien standar deviasi pada taraf 95% ỹ : nilai kesukaan rata-rata

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Lampiran 19. Perhitungan nilai kesukaan dispersi granul effervescent Formula 1

ỹ=∑ xi n i=1

n

ỹ=4+5+2+5+3+…+5 30

ỹ=121 30 ỹ=4,0333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(4-4,0333) 2

+(5-4,0333)2+(2-4,0333)2+…+(5-4,0333)2 30

S2=28,9667 30 S2=0,9656 S=�S2

S=�0,9656 S=0,9827

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (4,0333-(1,96×0,9827/√30)) ≤ μ ≤ (4,0333+(1,96×0,9827/√30)) P (4,0333-0,3517) ≤ μ ≤ (4,0333+0,3517)

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ỹ=∑ xi n i=1

n

ỹ=5+4+5+4+5+…+4 30

ỹ=117 30 ỹ=3,9000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(5-3,9000) 2

+(4-3,9000)2+(5-3,9000)2+…+(4-3,9000)2 30

S2=28,7000 30 S2=0,9567 S=�S2

S=�0,9567 S=0,9781

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,9000-(1,96×0,9781/√30)) ≤ μ ≤ (3,9000+(1,96×0,9781/√30)) P (3,9000-0,3500) ≤ μ ≤ (3,9000+0,3500)

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ỹ=∑ xi n i=1

n

ỹ=1+3+4+2+4+…+3 30

ỹ=111 30 ỹ=3,7000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(1-3,7000) 2

+(3-3,7000)2+(4-3,7000)2+…+(3-3,7000)2 30

S2=30,3000 30 S2=1,0100 S=�S2

S=�1,0100 S=1,0050

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,7000-(1,96×1,0050/√30)) ≤ μ ≤ (3,7000+(1,96×1,0050/√30)) P (3,7000-0,3596) ≤ μ ≤ (3,7000+0,3596)

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ỹ=∑ xi n i=1

n

ỹ=3+2+3+3+2+…+1 30

ỹ=67 30 ỹ=2,2333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(3-2,2333) 2

+(2-2,2333)2+(3-2,2333)2+…+(1-2,2333)2 30

S2=15,3667 30 S2=0,5122 S=�S2

S=�0,5122 S=0,7157

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,2333-(1,96×0,9827/√30)) ≤ μ ≤ (2,2333+(1,96×0,9827/√30)) P (2,2333-0,2561) ≤ μ ≤ (2,2333+0,2561)

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ỹ=∑ xi n i=1

n

ỹ=2+1+1+1+1+…+2 30

ỹ=34 30 ỹ=1,1333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(2-1,1333) 2

+(1-1,1333)2+(1-1,1333)2+…+(2-1,1333)2 30

S2=3,4667 30 S2=0,1156 S=�S2

S=�0,1156 S=0,3400

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (1,1333-(1,96×0,3400/√30)) ≤ μ ≤ (1,1333+(1,96×0,3400/√30)) P (1,1333-0,1217) ≤ μ ≤ (1,1333+0,1217)

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Lampiran 20. Perhitungan nilai kesukaan rasa granul effervescent Formula 1

ỹ=∑ xi n i=1

n

ỹ=4+3+1+2+3+…+1 30

ỹ=84 30 ỹ=2,8000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(4-2,8000) 2

+(3-2,8000)2+(1-2,8000)2+…+(1-2,8000)2 30

S2=60,8000 30 S2=2,0267 S=�S2

S=�2,0267 S=1,4236

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,8000-(1,96×1,4236/√30)) ≤ μ ≤ (2,8000+(1,96×1,4236/√30)) P (2,8000-0,5094) ≤ μ ≤ (2,8000+0,5094)

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ỹ=∑ xi n i=1

n

ỹ=5+5+3+5+5+…+2 30

ỹ=110 30 ỹ=3,6667

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(5-3,6667) 2

+(5-3,6667)2+(3-3,6667)2+…+(2-3,6667)2 30

S2=46,6667 30 S2=1,5556 S=�S2

S=�1,5556 S=1,2472

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,6667-(1,96×1,2472/√30)) ≤ μ ≤ (3,6667+(1,96×1,2472/√30)) P (3,6667-0,4463) ≤ μ ≤ (3,6667+0,4463)

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ỹ=∑ xi n i=1

n

ỹ=3+1+4+4+4+…+4 30

ỹ=111 30 ỹ=3,7000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(3-3,7000) 2

+(1-3,7000)2+(4-3,7000)2+…+(4-3,7000)2 30

S2=36,3000 30 S2=1,2100 S=�S2

S=�1,2100 S=1,1000

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,7000-(1,96×1,1000/√30)) ≤ μ ≤ (3,7000+(1,96×1,1000/√30)) P (3,7000-0,3936) ≤ μ ≤ (3,7000+0,3936)

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ỹ=∑ xi n i=1

n

ỹ=1+2+2+1+1+…+3 30

ỹ=76 30 ỹ=2,5333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(1-2,5333) 2

+(2-2,5333)2+(2-2,5333)2+…+(3-2,5333)2 30

S2=29,4667 30 S2=0,9822 S=�S2

S=�0,9822 S=0,9911

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,5333-(1,96×0,9911/√30)) ≤ μ ≤ (2,5333+(1,96×0,9911/√30)) P (2,5333-0,3547) ≤ μ ≤ (2,5333+0,3547)

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ỹ=∑ xi n i=1

n

ỹ=2+4+5+3+2+…+5 30

ỹ=69 30 ỹ=2,3000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(2-2,3000) 2

+(4-2,3000)2+(5-2,3000)2+…+(5-2,3000)2 30

S2=76,3000 30 S2=2,5433 S=�S2

S=�2,5433 S=1,5948

P (ỹ-(1,96×s/√n)) ≤ μ ≤ (ỹ+(1,96×s/√n))≅95%

P (2,3000-(1,96×1,5948/√30)) ≤μ≤ (2,3000+(1,96×1,5948/√30)) P (2,3000-0,5707) ≤μ≤ (2,3000+0,5707)

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Lampiran 21. Perhitungan nilai kesukaan warna granul effervescent Formula 1

ỹ=∑ xi n i=1

n

ỹ=3+2+2+2+3+…+2 30

ỹ=98 30 ỹ=3,2667

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(3-3,2667) 2

+(2-3,2667)2+(2-3,2667)2+…+(2-3,2667)2 30

S2=49,8667 30 S2=1,6622 S=�S2

S=�1,6622 S=1,2893

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,2667-(1,96×1,2893/√30)) ≤ μ ≤ (3,2667+(1,96×1,2893/√30)) P (3,2667-0,4614) ≤ μ ≤ (3,2667+0,4614)

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ỹ=∑ xi n i=1

n

ỹ=4+4+5+5+4+…+4 30

ỹ=120 30 ỹ=4,0000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(4-4,0000) 2

+(4-4,0000)2+(5-4,0000)2+…+(4-4,0000)2 30

S2=20,0000 30 S2=0,6667 S=�S2

S=�0,6667 S=0,8165

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (4,0000-(1,96×0,8165/√30)) ≤ μ ≤ (4,0000+(1,96×0,8165/√30)) P (4,0000-0,2922) ≤ μ ≤ (4,0000+0,2922)

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ỹ=∑ xi n i=1

n

ỹ=5+3+4+4+5+…+5 30

ỹ=121 30 ỹ=4,0333

S2₌∑ (xi-ӯ) 2 n i=1

n

S2=(5-4,0333) 2

+(3-4,0333)2+(4-4,0333)2+…+(5-4,0333)2 30

S2=26,9667 30 S2=0,8989 S=�S2

S=�0,8989 S=0,9481

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (4,0333-(1,96×0,9481/√30)) ≤ μ ≤ (4,0333+(1,96×0,9481/√30)) P (4,0333-0,3393) ≤ μ ≤ (4,0333+0,3393)

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ỹ=∑ xi n i=1

n

ỹ=2+1+3+1+2+…+1 30

ỹ=59 30 ỹ=1,9667

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(2-1,9667) 2

+(1-1,9667)2+(3-1,9667)2+…+(1-1,9667)2 30

S2=20,9667 30 S2=0,6989 S=�S2

S=�0,6989 S=0,8360

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (1,9667-(1,96×0,8360/√30)) ≤ μ ≤ (1,9667+(1,96×0,8360/√30)) P (1,9667-0,2992) ≤ μ ≤ (1,9667+0,2992)

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ỹ=∑ xi n i=1

n

ỹ=1+5+1+3+1+…+3 30

ỹ=52 30 ỹ=1,7333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(1-1,7333) 2

+(5-1,7333)2+(1-1,7333)2+…+(3-1,7333)2 30

S2=37,8667 30 S2=1,2622 S=�S2

S=�1,2622 S=1,1235

P (ỹ-(1,96×s/√n)) ≤ μ ≤ (ỹ+(1,96×s/√n))≅95%

P (1,7333-(1,96×1,1235/√30)) ≤μ≤ (1,7333+(1,96×1,1235/√30)) P (1,7333-0,4020) ≤μ≤ (1,7333+0,4020)

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Lampiran 22. Perhitungan nilai kesukaan aroma granul effervescent Formula 1

ỹ=∑ xi n i=1

n

ỹ=2+4+1+2+3+…+1 30

ỹ=86 30 ỹ=2,8667

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(2-2,8667) 2

+(4-2,8667)2+(1-2,8667)2+…+(1-2,8667)2 30

S2=53,4667 30 S2=1,7822 S=�S2

S=�1,7822 S=1,3350

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,8667-(1,96×1,3350/√30)) ≤ μ ≤ (2,8667+(1,96×1,3350/√30)) P (2,8667-0,4777) ≤ μ ≤ (2,8667+0,4777)

(52)

ỹ=∑ xi n i=1

n

ỹ=3+3+3+5+5+…+2 30

ỹ=112 30 ỹ=3,7333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(3-3,7333) 2

+(3-3,7333)2+(3-3,7333)2+…+(2-3,7333)2 30

S2=29,8667 30 S2=0,9956 S=�S2

S=�0,9956 S=0,9978

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,7333-(1,96×0,9978/√30)) ≤ μ ≤ (3,7333+(1,96×0,9978/√30)) P (3,7333-0,3571) ≤ μ ≤ (3,7333+0,3571)

(53)

ỹ=∑ xi n i=1

n

ỹ=5+2+4+3+4+…+5 30

ỹ=118 30 ỹ=3,9333

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(5-3,9333) 2

+(2-3,9333)2+(4-3,9333)2+…+(5-3,9333)2 30

S2=31,8667 30 S2=1,0622 S=�S2

S=�1,0622 S=1,0306

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (3,9333-(1,96×1,0306/√30)) ≤ μ ≤ (3,9333+(1,96×1,0306/√30)) P (3,9333-0,3688) ≤ μ ≤ (3,9333+0,3688)

(54)

ỹ=∑ xi n i=1

n

ỹ=4+1+2+1+2+…+3 30

ỹ=65 30 ỹ=2,1667

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(4-2,1667) 2

+(1-2,1667)2+(2-2,1667)2+…+(3-2,1667)2 30

S2=30,1667 30 S2=1,0056 S=�S2

S=�1,0056 S=1,0027

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,1667-(1,96×1,0027/√30)) ≤ μ ≤ (2,1667+(1,96×1,0027/√30)) P (2,1667-0,3588) ≤ μ ≤ (2,1667+0,3588)

(55)

ỹ=∑ xi n i=1

n

ỹ=1+5+5+4+1+…+4 30

ỹ=69 30 ỹ=2,3000

S2₌∑ (xi-ỹ) 2 n i=1

n

S2=(1-2,3000) 2

+(5-2,3000)2+(5-2,3000)2+…+(4-2,3000)2 30

S2=76,3000 30 S2=2,5433 S=�S2

S=�2,5433 S=1,5948

P (ỹ-(1,96×s/√n)) ≤μ≤ (ỹ+(1,96×s/√n))≅95%

P (2,3000-(1,96×1,5948/√30)) ≤ μ ≤ (2,3000+(1,96×1,5948/√30)) P (2,3000-0,5707) ≤ μ ≤ (2,3000+0,5707)

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Lampiran 23. Gambar volunter uji kesukaan

Gambar volunter uji kesukaan

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Lampiran 24. Gambar alat-alat yang digunakan

Gambar alat waktu alir dan sudut diam Gambar beaker glass

Gambar labu tentukur

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Lampiran 25. Gambar bahan yang digunakan

Gambar pewarna orange pasta Gambar laktosa