Beef Ball Combine

FST 4822

LABORATORY FOR CHEMISTRY AND TECHNOLOGY OF PLANT AND ANIMAL PRODUCTS

EXPERIMENT 2: PROCESSING OF BEEF BALLS

LECTURER : DR.SEYED HAMED MIRHOSSEINI DATE : 10TH AUGUST 2010GROUP : 1 (TUESDAY 12pm-3pm; 4-7pm)GROUP MEMBERS : MATRIC NO.

1. SITI HAJAR BT MD ALI 1469562. SALEHA BT HASSAN 1472233. WOO CHYE LIN 1478354. NURUL FARADINA AZWA BINTI ROSMAN 1481295. LIM KANG XIANG 1486266. GOH KONG SONG 1489977. LEE KAR LENG 1496998. SITI ALYANI BINTI MAT 150871

TITLE: Processing of Beef Ball

OBJECTIVES:

1. To know the way of making or produce the beef ball.

2. To examine the influence of processing condition and starch addition on sensory attributes,

overall quality and acceptability of beef balls.

INTRODUCTION:

Nowadays processing of meat has been world wide. Advances in technology have created

quality in processing method and also variety in meat products. The purposes of meat processing

are to prevent spoilage and also to provide flavorful and nutritious products (Pearson and Gillet,

1996).

One of the meat products is beef ball. Beef ball which is known as meat ball is a minced

meat product. Technically the production of beef balls is almost similar with the production of

fish ball. Whereby, the texture of the product is contributed by formation of a stable network

between protein and starch molecules. The solubilization of the protein occurs in the presence of

sodium chloride. Some of the restricted meat products are convenience foods which are prepared

from a mixture of comminuted mat, protein fat particles, water and carbohydrate (Barbut, 1995).

During the processing the meat is mixed with salt, sugar, black pepper, garlic

Tripolyphospate (TPP) and corn starch that will mix directly. The addition of salt in beef ball

processing will contribute to ionic strength of the system and also alter the pH of meat.

Furthermore, sugar also is added to serve as preservatives and flavor effects. Spices which are

black pepper can not be depended upon for any preservatives action (Pricel and Schweigert,

1987). In this experiment, the effect of corn starch towards meat during processing of beef ball

will be determined. Corn starch which act as thickening agents will give different result in the

quality and texture of beef ball. The mixture is then formed to the desired shape and this shape is

retained after freezing and cooking Fischer (1996) noted that beef ball is prepared from ground

meat formed in balls and boiled, and it is very important to consider the formula of the batter.

Beef balls are processed comminuted meat which can be classified as restricted meat and

it is very popular among some countries within the Asian region and certain European Countries.

The Asian type of beef balls are commonly produced by emulsifying fine ground meat with

starch of some sort, mixing salt and certain herbs specific to the ethnic cuisine and finally

shaping into balls.

MATERIALS AND PROCEDURES:

Materials:

Ingredients Weight (g)

Beef

Salt

Sugar

TPP

Garlic

Black pepper

Ice

1000.0

20.0

5.0

25.0

2.5

2.5

50.0

Apparatus:

Meat cutter

Meat grinder

Bowls

Balance

Mixer

Meatball machine

Spatula and spoon

PROCEDURES:

Meat was cut into cubes.

Beef cubes were washed before grinding.

Salt is added and mixed for 5 minutes

The rest of the ingredients were added and mixed for 10 minutes

The total weight of mixture is determined

Mixture is weighted into 30g for each beef balls and cooled for a while

Mixture was shaped into beef balls using hand and freeze.

Beef balls are then boiled.

The beef balls are weighed again and the texture determined

Analysis of pH, moisture, ash, protein and fats of beef ball and sensory evaluation are carried out

Chemical Analysis (Procedure)

a) Determination of pH:

The pH meter firstly needs to be calibrated by using buffer solution which is pH 4 and pH 7

The pH value will be determine when the reading is constant

b) Determination of moisture content via oven method:

The oven heat to 105°C and keep the temperature constant

Crucible dried and cleaned in the oven

By using thong, the crucible transferred into dessicator to cool

The crucible placed on a balance and weigh rapidly and accurately

The crucible returned in the dessicator

The weighing procedure repeated at least twice until a constant reading is obtained

2 to 5 g of sample weighed into the crucible mentioned above

The crucible together with the sample placed into the oven and leaved for at least 7 hours

The crucible was took out by using a thong and placed directly in the dessicator

After cool, the crucible and sample is weight

Step 10 was repeated to constant weight

c) Determination of ash via drying method:

The muffle furnace to 105°C and keep the temperature constant

Crucible dried and cleaned in the oven

By using thong, the crucible transferred into dessicator to cool

The crucible placed on a balance and weigh rapidly and accurately

The crucible returned in the dessicator

The weighing procedure repeated at least twice until a constant reading is obtained

3 to 5 g of sample weighed into the crucible mentioned above

Crucible is inserted together with sample into the muffle furnace

The crucibles and sample is left burned for 1 hour until no black particles present to obtain

permanent weight

The crucibles and sample is left cooled in the dessicator

Both crucible and ash is weighed for the record.

d) Determination of crude protein:

0.15g of dry sample is weighed accurately. The sample is being placed in micro Kjeldhal test

tube. For blank, no sample is used.

0.8g of mixed catalyst is added.

2.5ml of concentrated sulphuric acid is added and the tube is swirled gently to mix the content.

The tube is being heated slowly on the heating coil under fume hood. The content is being boiled

until the solution becomes clear and gives blue-green colour. The boiling is being continued for

another 10 minutes.

The flask is cooled.

A few ml of distilled water is added and the digested product is being transferred into the

distillation tube.

10ml of 45% NaOH solution is added slowly to separate the two layers of solution. The

distillation tube is being fixed neatly to the condenser.

In a conical flask, 10ml of 0.05N boric acid and a few drops of indicator are added.

The conical flask is placed at the distillate platform and the tip of distillate tube is immersed into

the acid solution.

The content of distillation flask is mixed by swirling it gently. Steam is being purged into the

flask.

The ammonia solution is being distilled into the conical flask for about 120 ml.

After mixing the distillation product by swirling the flask gently, the unreacted boric acid is

being titrated with 0.05N H2SO4 until neutral.

The same procedure is repeated for blank.

e) Determination of crude fat

Distillation/round bottom flask is dried in the oven for about 30 minute at 105°C

The dry sample following previous procedure is used and the dry weight of sample is recorded

The thimble is weighted and the dry sample is placed in the thimble. Then, the thimble is

inserted into Soxhlet apparatus

The round bottom flask is weighted accurately and 200 mL petroleum is poured into it

The Soxhlet is connected to the reflux and round bottom flask and then the sample is refluxed

continuously for about 8 hours

For the caution, the water content in the bath is ensured sufficiently for the refluxing process and

added if necessary and the temperature of heater is controlled

The petroleum ether is ensured not evaporated or dry during refluxing and if so, the petroleum

ether is added

After 8 hours, the petroleum ether in the round bottom flask is evaporated using rotary

evaporator

After that, the round bottom flask is placed in the oven for about 15 minutes at 105°C

The round bottom flask is placed into desiccators for about 30 minutes

The round bottom flask with the fat extracted is weighted together

RESULT

Table 1: Analysis of Beef Meat Ball

Samples Test

1 2 3 4

pH6.44 6.45 6.68 6.65 6.61 6.60 6.63 6.61

6.445±7.07 x10−3 6.665±0.0212 6.605±7.07 x10−3 6.620±0.014

Moisture (%)

28.78 29.31 59.76 59.83 63.93 63.46 66.38 65.2529.045±0.3748 59.795±0.049

563.695±0.3323 65.815±0.799

0

Ash (%)4.82 4.28 2.49 2.51 4.71 4.88 4.06 4.024.550±0.3818 2.500±0.0141 4.795±0.1202 4.040±0.0283

Fat (%)0.20 0.24 0.04 0.04 0.20 0.13 0.23 0.290.220±0.0283 0.040±0.0028 0.165±0.0495 0.260±0.0424

Protein (%)

0.54 0.54 14.88 15.46 12.60 12.07 14.73 11.490.540±0.0 15.170±0.410

112.335±0.3748 13.110±2.291

0Yield of beef ball 48 48 51 55

Table 2: Sensory Evaluation on Beef Meat Ball

SamplesWithout Corn Starch(control) With Corn Starch

1 2 3 4 Group

SensoryA B C D A B C D A B C D A B C D

Texture

4 4 2 3 4 3 2 3 2 1 2 1 2 2 3 4

2 1 2 3 3 3 3 2 3 3 4 2 3 4 2 3

2 3 1 3 2 3 2 3 2 4 4 4 2 3 4 4

2 3 1 3 2 3 2 2 3 4 3 4 3 4 3 4

2.44±0.964 2.63±0.619 2.88±1.088. 3.13±0.806

Flavour

2 2 1 1 2 3 2 1 3 3 2 2 3 2 2 4

1 2 1 1 3 2 3 1 3 4 4 4 3 3 3 4

1 2 1 2 1 2 2 2 1 4 3 3 2 3 4 4

3 3 2 3 1 2 2 2 3 3 2 3 3 3 3 3

1.75±0.775 1.94±0.680 2.94±0.854 3.06±0.680

Overall

3 3 2 2 3 3 2 2 4 2 2 2 4 3 3 4

2 2 2 1 3 2 2 2 3 4 2 2 3 3 3 4

2 2 2 3 1 2 2 3 1 3 2 3 2 3 4 4

3 2 2 3 1 2 2 2 3 3 2 2 3 3 4 3

2.25±0.577 2.13±0.619 2.50±0.816 3.31±0.602

Key for sensory evaluation: 1 – Dislike extremely

2 – Dislike slightly

3 – Neither like nor dislike

4 – Like slightly

5 – Like extremely

From the ANOVA analysis, we can conclude that:-

i. There are no any significant differences among the four samples on the sensory of texture of the beef ball.

ii. There are significant differences among the four samples on the sensory of flavour of the beef ball.

Sample A B C D

Mean ± SD 1.75b ± 0.775 1.94b ± 0.680 2.94a ± 0.854 3.06a ± 0.680

iii. There are significant differences among the four samples on the overall acceptances of the beef ball.

Sample A B C D

Mean ± SD 2.25b ± 0.577 2.13b ±0.619 2.50b ± 0.816 3.31a ± 0.602

DISCUSSION:

Beef ball is a minced meat product. In this experiment, beef meat is produced by mixing

meat with some nonmeat ingredients such as salt, black pepper, sugar, garlic, MSG, corn starch

and Tripolyphosphate (TPP). These nonmeat ingredients such as salt, black pepper, sugar and

garlic help to improve the eating quality of the final product. The eating quality is defined as the

juiciness, tenderness and flavor of the beef ball. The spice of black pepper not only function to

impart the flavor of beef ball, it also known to have great amount of antioxidant properties which

can preserve the beef ball by retarding oxidative changes. Oxidation reaction can result the

destruction of lipids and essential amino acids content in the beef ball. Therefore, the nutritive

values of beef balls decreases. Besides, MSG is a type of flavor enhancer which can potentiate

the flavor in beef ball. In addition, ices play an important role in making beef ball. Ices is used to

cool the machine before grinding of meat to avoid the fat render out or protein denatured due to

elevated temperature when grinding (Canhill and others, 1976). Besides, ice also used to cool our

hands before shaping the beef ball. Thus, the meat will not stick on our hand, it can also help to

improve the composition of meat and increase palatability of the beef balls.

Corn starch is also added in bee ball making. It acts as thickening agent which can

thicken or modify the texture and consistency of beef ball. It helps to improve the water holding

capacity and thus produce more juicy beef ball. Corn starch contributes to the stable formation of

protein-starch networks. This stable network can entrap water and thus the beef ball shown to be

juicier (Park and others 1993). In addition, Tropolyphospahe such as Sodium Tripolyphosphate

also is added while making beef ball to enhance water holding capacity. However, the addition

of STPP increases the pH of meat. Increasing the meat pH improves water-holding capacity by

moving the meat pH further from the meat protein isoelectric point. As the pH of the meat has

increased, subsequently changes in meat color should be expected. The minced meat becomes

darker in color (Banks and others, 1998). Salt is also added while making beef ball. Salt is used

to increase shelf life and enhance flavor of the beef ball. It also used to increase water holding

capacity and it works without changing the meat pH. In addition, adding of salt contributes to the

swelling of the protein. The swelling of the meat proteins enable them to bind more water

(Detienne and Wicker, 1999). Lastly, sugar is added to develop the surface color of meat ball

through browning reaction. Browning reaction occurs during the heating of sugar in mixture

contains amino acid or protein.

During this laboratory, first and second groups made beef balls without adding corn

starch whereas third and fourth groups made beef balls by adding corn starch. After that, group 1

and group 3 did analysis on raw beef balls whereas group 2 and group 4 did analysis on cooked

beef balls. According to the results obtained from this experiment, the pH values for raw and

cooked beef balls with corn starch added or without adding corn starch are almost the same

which are 6.445±0.007 for group 1, 6.665±0.021 for group 2, 6.605±0.007 for group 3 and

6.620±0.014 for group 4. The pH values are more alkaline, this is due to the addition of TPP

which is alkaline in nature. This also indicates that the addition of corn starch in the beef ball

making will not affect the pH values of beef ball. The ash content of raw beef ball without starch

is 4.550±0.3818% whereas cooked one is 2.500±0.0141%. The ash content for cooked beef ball

with starch added is 4.795±0.1202% whereas cooked one is 4.040±0.0283%. By comparing the

ash content for raw and cooked beef balls, the ash content of raw beef balls is higher compared

to the cooked beef ball. This is because a lot of nutrients might have dissolved in the hot water

bath which used to cook the beef balls.

Moisture content in the raw beef ball without corn starch is 29.045±0.3748 whereas with

corn starch one is 63.695±0.3323. Moisture content in the cooked beef balls without corn starch

is 59.795±0.0495 whereas with corn starch one is 65.81±0.7990. The results show that the raw

meat balls with starch added have higher moisture content compared to the beef balls without

starch. The results also show that the cooked beef balls have higher moisture content compared

to the raw beef balls. These results are incorrect theoretically. Corn starch added to beef ball

making to enhance the water holding capacity of beef ball (Schmidt, 1998). Thus, the available

water or free water in the beef ball might decrease and the moisture content should be lower.

Besides, the moisture content in the cooked beef ball should be lower due to the evaporation of

free water from beef balls during cooking. Hence, the cooked beef balls taste salty. However, the

moisture content of raw beef ball without corn starch for group 1 has very low moisture content

if compared to other group. This might be caused by some experimental errors or personal errors.

The length of time taken for evaporation process is too long.

Protein content for raw beef balls without corn starch is 0.540%, cooked beef balls

without corn starch is 15.170%, raw beef balls with starch added is 12.335% whereas cooked

beef balls with starch added is 13.110%. The protein content in raw or cooked beef balls with or

without corn starch is about the same, which is in range of 12.0-15.0%. However, protein content

of raw beef balls without corn starch is very low compared to others. This is also due to

experimental errors or personal errors. Students might have added wrong solution while carried

out the Kjeldahl method. For example, student added distilled water instead of sulfuric acid for

digestion of organic nitrogenous compound. Thus, no digestion occurs during the process. The

result obtained is near to zero.

Subsequently, sensory evaluation has been carried out to evaluate the texture, flavor and

overall acceptability of beef balls (without corn starch and with corn starch) among the students.

The results for texture evaluation are above 2.50 for beef balls that made by four different

groups. This indicates that students are neither like nor dislike the both types of beef balls. The

analysis of ANOVA also concludes that there are no any significant differences on the texture

among the four samples. Theoretically, beef balls with corn starch will be more tender and nicer

to eat. This is due to the function of corn starch that helps to enhance the moisture holding

capacity of beef balls, thus the beef balls produced are more juicy and nicer to eat. However,

students can accept both types of beef balls by giving same rate of scores. On the other hand, the

results for flavor evaluation for beef balls without corn starch is in range of 1.70-2.00 where as

for beef balls with corn starch is about 3.00. These values show that students prefer the flavor of

beef balls with corn starch added. By comparing the scores of overall acceptability for both types

of beef balls, students like to eat beef balls with corn starch which have been rated higher if

compared to beef balls without corn starch.

CONCLUSION

From the experiment, we can conclude that making beef balls by adding corn starch can

change its’ pH values, moisture content and ash values. Besides, it is inferred that beef balls with

corn starch added will be having a better sensory attributes than beef balls without corn starch

added. From the ANOVA analysis, we can conclude that there are significant differences on

flavour and overall acceptability among four samples. From the sensory evaluation, we can

prove that beef ball with corn starch added gives a better sensory attributes thus it has rated

higher if compared with control beef ball. As a conclusion, it is highly recommended that

addition of corn starch while making beef balls is more suitable in order to obtain desired end

products and to increase commercial value of beef balls.

QUESTIONS:

1. What is the role of corn starch and TPP? Discuss their effects on the texture development

of the product.

Corn starch and TPP can be classified as the meat extenders and fillers which primarily used

with the objective of making meat products lower-cost. In the upmarket sector there was

traditionally less demand for highly extended products as their sensory properties could not

fully match “full-meat” products. Besides, it gives the desired texture in keeping the ball binds

to each other forming round-shaped mixture. Corn starch is used as fillers (additives) to gives

beef balls more cohesive texture with binding potential and gelling properties. It also act as a

replacer that gives mouthfeel attributes when consumed. Nevertheless, high amount of this

filler would result in an atypically pale color and lead to loss of beef flavor. Meanwhile,

Tripolyphosphate (TPP) is one of the common additives for raw-cooked meat product whereby

they assist in the development of protein network structures which it start absorbing increasing

amounts of moisture at the temperature range of 50-70°C, at which some of the loosely bound

water is expelled from the protein structure networks. Hence, liquid purge can be decreased or

avoided.

2. Can sugar and salt omitted from the formulations? Discuss

Sugar should not be omitted from the formulation because sugar can prevent the granules from

clumping together and also omit the formation of lumps. In addition, sugar also can contribute

to a sweet taste when a higher amount of sugar was added. Salt should also not be omitted

because salt is added in the processing to extract salt-soluble proteins, thus increasing the

binding, yield and juiciness of the product.

REFERENCES :

1. Banks, W.T, Wang, C and Brewer, M.S. 1998. Sodium lactate/ Sodium Tripolyphosphate

combination effects on aerobic plate counts, pH and color of fresh pork muscle. Meat Science.

50:499.

2. Barbut, S. 1995. Importance of fat emulsification and protein matrix characterization in meat

batter stability. Journal Muscle Foods, 6: pp 161-177 (retrieve from

http://psasir.upm.edu.my/745/1/101-108.pdf)

3. Canhill, V.R, Miller, J.C, Parrett, N.A. 1976. Meat Processing. P1-2.

4. Detienne, N.A. and Wicker, L. 1999. Sodium chloride and tripolyphosphate effects on physical

and quality characterictics of injected pork. Journal of Food Science. 64:1042.

5. Fischer, A.1996. Classification and quality aspects of German processed meat. Short Course

Manual. Malang: Faculty of Animal Husbandary, Brawijaya University (retrieve from

http://psasir.upm.edu.my/745/1/101-108.pdf)

6. Heinz, G., Hautzinger, P (2007). Meat processing technology for small to medium scale

producers. AI407/E

7. Park, J, Kim, K.S and Rhee, K.C. 1993. High Protein Texturized Products of Defatted Soy

Flour, Corn Starch and Beef. Journal of Food Science. 58(1): p21-27.

8. Pearson,A.M. Processed Meat,3rd Edition. United States of Amerika: Chapman and Hall,

1996 .

9. Price, James.F and Schweigert,Bernard.F. The Science of Meat and Meat Products,3rd Edition.

Westport: Food and Nutrition Press, Inc, 1987

10. Schmidt, G. R. 1998. Meat Science, Milk Science and Technology. P83.

11. Tseng, F.F., D.D. Liu and M.T. Chen, 2000. Evaluation of transglutaminase on the quality of

low-salt chicken meat-balls. Meat Sci., 55: 427-431.

APPENDICES:

http://psasir.upm.edu.my/745/1/101-108.pdf

http://psasir.upm.edu.my/745/1/101-108.pdf

I) Calculation of pH

Group 1:

Average, µ = 6.44+6.45

2

= 6.445Standard deviation

=√ Σ ( x−μ )2

N−1

= √ (6.44−6.445 )2+ (6.45−6.445 )2

2−1

= √ 0.000051

= 7.07 x10−3

Group 2:

Average, µ = 6.68+6.65

2


=√ Σ ( x−μ )2

N−1

= √ (6.68−6.665 )2+(6.65−6.665 )2

2−1

= √ 0.000451

= 0.0212

Group 3:

Average, µ = 6.61+6.60

2


=√ Σ ( x−μ )2

N−1

= √ (6.61−6.605 )2+(6.60−6.605 )2

2−1

= √ 0.000051

= 7.07 x10−3

Group 4:

Average, µ = 6.63+6.61

2


=√ Σ ( x−μ )2

N−1

= √ (6.63−6.62 )2+(6.61−6.62 )2

2−1

= √ 0.00021

= 0.0141

II) Calculation of Moisture

Example calculation:Sample 1: Fresh uncooked beef meat ball with no corn starch added.

Moisture content (%) =Weight of sample−weight of dried sample

W eight of samplex100 %

Sample A

Moisture content (%) =3.0517−0.8783

3.0517 x 100%

= 28.78%

Sample B

Moisture content (%) =3.1490−0.9231

3.1490 x 100%

= 29.31%

Average moisture content (%) = 28.78+29.31

2

= 29.045%

Group 1:

Average, µ = 28.78+29.31

2


=√ Σ ( x−μ )2

N−1

= √ (28.78−29.045 )2+(29.31−29.045 )2

2−1

= √ 0.140451

= 0.3748

Group 2:

Average, µ = 59.76+59.83

2


=√ Σ ( x−μ )2

N−1

= √ (59.76−59.7 95 )2+(59.83−59.795 )2

2−1

= √ 0.002451

= 0.0495

Group 3:

Average, µ = 63.93+63.46

2


=√ Σ ( x−μ )2

N−1

= √ (63.93−63.695 )2+(63.46−63.695 )2

2−1

= √ 0.110451

Group 4:

Average, µ = 66.38+65.25

2


=√ Σ ( x−μ )2

N−1

= √ (66.38−65.815 )2+(65.25−65.815 )2

2−1

= √ 0.638451

= 0.3323 = 0.7990

III. Calculation for Ash Content

Example calculation:

Sample 1: Fresh uncooked beef meat ball with no corn starch added.

Ash content (%) =Weight of cruciblewith ash−weight of crucible

Weight of samplex100 %

Sample A

Ash content (%) =53.4522−53.3049

3.0576 x 100%

= 4.82%

Sample B

Ash content (%) =56.5135−56. 3828

3.0508 x 100%

= 4.28%

Average ash content (%) = 4.82+4.28

2

= 4.55%

Group 1:

Average, µ = 4.82+4.28

2


=√ Σ ( x−μ )2

N−1

= √ ( 4.82−4.55 )2+(4.28−4.55 )2

2−1

Group 2:

Average, µ = 2.49+2.51

2


=√ Σ ( x−μ )2

N−1

= √ (2.49−2.50 )2+(2.51−2.50 )2

2−1

= √ 0.14581

= 0.3818

= √ 0.00021

= 0.0141

Group 3:

Average, µ = 4.71+4.88

2


=√ Σ ( x−μ )2

N−1

= √ ( 4.71−4.795 )2+(4.88−4.795 )2

2−1

= √ 0.014451

= 0.1202

Group 4:

Average, µ = 4.06+4.02

2


=√ Σ ( x−μ )2

N−1

= √ ( 4.06−4.04 )2+ (4.02−4.04 )2

2−1

= √ 0.00081

= 0.0283

IV) Calculation of Fat



Sample A

Weight of sample (g) = 3.0130g

Weight of round bottom flask (g) = 87.8526g

Weight of round bottom flask+ fat extracted (g) = 87.8586g

Weight of fat extracted (g) = 0.0060g

% Fat content = Weight of fat extracted

weight of samplex 100 %

% Fat content = 0.00603.0130

x100% = 0.20%

Sample B

Weight of sample (g) = 3.0719g

Weight of round bottom flask (g) = 89.8263g

Weight of round bottom flask+ fat extracted (g) = 89.8336g

Weight of fat extracted (g) = 0.0073g

% Fat content = 0.00733.0719

x100% = 0.24%

Average fat content (%) = 0.20+0.24

2

= 0.22%

Group 1:

Average, µ = 0.20+0.24

2


=√ Σ ( x−μ )2

N−1

= √ ( 0.20−0.22 )2+(0.24−0.22 )2

2−1

= √ 0.00081

= 0.0283

Group 2:

Average, µ = 0.042+0.038

2


=√ Σ ( x−μ )2

N−1

= √ ( 0.042−0.04 )2+ (0.038−0.04 )2

2−1

= √ 0.0000081

= 0.0028

Group 3:

Average, µ = 0.20+0.13

2


=√ Σ ( x−μ )2

N−1

= √ ( 0.20−0.165 )2+( 0.13−0.165 )2

2−1

Group 4:

Average, µ = 0.23+0.29

2


=√ Σ ( x−μ )2

N−1

= √ ( 0.23−0.26 )2+ (0.29−0.26 )2

2−1

= √ 0.002451

= 0.0495

= √ 0.00181

= 0.0424

V) Calculation of Protein



Sample A:

Weight of sample = 0.1613 g

Volume of H2SO4 to titrate Boric acid = 0.3 ml

Volume of H2SO4 to titrate blank = 0.1 ml

Normality of H2SO4 = 0.05 N

Therefore, % of nitrogen = (0.2 x 0.05 x 1.4) ÷ 0.1613

= 0.0868

% of protein = 0.0868 x 6.25

= 0.54 %

Sample B:

Weight of sample = 0.1630 g

Volume of H2SO4 to titrate Boric acid = 0.3 ml

Volume of H2SO4 to titrate blank = 0.2ml

Normality of H2SO4 = 0.05 N

Therefore, % of nitrogen = (0.2 x 0.05 x 1.4) ÷ 0.1630

= 0.8589

% of protein = 0.8589 x 6.25

= 0.54 %

Average protein content (%) = 0.54+0.54

2

= 0.54%

Group 1:

Average, µ = 0.54+0.54

2


=√ Σ ( x−μ )2

N−1

= √ ( 0.54−0.54 )2+(0.54−0.54 )2

2−1= 0.0

Group 2:

Average, µ = 14.88+15.46

2


=√ Σ ( x−μ )2

N−1

= √ (14.88−15.17 )2+(15.46−15.17 )2

2−1

= √ 0.16821

= 0.4101

Group 3:

Average, µ = 12.60+12.07

2


=√ Σ ( x−μ )2

N−1

= √ (12.60−12.335 )2+(12.07−12.335 )2

2−1

= √ 0.140451

= 0.3748

Group 4:

Average, µ = 14.73+11.49

2


=√ Σ ( x−μ )2

N−1

= √ (14.73−13.11)2+(11.49−13.11 )2

2−1

= √ 5.24881

= 2.2910

ANOVA analysis

Texture Analysis

Descriptives

Score

N MeanStd.

Deviation Std. Error

95% Confidence Interval for Mean

Minimum MaximumLower Bound Upper Bound

A 16 2.44 .964 .241 1.92 2.95 1 4

B 16 2.63 .619 .155 2.30 2.95 2 4

C 16 2.88 1.088 .272 2.30 3.45 1 4

D 16 3.13 .806 .202 2.70 3.55 2 4

Total

64 2.77 .904 .113 2.54 2.99 1 4

The average texture of the samples is between 2.54 and 2.99 at 95% of the time.

Test of Homogeneity of Variances

Score

Levene Statistic df1 df2 Sig.

2.082 3 60 .112

Since the p > 0.05 (p=0.112)So, the variances of the four samples based on its texture on beef ball are not significantly different.

ANOVA

Score

Sum of Squares Df Mean Square F Sig.

Between Groups

4.297 3 1.432 1.821 .153

Within Groups 47.188 60 .786

Total 51.484 63

Robust Tests of Equality of Means

Score

Statistica df1 df2 Sig.

Welch 1.923 3 32.610 .145

Brown-Forsythe

1.821 3 52.397 .155

a. Asymptotically F distributed.

Since the p > 0.05 (p = 0.153) in ANOVA table,Therefore, there are no significant differences among the four samples of beef ball on the sensory of texture.

Flavour Analysis

Descriptives

Score

N Mean Std. Deviation Std. Error


Minimum MaximumLower Bound Upper Bound

A 16 1.75 .775 .194 1.34 2.16 1 3

B 16 1.94 .680 .170 1.58 2.30 1 3

C 16 2.94 .854 .213 2.48 3.39 1 4

D 16 3.06 .680 .170 2.70 3.42 2 4

Total 64 2.42 .940 .117 2.19 2.66 1 4

The average flavour of the samples is between 2.19 and 2.66 at 95% of the time.


score


.592 3 60 .623

Since the p > 0.05 (p = 0.623)So, the variances of the four samples based on its flavour are no significant different.

ANOVA

score

Sum of Squares df Mean Square F Sig.

Between Groups

21.797 3 7.266 12.893 .000


Total 55.609 63


score


Welch 12.874 3 33.209 .000

Brown-Forsythe

12.893 3 57.764 .000


Since the p < 0.05 (p = 0.000)Therefore, there are significant differences among the four samples on the sensory of flavour of beef ball.Since there is a significant difference among the samples, the ones that are different can be determined using Tukey’s Test.

Multiple Comparisons

scoreTukey HSD

(I) (J) Mean Std. Error Sig. 95% Confidence Interval

group groupDifference (I-

J) Lower Bound Upper Bound

A B -.188 .265 .894 -.89 .51

C -1.188* .265 .000 -1.89 -.49

D -1.313* .265 .000 -2.01 -.61

B A .188 .265 .894 -.51 .89

C -1.000* .265 .002 -1.70 -.30

D -1.125* .265 .000 -1.83 -.42

C A 1.188* .265 .000 .49 1.89

B 1.000* .265 .002 .30 1.70

D -.125 .265 .965 -.83 .58

D A 1.313* .265 .000 .61 2.01

B 1.125* .265 .000 .42 1.83

C .125 .265 .965 -.58 .83

*. The mean difference is significant at the 0.05 level.

From the comparisons, we can conclude that flavour of sample A is not significantly different than B but is significantly different than sample C and D.

Score

Tukey HSDa

group N

Subset for alpha = 0.05

1 2

A 16 1.75

B 16 1.94

C 16 2.94

D 16 3.06

Sig. .894 .965

Means for groups in homogeneous subsets are displayed.

a. Uses Harmonic Mean Sample Size = 16.000.

Sample A B C DMean ± SD 1.75b ± 0.775 1.94b ± 0.680 2.94a ± 0.854 3.06a ± 0.680

Overall Acceptance

Descriptives

score

N MeanStd.

Deviation Std. Error


Minimum MaximumLower BoundUpper Bound

A 16 2.25 .577 .144 1.94 2.56 1 3

B 16 2.13 .619 .155 1.80 2.45 1 3

C 16 2.50 .816 .204 2.06 2.94 1 4

D 16 3.31 .602 .151 2.99 3.63 2 4

Total 64 2.55 .795 .099 2.35 2.75 1 4

The average overall acceptance of sample 1 is between 2.35 and 2.75 at 95% of the time.


score


1.531 3 60 .216

Since p > 0.05 (p = 0.216)So, the variances of the four samples based on its overall acceptance are no significant different.

ANOVA

score

Sum of Squares df Mean Square F Sig.

Between Groups

13.672 3 4.557 10.442 .000


Total 39.859 63


score


Welch 12.042 3 33.133 .000

Brown-Forsythe

10.442 3 54.825 .000


Since p < 0.05 (p = 0.000) in ANOVA table

Therefore, there are significant differences among the four samples on the overall acceptance of the beef ball.Since there is a significant difference among the samples, the ones that are different can be determined using Tukey’s Test.

Multiple Comparisons

scoreTukey HSD

(I) group

(J) group

Mean Difference (I-

J) Std. Error Sig.

95% Confidence Interval

Lower Bound Upper Bound

A B .125 .234 .950 -.49 .74

C -.250 .234 .709 -.87 .37

D -1.063* .234 .000 -1.68 -.45

B A -.125 .234 .950 -.74 .49

C -.375 .234 .383 -.99 .24

D -1.188* .234 .000 -1.80 -.57

C A .250 .234 .709 -.37 .87

B .375 .234 .383 -.24 .99

D -.813* .234 .005 -1.43 -.20

D A 1.063* .234 .000 .45 1.68

B 1.188* .234 .000 .57 1.80

C .813* .234 .005 .20 1.43

*. The mean difference is significant at the 0.05 level.

From the comparisons, we can conclude that overall acceptance of sample D is significantly different than sample A, B and C.

score

Tukey HSDa

group N

Subset for alpha = 0.05

1 2

B 16 2.13

A 16 2.25

C 16 2.50

D 16 3.31

Sig. .383 1.000

Means for groups in homogeneous subsets are displayed.

a. Uses Harmonic Mean Sample Size = 16.000.

Sample A B C DMean ± SD 2.25b ± 0.577 2.13b ±0.619 2.50b ± 0.816 3.31a ± 0.602

Documents

Beef Ball Combine