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 2010
GROUP : 1 (TUESDAY 12pm-3pm; 4-7pm)
GROUP MEMBERS : MATRIC NO.
1. SITI HAJAR BT MD ALI 146956
2. SALEHA BT HASSAN 147223
3. WOO CHYE LIN 147835
4. NURUL FARADINA AZWA BINTI ROSMAN 148129
5. LIM KANG XIANG 148626
6. GOH KONG SONG 148997
7. LEE KAR LENG 149699
8. 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 1000.0
Salt 20.0
Sugar 5.0
TPP 25.0
Garlic 2.5
Black pepper 2.5
Ice 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 1 2 3 4
Test
6.44 6.45 6.68 6.65 6.61 6.60 6.63 6.61
pH 6.445±7.07 x 10−3 6.665±0.0212 6.605±7.07 x 10−3 6.620±0.014
28.78 29.31 59.76 59.83 63.93 63.46 66.38 65.25
Moistur
29.045±0.3748 59.795±0.049 63.695±0.3323 65.815±0.799
e (%)
5 0
4.82 4.28 2.49 2.51 4.71 4.88 4.06 4.02
Ash (%)
4.550±0.3818 2.500±0.0141 4.795±0.1202 4.040±0.0283
0.20 0.24 0.04 0.04 0.20 0.13 0.23 0.29
Fat (%)
0.220±0.0283 0.040±0.0028 0.165±0.0495 0.260±0.0424
0.54 0.54 14.88 15.46 12.60 12.07 14.73 11.49
Protein
0.540±0.0 15.170±0.410 12.335±0.3748 13.110±2.291
(%)
1 0
Yield of
beef ball
48 48 51 55

Table 2: Sensory Evaluation on Beef Meat Ball

 Without Corn Starch(control) With Corn Starch
Samples
1 2 3 4
Group
A B C D A B C D A B C D A B C D
Sensory
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
Texture 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
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
Flavour 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
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
Overall 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:

I) Calculation of pH

Group 1: Group 2:
6.44+6.45 6.68+6.65
Average, µ = Average, µ =
2 2
 = 6.445 = 6.665
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 6.44−6.445 + 6.45−6.445 = 6.68−6.665 + 6.65−6.665
√ (

0.00005
2−1 √
(

0.00045
2−1
=
√ 1
= 7.07 x 10−3
=
√ 1
= 0.0212

Group 3: Group 4:
6.61+ 6.60 6.63+6.61
Average, µ = Average, µ =
2 2
= 6.605 = 6.62
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 6.61−6.605 + 6.60−6.605 = 6.63−6.62 + 6.61−6.62
√ (

0.00005
2−1 √
(

0.0002
2−1
=
√ 1
= 7.07 x 10−3
=
√ 1
= 0.0141

II) Calculation of Moisture

Example calculation:
Sample 1: Fresh uncooked beef meat ball with no corn starch added.
Weight of sample−weight of dried sample
Moisture content (%) = x 100 %
W eight of sample

Sample A

3.0517−0.8783
Moisture content (%) = x 100%
3.0517

= 28.78%
Sample B
 3.1490−0.9231
Moisture content (%) = x 100%
3.1490

= 29.31%
28.78+29.31
Average moisture content (%) =
2
= 29.045%

Group 1: Group 2:
28.78+29.31 59.76+59.83
Average, µ = Average, µ =
2 2
= 29.045 = 59.795
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
( 28.78−29.045 )2 + ( 29.31−29.045 )2
√ (
N −1
( 59.76−59.7 95 )2 + ( 59.83−59.795 )2
=
√ 0.14045
2−1
=
√ 0.00245
2−1
=
√
= 0.3748
1
=
√
= 0.0495
1

Group 3: Group 4:
63.93+63.46 66.38+65.25
Average, µ = Average, µ =
2 2
= 63.695 = 65.815
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 63.93−63.695 + 63.46−63.695 = 66.38−65.815 + 65.25−65.815
√ (

0.11045
2−1 √ (

0.63845
2−1
=
√ 1
= 0.3323
=
√ 1
= 0.7990

III. Calculation for Ash Content

Example calculation:

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

Weight of crucible with ash−weight of crucible

Ash content (%) = x 100 %
Weight of sample

Sample A

53.4522−53.3049
Ash content (%) = x 100%
3.0576

= 4.82%

Sample B

56.5135−56. 3828
Ash content (%) = x 100%
3.0508

= 4.28%

4.82+4.28
Average ash content (%) =
2

= 4.55%

Group 1: Group 2:
4.82+4.28 2.49+2.51
Average, µ = Average, µ =
2 2
= 4.55 = 2.50
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
( 4.82−4.55 )2+ ( 4.28−4.55 )2
√ (
N −1
( 2.49−2.50 )2 + ( 2.51−2.50 )2
=
√ 0.1458
2−1
=
√0.0002
2−1
=
√
= 0.3818
1
=
√
= 0.0141
1

Group 3: Group 4:
 4.71+4.88 4.06+ 4.02
Average, µ = Average, µ =
2 2
= 4.795 = 4.04
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 4.71−4.795 + 4.88−4.795 = 4.06−4.04 + 4.02−4.04
√ (

0.01445
2−1 √ (

0.0008
2−1
=
√ 1
= 0.1202
=
√ 1
= 0.0283

IV) Calculation of Fat

Example calculation:

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

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

Weight of fat extracted

% Fat content = x 100 %
weight of sample

0.0060
% Fat content = x100% = 0.20%
3.0130

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

 0.0073
% Fat content = x100% = 0.24%
3.0719

0.20+0.24
Average fat content (%) =
2

= 0.22%

Group 1: Group 2:
0.20+0.24 0.042+ 0.038
Average, µ = Average, µ =
2 2
= 0.22 = 0.04
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 0.20−0.22 + 0.24−0.22 = 0.042−0.04 + 0.038−0.04
√ (

0.0008
2−1 √ (

0.000008
2−1
=
√ 1
= 0.0283
=
√ 1
= 0.0028

Group 3: Group 4:
0.20+0.13 0.23+0.29
Average, µ = Average, µ =
2 2
= 0.165 = 0.26
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 0.20−0.165 + 0.13−0.165 = 0.23−0.26 + 0.29−0.26
√ (

0.00245
2−1 √ (

0.0018
2−1
=
√ 1
= 0.0495
=
√ 1
= 0.0424

V) Calculation of Protein
Example calculation:

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

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 %

0.54+0.54
Average protein content (%) =
2
= 0.54%
 Group 1: Group 2:
0.54+0.54 14.88+15.46
Average, µ = Average, µ =
2 2
= 0.54 = 15.17
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 0.54−0.54 + 0.54−0.54 = 14.88−15.17 + 15.46−15.17
√ (

= 0.0
2−1 √ (

0.1682
2−1
=
√ 1
= 0.4101

Group 3: Group 4:
12.60+12.07 14.73+ 11.49
Average, µ = Average, µ =
2 2
= 12.335 = 13.11
Standard deviation Standard deviation
)2 )2
= Σ x−μ = Σ x−μ
√ (
N −1
)2 ( )2
√ (
N −1
)2 ( )2
= 12.60−12.335 + 12.07−12.335 = 14.73−13.11 + 11.49−13.11
√ (

0.14045
2−1 √ (

5.2488
2−1
=
√ 1
= 0.3748
=
√ 1
= 2.2910

ANOVA analysis

Texture Analysis
 Descriptives
Score
95% Confidence Interval for
Std. Mean
N Mean Deviation Std. Error Lower Bound Upper Bound Minimum Maximum
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
Tot 64 2.77 .904 .113 2.54 2.99 1 4
al

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 4.297 3 1.432 1.821 .153
Groups
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- 1.821 3 52.397 .155
Forsythe
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

95% Confidence Interval for

Mean

N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum
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.
 Test of Homogeneity of Variances
score
Levene
Statistic df1 df2 Sig.
.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 21.797 3 7.266 12.893 .000
Groups
Within Groups 33.813 60 .564
Total 55.609 63

Robust Tests of Equality of Means

score
Statistica df1 df2 Sig.
Welch 12.874 3 33.209 .000
Brown- 12.893 3 57.764 .000
Forsythe
a. Asymptotically F distributed.

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
score
Tukey HSD
Mean 95% Confidence Interval
(I) (J) Difference (I-
group group J) Std. Error Sig. 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
Subset for alpha =
0.05

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

Overall Acceptance
 Descriptives

score

95% Confidence Interval for

Mean

Std. Upper
N Mean Deviation Std. Error Lower Bound Bound Minimum Maximum
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.

Test of Homogeneity of Variances

score
Levene
Statistic df1 df2 Sig.
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 13.672 3 4.557 10.442 .000
Groups
Within Groups 26.188 60 .436
Total 39.859 63
 Robust Tests of Equality of Means
score
Statistica df1 df2 Sig.
Welch 12.042 3 33.133 .000
Brown- 10.442 3 54.825 .000
Forsythe
a. Asymptotically F distributed.
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
score
Tukey HSD
Mean 95% Confidence Interval
(I) (J) Difference (I-
group group J) Std. Error Sig. 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
a
Tukey HSD
Subset for alpha =
0.05
group N 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 D
Mean ± SD 2.25b ± 0.577 2.13b ±0.619 2.50b ± 0.816 3.31a ± 0.602