Lab 4 - Sensory Evaluation Of Food Products

FST 4822 LABORATORY FOR CHEMISTRY AND TECHNOLOGY OF PLANT AND ANIMAL PRODUCTS SEAFOOD BASED LAB: SENSORY EVALUATIONS OF FOOD PRODUCTS GROUP : GROUP 6 GROUP MEMBERS : TEY CHEE SENG SITI FARHIAH BINTI ABDUL MANAN FARHANA YUSOF NUR BAITI SOFYUDDIN AMINAH LET FARHAH IZZATI SHUKOR YONG XIANGPEI SEE HUI YONG PROGRAM : LAB : DATE : LECTURER:

137999 136430 136499 136786 137176 137259 138713 138835

BACHELOR OF FOOD SCIENCE &TECHNOLOGY MAKMAL PEMPROSESAN & MAKMAL BIOKIMIA 30TH JULY 2008 PROFESSOR MADYA DR. AZIS ARIFFIN

Introduction: According to Elizabeth (1977), a sensory evaluation was made by senses of taste, smell, touch, and hearing when the food product was consumed. The complex sensation from the interaction of our senses was used to measure the food quality and for new product development. Many developments and advances in this area have been made since Methods for Sensory Evaluation of Food was published in 1967 (Elizabeth, 1977). Research, product development, and quality control are the three main areas that sensory testing is being utilized (Meilgaard et al., 1991). There are many different types of sensory tests. Attribute difference tests usually determine in what sense a certain quality or trait differ between samples. Meanwhile, affective sensory tests evaluate the consumer acceptance of products. Overall difference tests determine the sensory difference which may exist between samples. A triangle test is a type of difference test to determine if there is a sensory difference between two products. For example, a researcher may want to see if changing one ingredient to make a food product will affect the taste of the final product. (Meilgaard et al., 1999). The sensory test measures if any differences detected are truly significant by analyzing the sensory data for statistical significance. After statistical analysis, the researchers can make a meaningful interpretation from the results of the sensory data (Meilgaard et al., 1999). The simplest form of sensory evaluation that can be performed is the evaluation that is done at the bench by research worker who develops the new food products. The research worker usually relies on his or her own evaluation to determine the differences in the food products. The more formal manner of sensory evaluation is conducted by laboratory and consumer panels. Sensory evaluation that is performed in the presence of large scale panelists can give clear prediction of consumer reactions. Different food products have different composition and ratio of ingredients used in them. In order to determine the best composition and ratio that suits the consumer needs, sensory evaluation is done so that food industry can improvise their products to be sold to the consumers. For this experiment, we were asked to prepare three fish-based products which are fish ball, fish cake and fish finger. All of these were made using surimi block into different end products. The final products were then undergone sensory test to determine the acceptance among us. During the preparation of the products, different processing methods were used although the ingredients added to make up the mixture was similar. For fish ball, the mould machine was used to shape the fish ball into round shape before boiling it in a boiler. On the other hand, fish cake was cut into shape by using cutting utensils before putting it into steamer and followed by

frying. While for fish finger, it was also put into steamer followed by battered and coated with breadcrumbs before frying it. These different processing methods gave various sensory evaluations on the products although the basic ingredients were the same. The processing methods give effects on the end products in terms of texture, overall appearance and also acceptability among consumers. As for fish cake and fish finger, both of them were fried in hot oil temperatures. Frying operations that are actually applied to dry the food and to increase shelf life may finally cause losses of nutrients, especially fat soluble vitamins. For example vitamin E which is absorbed from oil by crisps during frying, is oxidized during storage. Bunnell et al. (1965) found that 77% loss of vitamin E after eight weeks at ambient temperature. As mentioned above, fish cake and fish finger were fried in hot oil which will produce crust formation and seal the food surface. The objective of the experiment was to introduce the students the proper method to evaluate the product characteristics which have undergone different processing methods. Apart of that, this experiment was also allow the students to learn more about hypothesis testing and statistical significance and to determine the most important sensory attributes for fried products. Equipments and Utensils: •

Cooking oil

•

Cutting utensils

•

Frying pan

•

Serving plates

•

Moulding machine

•

Bowl cutter mixer

Ingredients Ingredient Surimi Salt MSG Wheat flour STPP (Sodium Tripolyphosphate) Fish Flavour

Formulation 6 kg 50g 30g 600g 120g 40g

Ice

300g

Batter ingredient Water

300g

Wheat flour Corn starch

150g 50g

Procedure a) Preparation of food product (fish ball, fish cake and fish finger) All the ingredients were weighed accurately Surimi and salt were mixed by using bowl cutter mixer for 5 minutes Wheat flour, fish flavour, salt, MSG and ice were added. The mixture was mixed for 10 minutes The mixture was divided into three parts 1) Fish Ball

2) Fish Cake

3) Fish Finger

1) Fish Ball The mixture was extruded in the moulding machine ↓ Fish ball ↓ Boiled in boiling water *until the fish balls were floating ↓ The fish ball was cooled for 20 minutes 2) Fish Cake The mixture was compressed into sheets by using fish cake Molding ↓ The compressed mixture was steamed by using steamer for 20minutes ↓ After that, the steamed sheets was cut into a uniform pieces ↓ Fish cake 3) Fish finger The mixture was compressed into sheets by using fish cake moulding

The compressed sheets was steamed by using steamer for 20 minutes After that, the steamed sheets was cut into a uniform pieces

Fish finger All the ingredients were weighed accurately for the preparation of batter The fish finger was coated with the batter and bread crumb b) Preparation for the sensory analysis The prepared foods were cooked *fish ball-boiling method *fish cake and fish finger -deep frying method The cooked samples were cut into bite size and placed them on the serving plates To ensured that product identity was not disclosed, the cooked samples were coded with 3 digit random code such as 123, 456 etc Each product was evaluated by each group. The average and standard deviation of the results was computed by combining the results of evaluations from the other groups in the session The texture, taste, color, odor, overall appearance and overall acceptability were evaluated according to the scale provided below: a) Texture- very hard-1; slight hard-2; soft-3 b) Taste- not acceptable-1; moderately acceptable-2; very acceptable-3 c) Color- not acceptable-1; moderately acceptable-2; very acceptable-3 d) Odor- not acceptable-1; moderately acceptable-2; very acceptable-3 e) Overall appearance- not acceptable-1; moderately acceptable-2; very acceptable-3 f) Overall acceptability- not acceptable-1; moderately acceptable-2; very acceptable-3

Figure 1: Operating mixer with ingredient inside

Figure 2: Operating fish ball Making

Figure 3: Fish ball

Results and analysis: Sensory Attributes Texture

ANOVA Test Tukey’s Test No significant different at 1% level. Significant different at 5% Fish cake is significantly more acceptable compared to fish level. ball and fish finger at 5% level. No significant different at 1% -

Taste

Color Odor Overall Appearance Overall Acceptability

level. Significant different at 1% Fish ball is significantly unacceptable than fish cake level. and fish finger at 5% level. Significant different at 1% Fish ball is significantly unacceptable than fish cake level. and fish finger at 5% level. Significant different at 1% Fish cake is significantly more acceptable than fish ball at 5% level. level.

Discussion : Beef balls and fish fingers are both categorized as meat products. To make meat products to be consumable, various methods of cooking have been developed. With the application of different cooking methods, such as boiling, frying, steaming, grilling or baking, an internal altering of texture, taste, odor, flavor have resulted. Even though both are meats, the cooking methods can have different effect on both these meat products. The beef balls can have a more distinct effect as rather than the fish fingers. At cooking temperature, proteins and sugars of meat products can produce brown pigments. This is the result of interaction of deposits from smoke on the meat surfaces and the endogenous compounds. On heating, lipid in meat can undergo thermally induced oxidation, where a range of volatiles which contribute to the formation of flavor. In experiment, we had steamed fish fingers and also fish cakes. Steaming did not change the color of both the end products much and remained in their almost natural color. Since steaming is a wet heat and water is available, steaming makes the fish fingers and fish cake even tender. The tenderness of fish increases as the temperature increases. This effect also happens on the beef balls. Frying is a way of cooking. By frying, a considerable amount of oils can be migrated to fried food. This has indirectly contributed to the absorption of oil by food and caused a change in the flavor of meat. The meats which have been fried usually absorb the fatty acid profile of the

oil, which act as the frying medium. The fatty acid which acts as the flavor carrier allows the solubility of the fat soluble compound, such as flavor and thus enhances the flavor of the fried foods. Hence, frying increases the adsorption of fat content on foods and the higher contents of fats or oils on the fried foods can enhance the flavor better. Moreover, the fried foods are oily and give us a smooth and oily mouth feel. Upon frying, protein, carbohydrates and fats also undergo changes. Maillard reaction (browning and caramelization) takes place in the fried food due to the reaction of amino acid as well as sugars to give a desirable brown color. This reaction also contributes the fried foods a richer flavor. A golden brown color can be formed on the surface of the meat. As for texture, frying impart crispiness on the fried foods. Nonetheless, frying makes the fried foods softer if the foods are kept and stored for a longer period of time. In terms of taste, the deep-fried foods are less palatable if the deep-fried foods are soaked in the oil for a period of time. Moreover, the oil used in deep frying has an effect on the sensory attributes of fried foods. The age and thermal history of oil can affect the sensory attributes of fried foods. The prolonged frying at high temperature can cause the oxidation of oil to form unpleasant flavors and odors, such as volatile carbonyls, hydroxyl acid, keto acid and epoxy acid. The formation of these unpleasant flavors and odors are the results of oxidation of the oil when it is exposed to long time frying. Apart from that, prolonged frying also allows the formation of dark color oil and this indirectly will affect the fried food to obtain a darker and undesirable color. The above discussed attributes are the effects of frying on meat products. When comparing meat products, such as fish fingers and meat balls, meat balls have the more distinct effects as compared to the fish fingers. Fish finger obtains a lesser effect and color of fish meat does not change after frying. For fried products, the most important sensory attributes of fried breaded fish fingers are texture, flavor and appearance. Breading is a system where it provides a coating and protection to the meat products. This system can prevent the moisture inside the meat to be lost in the form of water vapor. With the presence of this coat, the moisture contents of the meat continue to retain inside meat and thus free from water diffusion. As a result, fried meat product becomes juicier and do not harden or toughen during deep frying. The deep-fried breaded fingers also have enhanced flavours. This is because, in frying, the bread coating carries flavor as well as seasonings added. Thus, this has enhanced the sensory characteristics of the particular products and also the acceptability of consumers. Apart from that, texture of the breaded fingers is also important. With the suitable temperature and also pressure, softness and crispiness can result. The bread coating can impart crispiness on the fried product when atmospheric pressure is applied. On the other hand, the breaded fingers can be soft when the positive pressure is used.

Smoking is also another type of way of cooking meat. Smoked meat can be eaten straight without cooking. Smoking can preserve the meat and thus extend the shelf life of meat products. By smoking, varieties of meat products in market can be achieved. With the smoking of meat, a desirable color can be developed. As a result, smoking has enhanced the brown color of the meat surface and it gives a more acceptable and desirable color to consumers. Furthermore, the smoked meats also possess a developed aroma and flavor. The Phenols from the wood smoke, which is used in smoking has an effect on flavor and aroma development of smoked meat. Two of the examples of phenols in wood smoke are guaiacol and syringol with the former has more contribution on flavor and the latter enhances the aroma. Sensory attributes of a food item are typically perceived in the following order such as appearance, odor/ aroma/ fragrance, consistency and texture, and flavor. In this experiment, we have done the sensory evaluation of the fish surimi which we had previously processed. Through this experiment, the sensory evaluation of fish balls, breaded fish fingers and fish cake had been done. Based on the Results and Analysis, there was no significance difference among the three samples in terms of texture, odor and color. These three samples have similar acceptable appearance. However, fish cake has the best taste as compared to the breaded fish fingers and fish balls. The delicious taste of fish cake could be due to the proper frying temperature and time had been carried out. The breaded fingers had a lesser preference amongst the judges could be due to the lack of frying time on the minced fish meat. People handled the frying of the breaded fish fingers might be affected by the colour of the golden orange of the crumb. This is because the crumbs which acted as the coat were fried completely, whereas the minced fish meats inside were blocked by the bread system. As a result, softer texture of minced fish meats was formed inside and affected the sensory evaluation of judges. Fish ball has the most unacceptable overall appearance after frying. This could be due to the fish balls were too soft and not elastic. Besides that, the fish ball also had an undesirable smell which decreased the appetite of judges. Hence, with all the results above, it can be concluded that fish cake had the highest overall acceptability. Inferences: 1. There are many type of fish product that is available in the market which is differing in their sensory characteristics. 2. Different types of processing equipment, yielding products with different quality parameters. 3. The sensory evaluation is crucial since the determination cannot be duplicated in any other analysis. 4. The results of sensory evaluation can be analyzed mathematically. 5. The sensory evaluation can be used to identify consumer’s preference.

6. The sensory evaluation is important in evaluating the characteristic of the product. Conclusion Based on the results, fish cake is the most preferable product compared to fish balls and breaded fish fingers. This is due to the type of cooking that have been conducted to the three types of products which are frying, boiling and steaming and is much affected by the time and temperature of cooking. References: 1. Elizabeth L., Laboratory Methods for Sensory Evaluation of Food, Ottawa, Canada, 5 & 6, 1977. 2. P. J. Fellows, Food Processing Technology: Principles and Practice, 362, 1999 3. Meilgaard, M., Civille, G.V., and Carr, B.T., Sensory Evaluation Techniques, 3rd Edition, CRC Press LLC, Boca Raton, FL, 1999 4. Pearson, A.M. and Gillett, T.A. (1999). Meat Cookery and Cooked Meat Products. In

Processed Meats.( pp. 111).Maryland : An Aspen Publication, Aspen Publishers, Inc. Galthersburg. 5. Toldra, F., Meat: Chemistry and Biochemistry. In Y. H., Hui. (2006). Handbook of Food Science, Technology, and Engineering (pp 28-3). New York: CRC Press. 6. FST 4822 Laboratory Manual 7. FST 4822 Lecture Note-Meat Technology 8. FST 4829 Sensory Evaluation Lecture Note 9. http://food.oregonstate.edu/sensory/theresa.html (accessed: 15 August 2008) 10. http://www.dietriot.com/stories/fry-light-qa.asp (accessed: 15 August 2008)

Index: Results: a) Texture Group 1 2 3 4 5 6 Total Mean

Fish Finger 3 2 3 2 2 2 14 2.33

Products Fish Cake 3 2 2 2 3 2 14 2.33

Total Fish Ball 2 2 2 3 3 3 15 2.50

8 6 7 7 8 7 43

Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the texture attribute of the three samples is done by using ANOVA:

Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 43 2 / (3 × 6) = 1849 / 18 = 102.72 Sum of squares, ss Sum of squares, samples

= (Sum of the squares of the total for each sample / number of judgments for each sample) − CF = (152 + 142 + 142) / 6 − 102.72 = 617 / 6 – 102.72 = 0.11

Sum of squares, judges

= (Sum of the squares of the total for each judge / number of judgments by each judge) − CF = (82 + 62 + 72 + 72+ 82 + 72) / 3 − 102.72 = 311 / 3 − 102.72 = 0.95

Sum of squares, total

= Sum of the squares of each judgment − CF = (22 + 22 + …..32 + 22) − 102.7

= 107 − 102.72 = 4.28 Sum of squares, error

= ss total − ss judges − ss samples = 4.28 – 0.11 - 0.95 = 3.22

Degrees of freedom, df df, samples = 3 – 1 = 2 df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 0.11/2 = 0.06 ms, judges = 0.95/5 = 0.19 ms, error = 3.22/10 = 0.32 Variance ratio, F F, samples = 0.06/0.32 = 0.19 F, judges = 0.19/0.32 = 0.59 Analysis of variance table (ANOVA table): Source of variation Df Samples 2 Judges 5 Error 10 Total 17 1% level of significance Samples: F calculated (0.19) < F table (7.56)

ss 0.11 0.95 3.22 4.28

ms 0.06 0.19 0.32

F 0.19 0.59

5% level of significance F calculated (0.19) < F table (4.10)

Conclusion: There is no significant difference in the texture attribute between the samples at both 5% level and 1% level. Judges: F calculated (0.59) < F table (5.64)

F calculated (0.59) < F table (3.33)

Conclusion: There is no significant difference in the texture attribute between the samples at both 5% level and 1% level. b) Taste Group 1 2 3

Fish Finger 3 2 2

Products Fish Cake 3 3 3

Total Fish Ball 2 2 3

8 7 8

4 5 6 Total Mean

2 2 2 13 2.17

2 3 3 17 2.83

1 2 2 12 2.00

5 7 7 42

Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the taste attribute of the three samples is done by using ANOVA: Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 42 2 / (3 × 6) = 1764 / 18 = 98.00 Sum of squares, ss Sum of squares, samples

= (Sum of the squares of the total for each sample / number of judgments for each sample) − CF = (132 + 172 + 122) / 6 − 98.0 = 602 / 6 – 98.0 = 2.33

Sum of squares, judges

= (Sum of the squares of the total for each judge / number of judgments by each judge) − CF = (82 + 72 + 82 + 52+ 72 + 72) / 3 − 98.0 = 300 / 3 − 98.0 = 2.00

Sum of squares, total

= Sum of the squares of each judgment − CF = (32 + 32 + …..32 + 22) − 98.0 = 104 − 98.0 = 6.00

Sum of squares, error

= ss total − ss judges − ss samples = 6.00 − 2.00 − 2.33 = 1.67

Degrees of freedom, df df, samples = 3 – 1 = 2 df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 2.33/2 = 1.17 ms, judges = 2.00/5 = 0.40 ms, error = 1.67/10 = 0.17

Variance ratio, F F, samples = 1.17/0.17 = 6.88 F, judges = 0.40/0.17 = 2.35

Analysis of variance table (ANOVA table): Source of variation df Samples 2 Judges 5 Error 10 Total 17 1% level of significance Samples: F calculated (6.88) < F table (7.56)

ss 2.33 2.00 1.67 6.00

ms 1.17 0.40 0.17

F 6.88 2.35

5% level of significance F calculated (6.88) > F table (4.10)

Conclusion: There is significant difference in the taste attribute between the samples at 5% level. Judges: F calculated (2.35) < F table (5.64)

F calculated (2.35) < F table (3.33)

Conclusion: There is no significant difference among the judges at both 5% level and 1% level. Since there is a significant difference among the samples at 5% level, Tukey’s Test is used to determine which sample are significant different from the others. The standard error of the sample mean is calculated. SE =

0.17 6

= 0.028 = 0.17 For 3 samples and 10 degrees of freedom, the value obtained from table is 3.88 Least significant difference = 3.88 × 0.17 = 0.66 Any two sample means that differ by 0.66 or more are significantly different at the 5% level. Sample scores

:

Sample means

:

Fish Finger 13 2.17

Fish Cake 17 2.83

The sample means are arranged according to magnitude: Fish Cake Fish Finger 2.83 2.17

Fish Ball 12 2.00 Fish Ball 2.00

Each mean is compared with the others to see if the difference is 0.66 or more Fish Cake – Fish Finger = 2.83 – 2.17 = 0.66 = 0.66 Fish Cake – Fish Ball = 2.83 – 2.00 = 0.83 > 0.66 Fish Finger – Fish Ball = 2.17 – 2.00 = 0.17 < 0.66 The results are shown using letters to indicate differences: Fish Cake Fish Finger 2.83a 2.17b

Fish Ball 2.00b

Any two values not followed by the same letter are significantly different at the 5% level. Conclusion: Fish cake is significantly more acceptable in taste attribute compared to fish ball and fish finger at 5% level. c) Color Group 1 2 3 4 5 6 Total Mean

Fish Finger 3 3 3 3 3 3 18 3.00

Products Fish Cake 3 2 3 2 3 2 15 2.50

Total Fish Ball 2 2 2 3 3 2 14 2.33

8 7 8 8 9 7 47

Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the color attribute of the three samples is done by using ANOVA: Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 47 2 / (3 × 6) = 2209 / 18 = 122.72 Sum of squares, ss Sum of squares, samples

Sum of squares, judges

= (Sum of the squares of the total for each sample / number of judgments for each sample) − CF = (182 + 152 + 142) / 6 − 122.72 = 745 / 6 – 122.72 = 1.45 = (Sum of the squares of the total for each judge / number of judgments by each judge) − CF = (82 + 72 + 82 + 82+ 92 + 72) / 3 − 122.72 = 371 / 3 − 122.72 = 0.95

Sum of squares, total

= Sum of the squares of each judgment − CF = (32 + 32 + …..22 + 22) − 122.72 = 127 − 122.72 = 4.28

Sum of squares, error

= ss total − ss judges − ss samples = 4.28 − 0.95 − 1.45 = 1.88

Degrees of freedom, df df, samples = 3 – 1 = 2 df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 1.45/2 = 0.73 ms, judges = 0.95/5 = 0.19 ms, error = 1.88/10 = 0.19 Variance ratio, F F, samples = 0.73/0.19 = 3.84 F, judges = 0.19/0.19 = 1.00 Analysis of variance table (ANOVA table): Source of variation df Samples 2 Judges 5 Error 10 Total 17 1% level of significance Samples: F calculated (3.84) < F table (7.56)

ss 1.45 0.95 1.88 4.28

ms 0.73 0.19 0.19

F 3.84 1.00

5% level of significance F calculated (3.84) < F table (4.10)

Conclusion: There is no significant difference in the color attribute between the samples at both 5% level and 1% level. Judges: F calculated (1.00) < F table (5.64)

F calculated (1.00) < F table (3.33)

Conclusion: There is no significant difference among the judges at both 5% level and 1% level.

d) Odor Group 1 2 3 4 5 6 Total Mean

Fish Finger 3 2 2 3 3 2 15 2.50

Products Fish Cake 3 3 3 3 3 2 17 2.83

Total Fish Ball 2 1 1 1 1 1 7 1.17

8 6 6 7 7 5 39

Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the odor attribute of the three samples is done by using ANOVA: Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 39 2 / (3 × 6) = 1521 / 18 = 84.50 Sum of squares, ss Sum of squares, samples

Sum of squares, judges

= (Sum of the squares of the total for each sample / Number of judgments for each sample) − CF 2 = (15 + 172 + 72) / 6 − 84.50 = 563 / 6 – 84.50 = 9.33 = (Sum of the squares of the total for each judge / Number of judgments by each judge) − CF = (82 + 62 + 62 + 72+ 72 + 52) / 3 − 84.5 = 259 / 3 − 84.50 = 1.83

Sum of squares, total

= Sum of the squares of each judgment − CF = (32 + 32 + …..22 + 12) − 84.5 = 97 − 84.5 = 12.50

Sum of squares, error

= ss total − ss judges − ss samples = 12.50 − 1.83 − 9.33 = 1.34

Degrees of freedom, df df, samples = 3 – 1 = 2

df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 9.33/2 = 4.67 ms, judges = 1.83/5 = 0.37 ms, error = 1.34/10 = 0.13 Variance ratio, F F, samples = 4.67/0.13 = 35.92 F, judges = 0.37/0.13 = 2.85 Analysis of variance table (ANOVA table): Source of variation df ss Samples 2 9.33 Judges 5 1.83 Error 10 1.34 Total 17 12.50 1% level of significance

ms 4.67 0.37 0.13

F 35.92 2.85

5% level of significance

Samples: F calculated (35.92) > F table (7.56)

F calculated (35.92) > F table (4.10)

Conclusion: There is significant difference in the odor attribute between the samples at 1% level. Judges: F calculated (2.85) < F table (5.64)

F calculated (2.85) < F table (3.33)

Conclusion: There is no significant difference among the judges at 1% level. Since there is a significant difference among the samples at 1% level, Tukey’s Test is used to determine which sample are significant different from the others. The standard error of the sample mean is calculated. SE =

0.13 6

= 0.022 = 0.15 For 3 samples and 10 degrees of freedom, the value obtained from table is 3.88 Least significant difference

= 3.88 × 0.15 = 0.58

Any two sample means that differ by 0.58 or more are significantly different at the 5% level.

Sample scores

:

Sample means

:

Fish Finger 15 2.50

Fish Cake 17 2.83

The sample means are arranged according to magnitude: Fish Cake Fish Finger 2.83 2.50

Fish Ball 7 1.17 Fish Ball 1.17

Each mean is compared with the others to see if the difference is 0.66 or more Fish Cake – Fish Finger = 2.83 – 2.50 = 0.33 < 0.58 Fish Cake – Fish Ball = 2.83 – 1.17 = 1.66 > 0.58 Fish Finger – Fish Ball = 2.50 – 1.17 = 1.33 > 0.58 The results are shown using letters to indicate differences: Fish Cake Fish Finger 2.67a 2.17a

Fish Ball 1.67b

Any two values not followed by the same letter are significantly different at the 5% level. Conclusion: Fish ball is significantly unacceptable in odor attribute than fish cake and fish finger at 5% level. e) Overall Appearance Group Products Total Fish Finger Fish Cake Fish Ball 1 3 3 2 8 2 2 2 1 5 3 3 3 2 8 4 3 2 2 7 5 3 3 3 9 6 3 3 2 8 Total 17 16 12 45 Mean 2.83 2.67 2.00 Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the overall appearances of the three samples is done by using ANOVA: Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 45 2 / (3 × 6) = 2025 / 18 = 112.50 Sum of squares, ss Sum of squares, samples

= (Sum of the squares of the total for each sample / number of judgments for each sample) − CF 2 = (17 + 162 + 122) / 6 − 112.5 = 689 / 6 – 112.5 = 2.33

Sum of squares, judges

= (Sum of the squares of the total for each judge / number of judgments by each judge) − CF = (82 + 52 + 82 + 72+ 92 + 82) / 3 − 112.5 = 347 / 3 − 112.5 = 3.17

Sum of squares, total

= Sum of the squares of each judgment − CF = (32 + 32 + …..22 + 22) − 112.5 = 119 − 112.5 = 6.50

Sum of squares, error

= ss total − ss judges − ss samples = 6.50 – 3.17 - 2.33 = 1.00

Degrees of freedom, df df, samples = 3 – 1 = 2 df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 2.33/2 = 1.17 ms, judges = 3.17/5 = 0.63 ms, error = 1.00/10 = 0.10 Variance ratio, F F, samples = 1.17/0.10 = 11.70 F, judges = 0.63/0.10 = 6.30 Analysis of variance table (ANOVA table): Source of variation Df Samples 2 Judges 5 Error 10 Total 17 1% level of significance Samples: F calculated (11.70) > F table (7.56)

ss 2.33 3.17 1.00 6.50

ms 1.17 0.63 0.10

F 11.70 6.30

5% level of significance F calculated (11.70) > F table (4.10)

Conclusion: There is significant difference in the overall appearance between the samples at 1% level. Judges: F calculated (6.30) > F table (5.64)

F calculated (6.30) > F table (3.33)

Conclusion: There is significant difference among the judges at 1% level. Since there is a significant difference among the samples at 1% level, Tukey’s Test is used to determine which sample are significant different from the others.

The standard error of the sample mean is calculated. SE =

0.1 6

= 0.0167 = 0.13 For 3 samples and 10 degrees of freedom, the value obtained from table is 3.88 Least significant difference

= 3.88 × 0.13 = 0.50

Any two sample means that differ by 0.50 or more are significantly different at the 5% level. Sample scores

:

Sample means

:

Fish Finger 17 2.83

Fish Cake 16 2.67

Fish Ball 12 2.00

The sample means are arranged according to magnitude: Fish Cake Fish Finger 2.83 2.67

Fish Ball 2.00

Each mean is compared with the others to see if the difference is 0.66 or more Fish Cake – Fish Finger = 2.83 – 2.67 = 0.16 < 0.50 Fish Cake – Fish Ball = 2.83 – 2.00 = 0.83 > 0.50 Fish Finger – Fish Ball = 2.67 – 2.00 = 0.67 > 0.50 The results are shown using letters to indicate differences: Fish Cake Fish Finger Fish Ball 2.67a 2.17a 1.67b Any two values not followed by the same letter are significantly different at the 5% level. Conclusion: Fish ball is significantly unacceptable (overall appearance) than fish cake and fish finger at 5% level. f) Overall Acceptability Group Fish Finger 1 3 2 2 3 2 4 2 5 2 6 2 Total 13

Products Fish Cake 3 3 3 2 3 2 16

Total Fish Ball 2 1 2 1 2 2 10

8 6 7 5 7 6 39

Mean

2.17

2.67

1.67

Since there are three samples used in this test, the statistical analysis to determine whether there is significant difference between the overall acceptability of the three samples is done by using ANOVA: Analysis of variance (ANOVA): Correction factor (CF) = Total 2 / number of responses = 39 2 / (3 × 6) = 1521 / 18 = 84.50 Sum of squares, ss Sum of squares, samples

= (Sum of the squares of the total for each sample / number of judgments for each sample) − CF 2 = (13 + 162 + 102) / 6 − 84.50 = 525 / 6 – 84.50 = 3.00

Sum of squares, judges

= (Sum of the squares of the total for each judge / number of judgments by each judge) − CF = (82 + 62 + 72 + 52+ 72 + 62) / 3 − 84.50 = 259 / 3 − 84.50 = 1.83

Sum of squares, total

= Sum of the squares of each judgment − CF = (32 + 32 + …..22 + 22) − 84.50 = 91 − 84.50 = 6.50

Sum of squares, error

= ss total − ss judges − ss samples = 6.50 − 1.83 − 3.00 = 1.67

Degrees of freedom, df df, samples = 3 – 1 = 2 df, judges = 6 – 1 = 5 df, total = 18 – 1 = 17 df, error = 17 – 2 – 5 = 10 Mean square, ms ms, samples = 3.00/2 = 1.50 ms, judges = 1.83/5 = 0.37 ms, error = 1.67/10 = 0.17 Variance ratio, F F, samples = 1.50/0.17 = 8.82 F, judges = 0.37/0.17 = 2.18

Analysis of variance table (ANOVA table): Source of variation df Samples 2 Judges 5 Error 10 Total 17 1% level of significance

ss 3.00 1.83 1.67 6.50

ms 1.50 0.37 0.17

F 8.82 2.18

5% level of significance

Samples: F calculated (8.82) > F table (7.56)

F calculated (8.82) > F table (4.10)

Conclusion: There is significant difference in the overall acceptability between the samples at 1% level. Judges: F calculated (2.18) < F table (5.64)

F calculated (2.18) < F table (3.33)

Conclusion: There is no significant difference among the judges at 1% level. Since there is a significant difference among the samples at 1% level, Tukey’s Test is used to determine which sample are significant different from the others. The standard error of the sample mean is calculated. SE =

0.17 6

= 0.028 = 0.17 For 3 samples and 10 degrees of freedom, the value obtained from table is 3.88 Least significant difference

= 3.88 × 0.17 = 0.66

Any two sample means that differ by 0.66 or more are significantly different at the 5% level. Sample scores

:

Sample means

:

Fish Finger 13 2.17

Fish Cake 16 2.67

The sample means are arranged according to magnitude: Fish Cake Fish Finger 2.67 2.17

Fish Ball 10 1.67 Fish Ball 1.67

Each mean is compared with the others to see if the difference is 0.66 or more Fish Cake – Fish Finger = 2.67 – 2.17 = 0.50 < 0.66

Fish Cake – Fish Ball = 2.67 – 1.67 = 1.00 > 0.66 Fish Finger – Fish Ball = 2.17 – 1.67 = 0.50 < 0.66 The results are shown using letters to indicate differences: Fish Cake Fish Finger 2.67a 2.17ab

Fish Ball 1.67b

Any two values not followed by the same letter are significantly different at the 5% level. Conclusion: Fish cake is significantly more acceptable (overall acceptability) than fish ball at 5% level.