Sensory Evaluation

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EXPERIEMENT# 7
CONSUMER PREFERENCE TEST

Objectives:
* To know the principles of consumer preference tests
* To cite the uses of consumer preference tests
* To conduct properly paired preference and preference ranking tests * To analyze sensory data from preference tests

PAIREDPREFERENCE TEST
Materials:
Marshmallow – white (2 brands)
Saucers
Cups
Distilled water

Method:
A. Sample Preparation and Presentation
1. Prepare the Master Sheet.
a. Assign random 3-digit codes.
b. Since, there are only 2 coded samples per judge; toss a coin can decide which sample will be served first. 2. Prepare the Summary Sheet to serve as guide in presentation of samples. 3. Label saucers with the assigned sample codes.

4. Place each sample to properly labeled saucers.
B. Sample Evaluation
1. Present both samples to each panelist.
2. Follow the instruction written in the Answer Sheet.
3. Rinse with water to clear your palate every after tasting. Retasting is allowed. 4. Note down observation.
C. Data Evaluation
1. Refer to your Master Sheet to check the answers of your panelists. Record scores in the Data Sheet. Tabulate results.
2. Analyze results.
3. Compile results and submit written report.

Review of Related Literature:

Tabulation of Results:

Panelist No.| Sample|
1| B|
2| B|
3| A|
4| A|
5| B|
6| B|
7| B|
8| B|
9| A|
10| A|

The panelists are presented with 2 samples of same product but of different brands. They are to choose which marshmallow they prefer. 6 out of 10 panelists preferred Sample B over Sample A.

Computation and Analysis:
1. Quick Test

n=10

At 0.05 probability level, n of 10 corresponds to a tabular value of 9.

10>9 (p> 0.05)

Conclusion:
Observed number is greater than the tabular value. Therefore, Sample A is significantly preferred than Sample B. 2. Chi-square Test

x2 = (O1-E1)2-0.5 + (O2-E2)2-0.5 = (4-5)2-0.5 + (6-5)2-0.5 = 0.2 E1 E2 5 5 The tabular value at df=1 and p at 0.05 is 3.84. 0.2≤3.84 (p≥0.05). Conclusion:

Based on the Chi-square Test, Sample A is not significantly preferred over Sample B since the computed value is less than the tabular value. 3. Binomial probability in preference test

z= (Proportion observed – Proportion expected) – Continuity correction (Standard error of the proportion) = (4/10-1/2) -1/20
√(1/2)(1/2)/10
-0.15
0.158
= 0.949367088

0.95≤ 1.96.

Conclusion:
Sample A is not significantly preferred over Sample B since the computed value is less than the tabular z value (p≤0.05).

Summary:

Bibliography:

Appendix:

MASTER SHEET

TREATMENTS|
Panelist No.| Name| A| B|
1| Charlou| 862 2| 351 1|
2| Pamela| 653 1| 714 2|
3| Andrea| 742 2| 117 1|
4| Dolores| 975 2| 171 1|
5| Jenny| 113 1| 256 2|
6| Loren| 325 1| 663 2|
7| Rejoice| 461 2| 112 1|
8| Domini| 894 1| 442 2|
9| Clariz| 581 1| 991 2|
10| Cathy| 553 2| 655 1|

Legend:
SAMPLE| CODE|
A| Mello Max Marshmallow|
B| SM Bonus Marshmallow|

SUMMARY SHEET

TREATMENTS|
Panelist No.| Name| 1st| 2nd|
1| Charlou| 351| 862|
2| Pamela| 653| 714|
3| Andrea| 117| 742|
4| Dolores| 171| 975|
5| Jenny| 113| 256|
6| Loren| 325| 663|
7| Rejoice| 112| 461|
8| Domini| 894| 442|
9| Clariz| 581| 991|
10| Cathy| 655| 553|

DATA SHEET

TREATMENTS|
Panelist No.| Name| Answer| Sample Code| A = 4B = 6|
1| Charlou| 351| B| |
2| Pamela| 714| B| |
3| Andrea| 742| A| |
4| Dolores| 975| A| |
5| Jenny| 256| B| |
6| Loren| 663| B| |
7| Rejoice| 112| B| |
8| Domini| 442| B| |
9| Clariz| 581| A| |...
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