Need of a Sizing System for Indian Apparel Industry
Research study on the requirement of a Sizing System exclusive to Indian Apparel Industry
The research report here examines awareness among people about the Indian apparel sizing system in accordance with the universal sizing system. Also, a questionnaire provided at the end of the article would help people volunteering to collect anthropometric data on behalf of any organization.
HYPOTHESIS TESTING
TEST 1: CHI SQUARE GOODNESS OF FIT TEST
Question: Do you think India needs its own sizing system?
Response collected
RESEARCH HYPOTHESIS
In this study, the following hypothesis has been assumed to find out that whether the null hypothesis stand true or not.
- Null Hypothesis (H_{0}):
- There is an equal probability of people needing the Indian sizing system and satisfied with present sizing system.
- Alternate Hypothesis (H_{a}):
- People are not satisfied with present sizing system and need their own Indian sizing system.
Analysis
As we are assuming the null hypothesis, where the outcomes have equal probability so the expected result would be 1/3^{rd} of total sample (i.e., 165) as there are 3 choices.
Therefore, expected outcome can be
Yes | No | Maybe | |
Observed | 136 | 3 | 26 |
Expected | 55 | 55 | 55 |
where, O_{i} = Observed value
E_{i}= Expected value
Degree of freedom = (rows-1) (columns -1) = (2-1) (3-1) = 2
Observed | Expected | Difference | Difference Square | Difference Square/ Expected value | |
Yes | 136 | 55 | 81.00 | 6561.00 | 119.29 |
No | 3 | 55 | -52.00 | 2704.00 | 49.16 |
Maybe | 26 | 55 | -29.00 | 841 | 15.29 |
Chi squared
(χ^{2}) |
183.745 |
Significance level (α) = 0.05(indicates a 5% risk of concluding that a difference exists when there is no actual difference)
Chi squared is equal to 183.745, therefore its p-value is .00001.
- p-value > 0.1: No evidence
- p-value between 0.05 and 0.1: Weak evidence
- p-value between 0.01 and 0.05: Evidence
- p-value between 0.001 and 0.01: Strong evidence
- p-value < 0.001: Very strong evidence
TEST 2: Z test for proportion
Proportion of Gender
RESEARCH HYPOTHESIS
In this study, the following hypothesis has been assumed to find out that whether the null hypothesis stand true or not.
- Null Hypothesis:
- H_{0 }= Probability that number of male participants are equal to female participants.
- Alternate Hypothesis:
- H_{a} = Probability that number of male participants are not equal to female participants.
- z = calculated critical value.
- x = Observed success.
- n = Size of samples.
- p = Probability of success
- q or 1-p= Probability of failure
Here,
- n = 165
- X = 60 or 105
- p = 1/2
- 1-p = 1/2
Significance level = 0.05
At 0.05 Significance level Z is 1.96.
RESULT
Our observed value of Z is 3.51 which is greater than the critical value of 1.96. We therefore reject null hypothesis H_{0.}
Question: Which part of the body feels uncomfortable in your ready-made garments?
Response collected
RESEARCH HYPOTHESIS
In this study, the following hypothesis has been assumed to find out that whether the null hypothesis stand true or not.
- Null Hypothesis:
- H_{0 }= There is no difference between the size issue in formal, casual and party wear clothing.
- Alternate Hypothesis:
- H_{a} = There is difference between the size issue in formal, casual and party wear clothing.
Analysis
Formal wear (A) | Casual wear (B) | Party wear (C) | |
Bust | 56 | 32 | 42 |
Waist | 40 | 20 | 42 |
Hip | 20 | 7 | 24 |
Solution
Data table
Step-1: sum of squares between samples
Step-2: sum of squares within samples
Step-3: Total sum of squares
SST= SSB + SSW
=1069.5556+744.6667
=1814.2222
Step-4: variance between samples
MSB =SSB / k-1
=1069.5556/2
=534.7778
Step-5: variance within samples
MSW=SSW/ n–k
=744.6667/ 9-3
=744.6667/ 6
=124.1111
Step-6: test statistic F for one way ANOVA test
F=MSB / MSW
=534.7778 / 124.1111
=4.3089
- The degree of freedom between samples= k-1=2
- degree of freedom within samplesn–k = 9-3 = 6
ANOVA table
Source of Variation | Sums of Squares SS |
Degrees of freedom DF |
Mean Squares MS |
F | p-value |
Between samples | SSB = 1069.5556 | k-1 = 2 | MSB = 534.7778 | 4.3089 | 0.0692 |
Within samples | SSW = 744.6667 | n–k = 6 | MSW = 124.1111 | ||
Total | SST = 1814.2222 | n-1 = 8 |
Practical Garment Production Process – A Case Study
End-to-End detailed garment production process
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Apparel Manufacutring technological advancements adopted by different countries
RESULT
As calculated Fis 4.3089 which is smaller than 5.1433.
So, H_{0} is accepted, hence there is no difference between the size issue in formal, casual and party wear clothing.
F (2,6) at 0.05 level of significance=5.1433
TEST 4: Pearson Correlation Coefficient test
Question: If a sizing survey is initiated, will you be willing to volunteer?
Response collected
Age | No. of people willing to volunteer |
19 | 3 |
22 | 15 |
31 | 1 |
39 | 2 |
23 | 4 |
20 | 23 |
45 | 1 |
24 | 3 |
21 | 27 |
26 | 1 |
32 | 2 |
RESEARCH HYPOTHESIS
In this study, the following hypothesis has been assumed to find out that whether the null hypothesis stand true or not.
- Null Hypothesis:
- H_{0 }= There is no correlation between age of participants to number of participants willing to volunteer
- Alternate Hypothesis:
- H_{a} = There is correlation between age of participants to number of participants willing to volunteer.
Solution:
Let, X: Age of participants, Y: No. of people, M_{x}: Mean of X Values and M_{y}: Mean of Y Values
X | Y |
19 | 3 |
22 | 15 |
31 | 1 |
39 | 2 |
23 | 4 |
20 | 23 |
45 | 1 |
24 | 3 |
21 | 27 |
26 | 1 |
32 | 2 |
Sum = 302 | Sum= 82 |
Mean (M_{x})= 27.455 | M_{y}= 7.455 |
X – M_{x} & Y – M_{y}: Deviation scores
(X – M_{x})^{2} & (Y – M_{y})^{2}: Deviation Squared
(X – M_{x}) (Y – M_{y}): Product of Deviation Scores
X- M_{x} | Y-M_{y} | (X- M_{x})^{2} | (Y-M_{y})^{2} | (X- M_{x}) (Y-M_{y}) |
-8.455 | -4.455 | 71.479 | 19.843 | 37.661 |
-5.455 | 7.545 | 29.752 | 56.934 | -41.157 |
3.545 | -6.455 | 12.570 | 41.661 | -22.884 |
11.545 | -5.455 | 133.298 | 29.752 | -62.975 |
-4.455 | -3.455 | 19.843 | 11.934 | 15.388 |
-7.455 | 15.545 | 55.570 | 241.661 | -115.884 |
17.545 | -6.455 | 307.843 | 41.661 | -113.248 |
-3.455 | -4.455 | 11.934 | 19.843 | 15.388 |
-6.455 | 19.545 | 41.661 | 382.025 | -126.157 |
-1.455 | -6.455 | 2.116 | 41.661 | 9.388 |
4.545 | -5.455 | 20.661 | 29.752 | -24.793 |
Sum: 706.727 | Sum: 916.727 | Sum: -429.273 |
The value of r = -0.5333
Degree of correlation:
- Perfect: If the value is near ± 1, then it said to be a perfect correlation: as one variable increases, the other variable tends to also increase (if positive) or decrease (if negative).
- High degree: If the coefficient value lies between ± 0.50 and ± 1, then it is said to be a strong correlation.
- Moderate degree: If the value lies between ± 0.30 and ± 0.49, then it is said to be a medium correlation.
- Low degree: When the value lies below + .29, then it is said to be a small correlation.
RESULT
As r value is -0.5333, there is a negative correlation, which means there is a tendency for high X variable scores to go with low Y variable scores (and vice versa). Therefore, null hypothesis H_{0} is rejected.
TEST 5: Fisher’s Exact Method test
Question: Do you know that India does have their own sizing system?
Response collected
Response with respect to gender of participants
Male | Female | Total | |
Yes | 36 | 71 | 107 |
No | 24 | 34 | 58 |
Total | 60 | 105 | 165 |
RESEARCH HYPOTHESIS
In this study, the following hypothesis has been assumed to find out that whether the null hypothesis stand true or not.
- Null Hypothesis (H_{0}):
- The gender of the participants has no association with having awareness aboutIndian sizing system.
- Alternate Hypothesis (H_{a}):
- The gender of the participants has association with having awareness about Indian sizing system.
Assigning alphabets to each data
Male | Female | Total | |
Yes | a = 36 | b = 71 | a +b = 107 |
No | c = 24 | d = 34 | c +d = 58 |
Total | a +c = 60 | b +d = 105 | N = 165 |
For Fisher’s Exact Test we use a mathematical method called Combination which helps us find out the value of p (two tailed) so that it can be compared with significance level (α) of 0.05 to check our hypothesis.
Similarly, we need to calculate p_{1}, p_{2}, p_{3}, p_{4} 3-4 times in which the data is altered by +1 and -1 alternatively in each row which will give the p-value (p_{1}+p_{2}+p_{3}+p_{4}).
Table 2
Male | Female | Total | |
Yes | 32 | 75 | 107 |
No | 28 | 30 | 58 |
Total | 60 | 105 | 165 |
Assigning alphabets to each data
Male | Female | Total | |
Yes | a = 32 | b = 75 | a+b = 107 |
No | c = 28 | d = 30 | c+d = 58 |
Total | a+c = 60 | b+d = 105 | N = 165 |
Table 3
Male | Female | Total | |
Yes | 33 | 74 | 107 |
No | 27 | 31 | 58 |
Total | 60 | 105 | 165 |
Assigning alphabets to each data-
Male | Female | Total | |
Yes | a = 33 | b = 75 | a+b = 107 |
No | c = 27 | d = 31 | c+d = 58 |
Total | a+c = 60 | b+d = 105 | N = 165 |
Table 4
Male | Female | Total | |
Yes | 34 | 73 | 107 |
No | 26 | 32 | 58 |
Total | 60 | 105 | 165 |
Assigning alphabets to each data –
Male | Female | Total | |
Yes | a = 34 | b = 73 | a+b = 107 |
No | c = 26 | d = 32 | c+d = 58 |
Total | a+c = 60 | b+d = 105 | N = 165 |
Table 5
Male | Female | Total | |
Yes | 35 | 72 | 107 |
No | 25 | 33 | 58 |
Total | 60 | 105 | 165 |
Assigning alphabets to each data –
Male | Female | Total | |
Yes | a = 35 | b = 72 | a+b = 107 |
No | c = 25 | d = 33 | c+d = 58 |
Total | a+c = 60 | b+d = 105 | N = 165 |
Calculating p-value for the Fisher’s Exact Test: –
p-value = p_{1} + p_{2 +} p_{3 +} p_{4 + }p_{5}
= 0.165064951 + 0.018119212 + 0.037194863 + 0.068304538 +0.112244166
= 0.40092773
RESULT
As p value is greater than significance level of 0.05, H_{0 }is accepted. “The gender of the participants has no association with the knowledge of sizing system”, stands true.
This requires a systematic and scientific system for measuring and classifying human bodies, which is the purpose of creating a clothing scale. https://www.vitahaus.com/
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Hi Aryan Rathore, how can one reach you via email. Have quiries. You can write me through amaniwilliam@outlook.com
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