Need of a Sizing System for Indian Apparel Industry

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 (H₀):
  • 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 (χ2)					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:
  • H₀ = 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.

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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:
  • H₀ = 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

 

RESULT

As calculated Fis 4.3089 which is smaller than 5.1433.
So, H₀ 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

 

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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:
  • H₀ = 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 (Mx)= 27.455	My= 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- Mx	Y-My	(X- Mx)2	(Y-My)2	(X- Mx) (Y-My)
-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:

1. 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).
2. High degree: If the coefficient value lies between ± 0.50 and ± 1, then it is said to be a strong correlation.
3. Moderate degree: If the value lies between ± 0.30 and ± 0.49, then it is said to be a medium correlation.
4. 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 H₀ is rejected.

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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 (H₀):
  • 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 p₁, p₂, p₃, p₄ 3-4 times in which the data is altered by +1 and -1 alternatively in each row which will give the p-value (p₁+p₂+p₃+p₄).

 

 

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 = p₁ + p_{2 +} p_{3 +} p_{4 +} p₅

= 0.165064951 + 0.018119212 + 0.037194863 + 0.068304538 +0.112244166

= 0.40092773

 

RESULT

As p value is greater than significance level of 0.05, H₀ is accepted. “The gender of the participants has no association with the knowledge of sizing system”, stands true.