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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 (H0):
• There is an equal probability of people needing the Indian sizing system and satisfied with present sizing system.
• Alternate Hypothesis (Ha):
• 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 1/3rd 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, Oi = Observed value

Ei= 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:
• Ha = 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 H0.

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:
• Ha = 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-3: Total sum of squares

SST= SSB + SSW
=1069.5556+744.6667
=1814.2222

#### 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 samplesnk = 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

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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

## 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:
• Ha = There is correlation between age of participants to number of participants willing to volunteer.

Solution:

Let, X: Age of participants, Y: No. of people, Mx: Mean of X Values and My: 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 – Mx & Y – My: Deviation scores
(X – Mx)2 & (Y – My)2: Deviation Squared
(X – Mx) (Y – My): 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.

## 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 (Ha):
• 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 + p2 + p3 + p4 + 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.

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