**3. Reference statistical methodology in quality control**

Hypothesis testing is one of the useful tools of statistical methodology in quality control and improvement. In hypothesis testing, there are the null hypothesis (*H*0) and the alternative hypothesis (*H*1). While the null hypothesis *H*<sup>0</sup> indicates a certain point of view of the research question, the alternative hypothesis *H*<sup>1</sup> indicates the opposite of that point of view. The opposite can be stated as not equal, greater than, or less than. Not equal is a two-sided alternative hypothesis, and the latter two are one-sided alternative hypotheses. Therefore, the determination of the parameter values in a hypothesis to be tested is very important, as they may either come from past information, a theory or model, or conformity. A statistical inference is reached with correct determination.

When working with test results, it is assumed that the obtained test results are normally distributed. If the underlying distribution of the obtained results deviate moderately from normal distribution, *t*-tests perform reasonably well because of the robustness of the test. If the underlying distribution of the obtained results deviates substantially from normal distribution, when the sample size is large, because of the central limit theorem (CLT), they approximate normal distribution [14]. Especially in textile manufacturing, it is considered that the test results of properties of a product exhibit normal distribution.

In statistical inference, there may be errors, especially in hypothesis testing, wherein two kinds of errors exist. The first one is the null hypothesis is rejected even if it is true, which is the wrong decision. This is Type I Error and is symbolized by *α* which is also called the level of significance. In this case, the null hypothesis is unable to be rejected by 1*α* probability, or which is the right decision. The second kind of error is the null hypothesis is unable to be rejected even if it is false, which is the wrong decision. This is Type II Error and is symbolized by *β*. In this case, the null hypothesis is rejected by 1*β* probability, or which is the right decision. Hypothesis testing errors are shown in **Table 1**. The level of significance *α* would take values like 0.1, 0.05, 0.01, 0.001, etc.

By designing a test procedure in hypothesis testing, a value of the probability of Type I Error *α* is specified so that a small value of the probability of Type II Error *β* is obtained. The *α* risk can directly be controlled or chosen; the *β* risk can indirectly be controlled because it is the function of sample size; consequently, the larger the sample size, the smaller it is. In textiles production, Type I Error *α* is sufficient. The nature of textiles production for daily usage like apparel, home textiles (rugs, curtains, bedsheets, carpets, towels, etc.), Type I Error *α* is satisfactory, there is no requirement for Type II Error *β* in such cases. The important thing is to produce yarn, fabric, ready-wear, etc. with level of significance *α* = 0.05, which is usually used and is deemed enough. On the other hand, Type II Error *β* is strongly reasonable for technical textiles like medical, aerotextiles, geotextiles, etc.; even there are cases where 6*σ* is applied (such as in vivo medical textiles, aerotextiles). These special cases will not be studied in this manuscript; for the rest, only Type I Error *α* will be considered.

A hypothesis test can be conducted by different test statistics like the *z* test, *t*-test, *χ*<sup>2</sup> test, the appropriate one is selected in accordance with the purpose of the hypothesis test. The set of values of the test statistic which lead to the rejection of *H*<sup>0</sup> is named as the critical region or rejection region for the test.

Therefore, the procedures for a hypothesis test can be listed as:



#### **Table 1.**

*Hypothesis testing errors.*

Sampling is very important in hypothesis testing because an inference will be reached through the parameter information the samples contain and that conclusion will be applied to all of the rest of the population.

In a hypothesis testing, if *x* is a random variable with unknown mean *μ* and known variance *σ*2, then the hypothesis testing is that the mean is equal to a chosen value, *μ*0. The null hypothesis (*H*0) and the alternative hypothesis (*H*1) are stated as:

$$\begin{aligned} H\_0: \mu &= \mu\_0 \\ H\_1: \mu &\neq \mu\_0 \end{aligned} \tag{1}$$

Level of significance *α* is determined. *n* samples are taken from the random variable x and the *z* statistic is calculated:

$$Z\_0 = \frac{\overline{\mathfrak{X}} - \mu\_0}{\sigma / \sqrt{n}} \tag{2}$$

If j j *Z*<sup>0</sup> ≻*Zα=*<sup>2</sup> then *H*<sup>0</sup> is rejected, *Zα=*<sup>2</sup> is the upper *α=*2 percentage point of the standard normal distribution at a fixed significance level two-sided.

If *x* is a random variable with unknown mean *μ* and unknown variance *σ*<sup>2</sup> , then the hypothesis testing is that the mean is equal to a chosen value, *μ*0. The hypothesis is stated as:

$$\begin{aligned} H\_0: \mu &= \mu\_0 \\ H\_1: \mu &\neq \mu\_0 \end{aligned} \tag{3}$$

Since the variance is unknown, it is assumed that the *x* random variable has a normal distribution and deviations from normality will not affect the results much. Also, *σ*<sup>2</sup> is unknown and it is estimated by *s* 2. The level of significance *α* is determined. *n* samples are taken from the random variable *x* and the test statistic becomes a *t*-test:

$$t\_0 = \frac{\overline{\mathfrak{X}} - \mu\_0}{s / \sqrt{n}} \tag{4}$$

where instead of a normal distribution it becomes a *t* distribution with *n* � 1 degrees of freedom.

If j j ð Þ *t*<sup>0</sup> ≻*t<sup>α</sup>=*2,*n*�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*�<sup>1</sup> is the upper *α=*2 percentage point of the *t* distribution with *n* � 1 degrees of freedom at a fixed significance level two-sided.

Statistical tests on means are very little sensitive to normality assumptions but the tests on variances are sensitive. To test the variance of a normal distribution is equal to a chosen variance, *σ*<sup>2</sup> 0, then the hypothesis is stated as:

$$\begin{aligned} H\_0: \sigma^2 &= \sigma\_0^2\\ H\_1: \sigma^2 &\neq \sigma\_0^2 \end{aligned} \tag{5}$$

and the test statistic becomes a *χ*<sup>2</sup> test:

$$
\chi\_0^2 = \frac{(n-1)s^2}{\sigma\_0^2} \tag{6}
$$

where *s* <sup>2</sup> is the sample variance of *n* repeats. The level of significance *α* is determined. If *χ*<sup>2</sup> <sup>0</sup> ≻ *χ*<sup>2</sup> *<sup>α</sup>=*2,*n*�<sup>1</sup> or if *<sup>χ</sup>*<sup>2</sup> <sup>0</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α=*2,*n*�<sup>1</sup> then the null hypothesis *<sup>H</sup>*<sup>0</sup> is rejected for a fixed significance level, *χ*<sup>2</sup> *<sup>α</sup>=*2,*n*�<sup>1</sup> is the upper *<sup>α</sup>=*2 upper percentage point of the chi-square distribution with *<sup>n</sup>* � 1 degrees of freedom and *<sup>χ</sup>*<sup>2</sup> <sup>1</sup>�*α=*2,*n*�<sup>1</sup> is the lower 1 � ð Þ *α=*2 percentage. If a one-sided alternative is specified, then the hypothesis is:

$$\begin{aligned} H\_0: \sigma^2 &= \sigma\_0^2\\ H\_1: \sigma^2 &\prec \sigma\_0^2 \end{aligned} \tag{7}$$

and the null hypothesis is rejected if *χ*<sup>2</sup> <sup>0</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*�1. For the other one-sided alternative, the hypothesis is:

$$\begin{aligned} H\_0: \sigma^2 &= \sigma\_0^2\\ H\_1: \sigma^2 &\succ \sigma\_0^2 \end{aligned} \tag{8}$$

and the null hypothesis is rejected if *χ*<sup>2</sup> <sup>0</sup> ≻*χ*<sup>2</sup> *<sup>α</sup>*,*n*�1.

Chi-square testing is applied a lot in quality improvement by monitoring and control procedures. There may be a normal random variable with mean *μ* and variance *σ*2. If *σ*<sup>2</sup> ≤*σ*<sup>2</sup> 0, *σ*<sup>2</sup> <sup>0</sup> being a chosen value, then the natural inherent variability of the process will be within the requirements of the design and the production will mostly be within the specification limits. But if *σ*<sup>2</sup> ≻*σ*<sup>2</sup> 0, this means that the natural variability in the process is exceeding the specification limits. This case increases the percentage of non-conforming production items.

If there are two independent populations, as shown in **Figure 4**, then it will statistically be tested for the difference in means *<sup>μ</sup>*<sup>1</sup> � *<sup>μ</sup>*2. It is assumed that *<sup>μ</sup>*1, *<sup>x</sup>*1, *<sup>σ</sup>*<sup>2</sup> 1, and *n*<sup>1</sup> are known and belonging to Population 1; whereas *μ*2, *x*2, *σ*<sup>2</sup> 2, and *n*<sup>2</sup> are known and belonging to Population 2. Both samples of the populations are random, and both populations are normally distributed; if they are not normal, the conditions of the CLT applies.

The point estimator of *μ*<sup>1</sup> � *μ*<sup>2</sup> is the difference in sample means *x*<sup>1</sup> � *x*<sup>2</sup> and from the properties of expected values:

$$E(\overline{\mathfrak{x}}\_1 - \overline{\mathfrak{x}}\_2) = E(\overline{\mathfrak{x}}\_1) - E(\overline{\mathfrak{x}}\_2) = \mu\_1 - \mu\_2 \tag{9}$$

is obtained and the variance of *x*<sup>1</sup> � *x*<sup>2</sup> is:

$$V(\overline{\mathbf{x}}\_1 - \overline{\mathbf{x}}\_2) = V(\overline{\mathbf{x}}\_1) + V(\overline{\mathbf{x}}\_2) = \frac{\sigma\_1^2}{n\_1} + \frac{\sigma\_2^2}{n\_2} \tag{10}$$

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

**Figure 4.**

*Symbolization of two independent populations [1].*

From the assumptions and the preceding results, the quantity *Z* with N(0,1) distribution can be stated as:

$$Z = \frac{\overline{\mathbf{x}}\_1 - \overline{\mathbf{x}}\_2 - (\mu\_1 - \mu\_2)}{\sqrt{\frac{\sigma\_1^2}{n\_1} + \frac{\sigma\_2^2}{n\_2}}} \tag{11}$$

If it is tested that the difference in means *μ*<sup>1</sup> � *μ*<sup>2</sup> is zero, that they are equal, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_1 - \mu\_2 &= \mathbf{0} \\ H\_1: \mu\_1 - \mu\_2 &\neq \mathbf{0} \end{aligned} \tag{12}$$

Substituting 0 for *μ*<sup>1</sup> � *μ*2, becomes:

$$Z\_0 = \frac{\overline{\mathfrak{X}}\_1 - \overline{\mathfrak{X}}\_2}{\sqrt{\frac{\sigma\_1^2}{n\_1} + \frac{\sigma\_2^2}{n\_2}}} \tag{13}$$

If j j *Z*<sup>0</sup> ≻*Z<sup>α</sup>=*<sup>2</sup> then *H*<sup>0</sup> is rejected, *Z<sup>α</sup>=*<sup>2</sup> is the upper *α=*2 percentage point of the standard normal distribution at a fixed significance level two-sided.

If there are two independent populations, then the difference in means *μ*<sup>1</sup> � *μ*<sup>2</sup> will statistically be tested. It is assumed that *μ*1, *x*1, and *n*<sup>1</sup> are known belonging to Population 1; *μ*2, *x*2, and *n*<sup>2</sup> are known belonging to Population 2, but *σ*<sup>2</sup> <sup>1</sup> and *σ*<sup>2</sup> <sup>2</sup> are unknown. Both samples of the populations are random, and both populations are normally distributed; if they are not normal, the conditions of the CLT applies. The two *σ*<sup>2</sup> <sup>1</sup> and *σ*<sup>2</sup> <sup>2</sup> may be equal or not. In this manuscript, the condition that they are equal will be considered, becoming *σ*<sup>2</sup> <sup>1</sup> <sup>¼</sup> *<sup>σ</sup>*<sup>2</sup> <sup>2</sup> <sup>¼</sup> *<sup>σ</sup>*<sup>2</sup> . Since *<sup>σ</sup>*<sup>2</sup> <sup>1</sup> and *σ*<sup>2</sup> <sup>2</sup> are unknown, *t*-statistic will be used and sample variances of the two populations would be *s* 2 1, and *s* 2 2, respectively.

The expected value of the difference in sample means *x*<sup>1</sup> � *x*<sup>2</sup> which is an unbiased estimator of the difference in means is:

*Quality Control - An Anthology of Cases*

$$E(\overline{\mathbf{x}}\_1 - \overline{\mathbf{x}}\_2) = \mu\_1 - \mu\_2 \tag{14}$$

The variance of *x*<sup>1</sup> � *x*<sup>2</sup> is:

$$V(\overline{\infty}\_1 - \overline{\infty}\_2) = \frac{\sigma^2}{n\_1} + \frac{\sigma^2}{n\_2} = \sigma^2 \left(\frac{1}{n\_1} + \frac{1}{n\_2}\right) \tag{15}$$

Estimator of *σ*<sup>2</sup> is the combination of *s* 2 <sup>1</sup> and *s* 2 <sup>2</sup> it is the pooled estimator of *σ*2, denoted by *s* 2 *<sup>p</sup>*, which is:

$$s\_p^2 = \frac{(n\_1 - 1)s\_1^2 + (n\_2 - 1)s\_2^2}{n\_1 + n\_2 - 2} \tag{16}$$

*s* 2 *<sup>p</sup>* is the weighted average of the two sample variances *s* 2 <sup>1</sup> and *s* 2 2. The *z* test statistic for unknown *σ* is:

$$z = \frac{\overline{\mathbf{x}}\_1 - \overline{\mathbf{x}}\_2 - (\mu\_1 - \mu\_2)}{\sigma \sqrt{\frac{1}{n\_1} + \frac{1}{n\_2}}} \tag{17}$$

and then, for *t*-statistic *σ* is replaced by *sp*.

$$t = \frac{\overline{\mathbf{x}}\_1 - \overline{\mathbf{x}}\_2 - (\mu\_1 - \mu\_2)}{s\_p \sqrt{\frac{1}{n\_1} + \frac{1}{n\_2}}} \tag{18}$$

a *t* distribution with *n*<sup>1</sup> þ *n*<sup>2</sup> � 2 degrees of freedom, also called the pooled *t*-test.

If it is tested that the difference in means *μ*<sup>1</sup> � *μ*<sup>2</sup> is zero - meaning they are equalthe hypothesis is:

$$\begin{aligned} H\_0: \mu\_1 - \mu\_2 &= \mathbf{0} \\ H\_1: \mu\_1 - \mu\_2 &\neq \mathbf{0} \end{aligned} \tag{19}$$

Substituting 0 for *μ*<sup>1</sup> � *μ*2, it becomes:

$$t\_0 = \frac{\overline{\mathfrak{X}}\_1 - \overline{\mathfrak{X}}\_2}{s\_p \sqrt{\frac{1}{n\_1} + \frac{1}{n\_2}}}\tag{20}$$

If j j *t*<sup>0</sup> ≻ *t<sup>α</sup>=*2,*n*1þ*n*2�<sup>2</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ *n*<sup>2</sup> � 2 degrees of freedom at a fixed significance level two-sided.

If the variances of two independent normal distributions are tested if they are equal, *σ*<sup>2</sup> 1, *s* 2 <sup>1</sup> and *n*<sup>1</sup> for Population 1, and *σ*<sup>2</sup> 2, *s* 2 <sup>2</sup> and *n*<sup>2</sup> for Population 2, then the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_1^2 &= \sigma\_2^2\\ H\_1: \sigma\_1^2 &\neq \sigma\_2^2 \end{aligned} \tag{21}$$

F statistics is the ratio of the two sample variances:

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

$$F\_0 = \frac{s\_1^2}{s\_2^2} \tag{22}$$

*H*<sup>0</sup> is rejected if *F*<sup>0</sup> ≻*F<sup>α</sup>=*2,*n*1�1,*n*2�<sup>1</sup> or *F*<sup>0</sup> ≺ *F*1�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1�1,*n*2�1, which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> � 1 and *n*<sup>2</sup> � 1, respectively, at a fixed significance level two-sided.

#### **4. Proposed statistical approach**

In hypothesis testing, sampling is very important because an inference is reached from the values in the sample about the values in the population. Therefore sampling has to be done very carefully and samples should represent the population. Sampling is a wide subject in textile engineering. Regular sampling during production and acceptance sampling from a static lot are two grand different subjects. This broad topic of sampling in textile engineering can well be covered in a separate manuscript, so the details of sampling will not be dealt herein. Instead, the number of samples, which are repeats, will be indicated as *ni*.

In ring spinning yarn production the number of spindles per spinning frame is the determining factor for sampling. As a general application, at least five bobbins per 500 spindles per spinning frame are taken randomly for the tests of yarn properties. Different frame brands have different spindle numbers such as 576, 1008, 1296, and 1824. depending on the model of the frame. For example, at least 10 bobbins have to be chosen randomly for 576 spindles per frame, at least 15 bobbins have to be chosen randomly for 1008 spindles per frame, at least 15 bobbins again for 1296 spindles per frame, or at least 20 bobbins for 1824 spindles per frame.

When sampling for hypothesis testing in this chapter, the yarn lot is the population, and statistical inference and decisions will be made about the yarn lot from the samples selected from it. In order to conclude that the machine is adjusted correctly or to make a decision about its status, samples are randomly selected as different bobbins from independent, identical, and with equal probability of being chosen spindles on a spinning frame which are adjusted to produce a special yarn. Test results of the samples will give information about the yarn population. **Figure 5** shows the relationship between a population and a sample. In textile engineering, it is assumed that the property values of a textile material have a normal distribution, consequently in yarn spinning, yarn properties also exhibit normal distribution for property values.

The constant of variation (CV%) is a frequently used value in textile engineering. Starting from fibers to the end product, say apparel, fiber (fineness, length, breaking strength – breaking elongation, etc.), yarn (count, twist, irregularity, etc.), fabric (warp and weft density, fabric thickness, etc.), and apparel (measurements, weight, etc.) properties are all tested and the results are statistically analyzed; and the mean *x*, standard deviation *s*, and CV% *<sup>s</sup> <sup>x</sup>* � <sup>100</sup> are calculated. The value of CV% indicates much information about the property it was calculated from. Furthermore, comparisons and evaluations are done making it possible to have a comprehensible understanding of how production is continuing. The constant of variation of yarn count can be expressed as CV%YarnCount. The value CV%YarnCount has a close relationship with the technology of textile machinery. Technology of textile machinery developed considerably a lot when compared to the 1970s and 1980s. Textile machinery producers incorporate broad R&D departments and one of the main topics of their researches is

**Figure 5.** *Relationship between a population and a sample [1].*

on CV%YarnCount. As a general consideration, CV%YarnCount 5% was acceptable until the late 80's, whereas the CV%YarnCount decreased to 3% in the mid 90's, then to 1–1.5% in the late 90s, and since 2000s this value is acceptable if it is less than 1%. In order for the CV%YarnCount to be less than 1%, the variance of yarn count *s* has to be ≺ 1 also. Within the context of this paper, it will be considered that the spinning frames were produced after the year 2000; therefore, the proposed statistical approach will be explained by considering *s* as ≺ 1 in accordance with textile industry.

Yarn count is adjusted on the machine according to what the customer ordered. Yarn count will be indicated as *μ*<sup>0</sup> in this chapter.

The main aspect in both sampling, variation of yarn count, and yarn count is that every machine has to be adjusted to produce the yarn the customer ordered. The lot will be shipped as one and it does not make any difference for the customer which machine produced which yarn. The customer ordered the yarn lot and will regard it all the same at every single centimeter of yarn produced.

Suppose now a special yarn count of *μ*<sup>0</sup> in tex unit will be produced in twenty spinning frames in a spinning mill (**Figure 6**).

In this proposed statistical approach, the procedure starts with adjusting the Spinning Frame 1 (SF 1). The necessary adjustments to produce *μ*<sup>0</sup> tex yarn is done on the SF 1, the frame will run for a few minutes, the yarn will be produced a little bit, and *n*<sup>1</sup> bobbins from spindles are chosen as samples randomly from this normal distribution. The first thing is to test if the adjustments are correct and confront them with what the customer ordered. Since all the frames were produced after the year 2000, of the same brand, the same model, and the same technical specifications, the variance of yarn count has to be less than 1, with the latter being thus, the specified value for these hypothesis tests. In this manuscript it is argued that if the variance of yarn count is less than 1 it has to be tested before the yarn count. Then, the level of significance *α* is determined which is equal to 0.05 for ordinary textiles. *n*<sup>1</sup> bobbins from spindles of SF 1 are taken for yarn count tests done in the laboratory. The one-sided hypothesis is:

$$\begin{aligned} H\_0: \sigma\_1^2 &= \mathbf{1} \\ H\_1: \sigma\_1^2 &\prec \mathbf{1} \end{aligned} \tag{23}$$

and the *χ*<sup>2</sup> test statistic is:

$$\chi^2\_1 = \frac{(n\_1 - 1)s\_1^2}{1} \tag{24}$$

*s* 2 <sup>1</sup> is the sample variance of *n*<sup>1</sup> repeats from SF 1. The null hypothesis of variance of yarn count is rejected if *χ*<sup>2</sup> <sup>1</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*1�1. If it is unable to be rejected, then the procedure *Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

#### **Figure 6.**

*Representation of spinning frames in a spinning mill.*

continues by going back to the SF 1 and doing some more adjustments on the frame and repeating this test until the null hypothesis of variance of yarn count is rejected.

When it is guaranteed that the variance of yarn count is less than 1, then comes the yarn count statistics tests. The average of yarn count of SF 1 would be *μ*<sup>1</sup> and the twosided hypothesis of yarn count is:

$$\begin{aligned} H\_0: \mu\_1 &= \mu\_0 \\ H\_1: \mu\_1 &\neq \mu\_0 \end{aligned} \tag{25}$$

Variance is estimated by *s* 2 1, *x*<sup>1</sup> is the average of the *n*<sup>1</sup> repeats of yarn count from SF 1, the *t*-test statistic is:

$$t\_1 = \frac{\overline{\mathcal{X}}\_1 - \mu\_0}{s\_1 / \sqrt{n\_1}} \tag{26}$$

where instead of a normal distribution it is a *t* distribution with *n*<sup>1</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>1</sup> ≻ *t<sup>α</sup>=*2,*n*1�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*1�<sup>1</sup> is the upper *α=*2 percentage point of the *t* distribution with *n*<sup>1</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the null hypothesis of yarn count is rejected, then the procedure continues by going back to the SF 1 and doing some more adjustments on the frame and repeating these tests until the null hypothesis of yarn count is unable to be rejected.

Now the SF 1 is ready to produce what the customer ordered, so the procedure will continue with the statistics to make the SF 2 to produce what the customer ordered

and also the same as SF 1. The necessary adjustments to produce *μ*<sup>0</sup> tex yarn is done on the SF 2 and *n*<sup>2</sup> bobbins from spindles of SF 2 are chosen randomly. To test if the variance of yarn count of SF 2 is less than 1, the one-sided hypothesis is:

$$\begin{aligned} H\_0: \sigma\_2^2 &= \mathbf{1} \\ H\_1: \sigma\_2^2 &\prec \mathbf{1} \end{aligned} \tag{27}$$

and the *χ*<sup>2</sup> test statistic is:

$$
\chi^2\_2 = \frac{(n\_2 - 1)s\_2^2}{1} \tag{28}
$$

*s* 2 <sup>2</sup> is the sample variance of *n*<sup>2</sup> repeats from SF 2. The null hypothesis of variance is rejected if *χ*<sup>2</sup> <sup>2</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*2�1. If it is unable to be rejected, then the procedure continues by going back to the SF 2 and doing some more adjustments on the frame and repeating this test until the null hypothesis of variance of yarn count is rejected.

Both of the variances of yarn counts of SF 1 and SF 2 may be less than 1 but their equality has to be tested also. This is justified because they will all mix into one lot and it is not important from the view of point of customer which frame produced which yarn. To test their equality, the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_1^2 &= \sigma\_2^2\\ H\_1: \sigma\_1^2 &\neq \sigma\_2^2 \end{aligned} \tag{29}$$

and the F statistics is:

$$F\_{(1,2)} = \frac{s\_1^2}{s\_2^2} \tag{30}$$

*H*<sup>0</sup> is rejected if *F*ð Þ 1,2 ≻*Fα=*2,*n*1�1,*n*2�<sup>1</sup> or *F*ð Þ 1,2 ≺ *F*<sup>1</sup>�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1�1,*n*2�<sup>1</sup> which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> � 1 and *n*<sup>2</sup> � 1, respectively. If the null hypothesis is rejected, then the procedure continues by going back to the SF 2 and doing some more adjustments on the frame and repeating these tests until the null hypothesis of equality of variances of yarn counts is unable to be rejected.

When it is guaranteed that both the variance of yarn count is less than 1 for SF 2 and the two frames' variances are equal, then comes the yarn count statistics tests for SF 2. The average of yarn count of *n*<sup>2</sup> samples from SF 2 would be *μ*<sup>2</sup> and the two-sided hypothesis of yarn count is:

$$H\_0: \mu\_2 = \mu\_0$$

$$H\_1: \mu\_2 \neq \mu\_0$$

Variance is estimated by *s* 2 2, *x*<sup>2</sup> is the average of the *n*<sup>2</sup> repeats of yarn count from SF 2, the *t*-test statistic is:

$$t\_2 = \frac{\overline{\mathcal{X}}\_2 - \mu\_0}{s\_2 / \sqrt{n\_2}} \tag{32}$$

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

where instead of a normal distribution it is a *t* distribution with *n*<sup>2</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>2</sup> ≻*t<sup>α</sup>=*2,*n*2�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*2�<sup>1</sup> is the upper *α=*2 percentage point of the a *t* distribution with *n*<sup>2</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the null hypothesis of yarn count is rejected, then the procedure continues by going back to the SF 2 and doing some more adjustments on the frame and repeating these tests until the null hypothesis of yarn count is unable to be rejected.

Both of the yarn counts of SF 1 and SF 2 may be equal to what the customer ordered but their equality with each other has to be tested also because they will all mix into one lot and it is not important from the view of point of customer which frame produced which yarn. To test the yarn count equality of SF 1 and SF 2, even there is only one *μ*0, and for ease of reference, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_1 - \mu\_2 &= \mathbf{0} \\ H\_1: \mu\_1 - \mu\_2 &\neq \mathbf{0} \end{aligned} \tag{33}$$

and the pooled *t*-test statistic is:

$$t\_{(1,2)} = \frac{\overline{\mathfrak{X}}\_1 - \overline{\mathfrak{X}}\_2}{s\_{p(1,2)}\sqrt{\frac{1}{n\_1} + \frac{1}{n\_2}}}\tag{34}$$

$$s\_{p(1,2)}^2 = \frac{(n\_1 - 1)s\_1^2 + (n\_2 - 1)s\_2^2}{n\_1 + n\_2 - 2} \tag{35}$$

*s* 2 *<sup>p</sup>*ð Þ 1,2 is the pooled estimator of variance of SFs 1 and 2 with ð Þ *<sup>n</sup>*<sup>1</sup> <sup>þ</sup> *<sup>n</sup>*<sup>2</sup> � <sup>2</sup> degrees of freedom.

If *t*ð Þ 1,2 � � � �≻*tα=*2,*n*1þ*n*2�<sup>2</sup> then *H*<sup>0</sup> is rejected, *tα=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ *n*<sup>2</sup> � 2 degrees of freedom at a fixed significance level twosided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 2 and doing some more adjustments on the frame and repeating this and the above tests until the *H*<sup>0</sup> of yarn count is unable to be rejected. The operation steps can be summarized as below:

Step 1) Yarn count adjustment of SF 1.

Go to Step 1 and repeat until *H*<sup>0</sup> is unable to be rejected


Go to Steps 1 and 2, and repeat until *H*<sup>0</sup> is unable to be rejected Step 4) Yarn count adjustment of SF 2.

Go to Step 4 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 5) Testing the variance of yarn count of SF 2 to be less than 1. Go to Step 4 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 6) Testing the equality of variances of SF 1 and SF 2.

Go to Steps 4 and 5, and repeat until *H*<sup>0</sup> is unable to be rejected

Step 7) Testing the yarn count of SF 2 with *μ*0.

Go to Steps 4, 5, and 6, and repeat until *H*<sup>0</sup> is unable to be rejected. Step 8) Testing the equality of yarn counts of SF 1 and SF 2.

Go to Steps 4, 5, 6, and 7, and repeat until *H*<sup>0</sup> is unable to be rejected. The same will be repeated for the rest of the spinning frames until SF 20.

Now the SFs 1 and 2 are producing the same yarn having the same yarn count and same variance of yarn count. The SFs 1 and 2 can be considered as one machine producing the same product. A representation is given in **Figure 7**.

The procedure will continue with the statistics to make the SF 3 to produce what the customer ordered and also the same as SFs 1 and 2. The necessary adjustments to produce *μ*<sup>0</sup> tex yarn is done on the SF 3 and *n*<sup>3</sup> bobbins are chosen randomly, having *s* 2 3 variance and *x*<sup>3</sup> average yarn count. To test if the variance of yarn count of SF 3 is less than 1, the one-sided hypothesis is:

$$H\_0: \sigma\_3^2 = \mathbf{1} \tag{36}$$

$$H\_1: \sigma\_3^2 \prec \mathbf{1} \tag{36}$$

and the *χ*<sup>2</sup> test statistic is:

$$
\chi^2\_3 = \frac{(n\_3 - 1)s\_3^2}{1} \tag{37}
$$

*s* 2 <sup>3</sup> is the sample variance of yarn count of *n*<sup>3</sup> repeats from SF 3. The null hypothesis of variance of yarn count is rejected if *χ*<sup>2</sup> <sup>3</sup> ≺ *χ*<sup>2</sup> <sup>1</sup>�*α*,*n*3�1. If it is unable to be rejected, then the procedure continues by going back to the SF 3 and doing some more adjustments on the frame and repeating this test until the *H*<sup>0</sup> of variance of yarn count is rejected.

**Figure 7.** *Representation of SFs 1 and .2 producing the same yarn.*

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

Both of the variances of yarn count of (SFs 1 and 2) and SF 3 may be less than 1 but their equality has to be tested also because they will all mix into one lot. To test their equality, the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_{(1,2)}^2 &= \sigma\_3^2\\ H\_1: \sigma\_{(1,2)}^2 &\neq \sigma\_3^2 \end{aligned} \tag{38}$$

and the F statistics is:

$$F\_{(1-3)} = \frac{s\_{p(1,2)}^2}{s\_3^2} \tag{39}$$

*H*<sup>0</sup> is rejected if *F*ð Þ <sup>1</sup>�<sup>3</sup> ≻*Fα=*2,*n*1þ*n*2�2,*n*3�<sup>1</sup> or *F*ð Þ <sup>1</sup>�<sup>3</sup> ≺*F*<sup>1</sup>�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1þ*n*2�2,*n*3�<sup>1</sup>**,** which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> þ *n*<sup>2</sup> � 2 and *n*<sup>3</sup> � 1, respectively. If the *H*<sup>0</sup> of variance of yarn count is rejected, then the procedure continues by going back to the SF 3 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> of variance of yarn count is unable to be rejected.

When it is guaranteed that both the variance of yarn count is less than 1 for SF 3 and it is equal with the first two frames' pooled variance of yarn count, then come the yarn count statistics tests for SF 3. The average of yarn count of SF 3 would be *μ*<sup>3</sup> and the two-sided hypothesis:

$$\begin{aligned} H\_0: \mu\_3 &= \mu\_0 \\ H\_1: \mu\_3 &\neq \mu\_0 \end{aligned} \tag{40}$$

Variance is estimated by *s* 2 3, *x*<sup>3</sup> is the average of the *n*<sup>3</sup> repeats of yarn count from SF 3, the *t*-test statistic is:

$$t\_3 = \frac{\overline{\mathfrak{X}\_3} - \mu\_0}{s\_3 / \sqrt{n\_3}} \tag{41}$$

where instead of a normal distribution it is a *t* distribution with *n*<sup>3</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>3</sup> ≻*t<sup>α</sup>=*2,*n*3�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*3�<sup>1</sup> is the upper *α=*2 percentage point of the a *t* distribution with *n*<sup>3</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 3 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> of yarn count is unable to be rejected.

Both of the yarn counts of (SFs 1 and 2) and SF 3 may be equal to what the customer ordered but their equality with each other also has to be tested because they will all mix into one lot. To test the yarn count equality of (SFs 1 and 2) and SF 3, there is only one *μ*0, and for ease of reference, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_{(1,2)} - \mu\_3 &= \mathbf{0} \\ H\_1: \mu\_{(1,2)} - \mu\_3 &\neq \mathbf{0} \end{aligned} \tag{42}$$

and the pooled *t*-test statistic is:

The average of yarn counts of SF 1 and SF 2 is:

$$\frac{\overline{\mathfrak{X}}\_1 + \overline{\mathfrak{X}}\_2}{2} = \overline{\mathfrak{X}}\_{(1,2)}\tag{43}$$

then,

$$t\_{(1-3)} = \frac{\overline{\boldsymbol{\varpi}}\_{(1,2)} - \overline{\boldsymbol{\varpi}}\_{3}}{s\_{p(1-3)}\sqrt{\frac{1}{n\_1 + n\_2 - 2} + \frac{1}{n\_3}}}\tag{44}$$

$$s\_{p(1-3)}^2 = \frac{(n\_1 + n\_2 - 2)s\_{p(1,2)}^2 + (n\_3 - 1)s\_3^2}{n\_1 + n\_2 + n\_3 - 3} \tag{45}$$

*s* 2 *<sup>p</sup>*ð Þ <sup>1</sup>�<sup>3</sup> is the pooled estimator of variance of SFs 1�3 with ð Þ *<sup>n</sup>*<sup>1</sup> <sup>þ</sup> *<sup>n</sup>*<sup>2</sup> � <sup>3</sup> degrees of freedom.

If *t*ð Þ <sup>1</sup>�<sup>3</sup> � � � �≻*t<sup>α</sup>=*2,*n*1þ*n*2þ*n*3�<sup>3</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ *n*<sup>2</sup> þ *n*<sup>3</sup> � 3 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 3 and doing some more adjustments on the frame and repeating this and the above tests until the *H*<sup>0</sup> of yarn count is unable to be rejected. The operation steps can be summarized as below:

Step 1) Yarn count adjustment of SF 1.

Go to Step 1 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 2) Testing the variance of yarn count of SF 1 to be less than 1. Go to Step 1 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 3) Testing the yarn count of SF 1 with *μ*0**.**

Go to Steps 1 and 2, and repeat until *H*<sup>0</sup> is unable to be rejected Step 4) Yarn count adjustment of SF 2.

Go to Step 4 and repeat until *H*<sup>0</sup> is unable to be rejected


Go to Steps 4, 5, 6, and 7, and repeat until *H*<sup>0</sup> is unable to be rejected. Step 9) Yarn count adjustment of SF 3.

Go to Step 9 and repeat until *H*<sup>0</sup> is unable to be rejected


Go to Steps 9, 10, and 11, and repeat until *H*<sup>0</sup> is unable to be rejected Step 13) Testing the equality of yarn counts of (SFs 1 and 2) and SF 3.

Go to Steps 9, 10, 11, and 12, and repeat until *H*<sup>0</sup> is unable to be rejected The same will be done in a repeating pattern for the rest of the spinning frames until SF 20.

Now the SFs 1�3 are producing the same yarn having the same yarn count and the same variance of yarn count. The SFs 1�3 can be considered as one machine producing the same product. A representation is given in **Figure 8**.

The procedure will continue with the statistics to make the SF 4 to produce what the customer ordered and also the same as (SFs 1�3). The necessary adjustments to

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

**Figure 8.** *Representation of SFs 1*�*3 producing the same yarn.*

produce *μ*<sup>0</sup> tex yarn is done on the SF 4 and *n*<sup>4</sup> bobbins from spindles are chosen randomly, having *s* 2 <sup>4</sup> variance of yarn count and *x*<sup>4</sup> average yarn count. To test if the variance of yarn count of SF 4 is less than 1, the one-sided hypothesis is:

$$\begin{aligned} H\_0: \sigma\_4^2 &= \mathbf{1} \\ H\_1: \sigma\_4^2 &\prec \mathbf{1} \end{aligned} \tag{46}$$

and the *χ*<sup>2</sup> test statistic is:

$$
\chi^2\_4 = \frac{(n\_4 - 1)s\_4^2}{1} \tag{47}
$$

The *H*<sup>0</sup> of variance of yarn count is rejected if *χ*<sup>2</sup> <sup>4</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*4�1. If it is unable to be rejected, then the procedure continues by going back to the SF 4 and doing some more adjustments on the frame and repeating this test until the *H*<sup>0</sup> of variance of yarn count is rejected.

Both of the variances of yarn count of (SFs 1�3) and SF 4 may be less than 1 but their equality has to be tested also because they will all mix into one lot. To test their equality, the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_{(1-3)}^2 &= \sigma\_4^2\\ H\_1: \sigma\_{(1-3)}^2 &\neq \sigma\_4^2 \end{aligned} \tag{48}$$

and the F statistics is:

$$F\_{(1-4)} = \frac{s\_{p(1-3)}^2}{s\_4^2} \tag{49}$$

*H*<sup>0</sup> is rejected if *F*ð Þ <sup>1</sup>�<sup>4</sup> ≻*F<sup>α</sup>=*2,*n*1þ*n*2þ*n*3�3,*n*4�<sup>1</sup> or *F*ð Þ <sup>1</sup>�<sup>4</sup> ≺*F*1�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1þ*n*2þ*n*3�3,*n*4�<sup>1</sup>**,** which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> þ *n*<sup>2</sup> þ *n*<sup>3</sup> � 3 and *n*<sup>4</sup> � 1, respectively. If the *H*<sup>0</sup> is rejected, then the procedure continues by going back to the SF 4 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> is unable to be rejected.

When it is guaranteed that both the variance of yarn count is less than 1 for SF 4 and it is equal to the first three frames' pooled variance of yarn count, then come the yarn count statistics tests. The average of yarn count of SF 4 would be *μ*<sup>4</sup> and the twosided hypothesis is:

$$\begin{aligned} H\_0: \mu\_4 &= \mu\_0 \\ H\_1: \mu\_4 &\neq \mu\_0 \end{aligned} \tag{50}$$

Variance is estimated by *s* 2 4, *x*<sup>4</sup> is the average of the *n*<sup>4</sup> repeats of yarn count from SF 4, the *t*-test statistic is:

$$t\_4 = \frac{\overline{\mathcal{X}}\_4 - \mu\_0}{\mathfrak{s}\_4 / \sqrt{n\_4}} \tag{51}$$

where instead of a normal distribution it is a *t* distribution with *n*<sup>4</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>4</sup> ≻*tα=*2,*n*4�<sup>1</sup> then *H*<sup>0</sup> is rejected, *tα=*2,*n*4�<sup>1</sup> is the upper *α=*2 percentage point of the *t* distribution with *n*<sup>4</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 4 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> of yarn count is unable to be rejected.

Both of the yarn counts of (SFs 1�3) and SF 4 may be equal to what the customer ordered but their equality with each other has to be tested also because they will all mix into one lot. To test the yarn count equality of (SFs 1�3) and SF 4, there is only one *μ*0, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_{(1-3)} - \mu\_4 &= \mathbf{0} \\ H\_1: \mu\_{(1-3)} - \mu\_4 &\neq \mathbf{0} \end{aligned} \tag{52}$$

and the pooled *t*-test statistic is:

The average of yarn counts of (SFs 1 and 2) and SF 3 is:

$$\frac{\overline{\mathfrak{X}}\_{(1,2)} + \overline{\mathfrak{X}}\_{3}}{2} = \overline{\mathfrak{X}}\_{(1-3)}\tag{53}$$

then,

$$t\_{(1-4)} = \frac{\overline{\mathcal{X}}\_{(1-3)} - \overline{\mathcal{X}}\_{4}}{s\_{p(1-4)}\sqrt{\frac{1}{n\_1 + n\_2 + n\_3 - 3} + \frac{1}{n\_4}}}\tag{54}$$

$$s\_{p(1-4)}^2 = \frac{(n\_1 + n\_2 + n\_3 - 3)s\_{p(1-3)}^2 + (n\_4 - 1)s\_4^2}{n\_1 + \ldots + n\_4 - 4} \tag{55}$$

*s* 2 *<sup>p</sup>*ð Þ <sup>1</sup>�<sup>4</sup> is the pooled estimator of variance of SFs 1–4 with ð Þ *<sup>n</sup>*<sup>1</sup> <sup>þ</sup> *<sup>n</sup>*<sup>2</sup> � <sup>4</sup> degrees of freedom.

#### *Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

If *t*ð Þ <sup>1</sup>�<sup>4</sup> ≻*t<sup>α</sup>=*2,*n*1<sup>þ</sup> … <sup>þ</sup>*n*4�<sup>4</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ … þ *n*<sup>4</sup> � 4 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 4 and doing some more adjustments on the frame and repeating this and the above tests until the *H*<sup>0</sup> of yarn count is unable to be rejected. The operation steps can be summarized as below:

*(Continued)*

Step 14) Yarn count adjustment of SF 4.

Go to Step 14 and repeat until *H*<sup>0</sup> is unable to be rejected


Go to Steps 14, 15, 16, and 17, and repeat until *H*<sup>0</sup> is unable to be rejected The same will be repeated for the rest of the spinning frames until SF 20.

Now the SFs 1-4 are producing the same yarn having the same yarn count and same variance of yarn count. The SFs 1�4 can be considered as one machine producing the same product. A representation is given in **Figure 9**.

Suppose the same procedure is repeated for SFs 5, 6, and 7, and now the procedure will continue with the statistics to make the SF 8 to produce what the customer

#### **Figure 9.**

*Representation of SFs 1*�*4 producing the same yarn.*

ordered and also the same as (SFs 1–7). The necessary adjustments to produce *μ*<sup>0</sup> tex yarn is done on the SF 8 and *n*<sup>8</sup> bobbins from spindles are taken randomly, having *s* 2 8 variance of yarn count and *x*<sup>8</sup> average yarn count. To test if the variance of yarn count of SF 8 is less than 1, the one-sided hypothesis is:

$$\begin{aligned} H\_0: \sigma\_8^2 &= \mathbf{1} \\ H\_1: \sigma\_8^2 &\prec \mathbf{1} \end{aligned} \tag{56}$$

and the *χ*<sup>2</sup> test statistic is:

$$
\chi^2\_8 = \frac{(n\_8 - 1)\mathfrak{s}\_8^2}{1} \tag{57}
$$

*s* 2 <sup>8</sup> is the variance of yarn count of *n*<sup>8</sup> repeats from SF 8. The *H*<sup>0</sup> of variance of yarn count is rejected if *χ*<sup>2</sup> <sup>8</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*8�1. If it is unable to be rejected, then the procedure continues by going back to the SF 8 and doing some more adjustments on the frame and repeating this test until the *H*<sup>0</sup> of variance of yarn count is rejected.

Both of the variances of yarn count of (SFs 1�7) and SF 8 may be less than 1 but their equality has to be tested also because they will all mix into one lot. To test their equality, the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_{(1-\mathcal{T})}^2 &= \sigma\_8^2\\ H\_1: \sigma\_{(1-\mathcal{T})}^2 &\neq \sigma\_8^2 \end{aligned} \tag{58}$$

and the F statistics is:

$$F\_{(1-8)} = \frac{s\_{p(1-7)}^2}{s\_8^2} \tag{59}$$

*H*<sup>0</sup> is rejected if *F*ð Þ <sup>1</sup>�<sup>8</sup> ≻*Fα=*2,*n*1<sup>þ</sup> … <sup>þ</sup>*n*7�7,*n*8�<sup>1</sup> or *F*ð Þ <sup>1</sup>�<sup>8</sup> ≺*F*<sup>1</sup>�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1<sup>þ</sup> … <sup>þ</sup>*n*7�7,*n*8�<sup>1</sup> which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> þ … þ *n*<sup>7</sup> � 7 and *n*<sup>8</sup> � 1, respectively. If the *H*<sup>0</sup> of variance of yarn count is rejected, then the procedure continues by going back to the SF 8 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> of variance of yarn count is unable to be rejected.

When it is guaranteed that both the variance of yarn count is less than 1 for SF 8 and it is equal to the first seven frames' pooled variance, then come the yarn count statistics tests for SF 8. The average of yarn count of SF 8 would be *μ*<sup>8</sup> and the two-sided hypothesis is:

$$\begin{aligned} H\_0: \mu\_8 &= \mu\_0 \\ H\_1: \mu\_8 &\neq \mu\_0 \end{aligned} \tag{60}$$

Variance is estimated by *s* 2 8, *x*<sup>8</sup> is the average of the *n*<sup>8</sup> repeats of yarn count from SF 8, the *t*-test statistic is:

$$t\_8 = \frac{\overline{\mathfrak{X}}\_8 - \mu\_0}{\mathfrak{s}\_8 / \sqrt{n\_8}} \tag{61}$$

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

where instead of a normal distribution it is a *t* distribution with *n*<sup>8</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>8</sup> ≻*t<sup>α</sup>=*2,*n*8�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*8�<sup>1</sup> is the upper *α=*2 percentage point of the a *t* distribution with *n*<sup>8</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 8 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> of yarn count is unable to be rejected.

Both of the yarn counts of (SFs 1�7) and SF 8 may be equal to what the customer ordered but their equality with each other also has to be tested because they will all mix into one lot. To test the yarn count equality of (SFs 1�7) and SF 8, there is only one *μ*0, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_{(1-7)} - \mu\_8 &= \mathbf{0} \\ H\_1: \mu\_{(1-7)} - \mu\_8 &\neq \mathbf{0} \end{aligned} \tag{62}$$

and the pooled *t*-test statistic is:

The average of yarn counts of (SFs 1–6) and SF 7 is:

$$\frac{\overline{\mathfrak{X}}\_{\text{(1-6)}} + \overline{\mathfrak{X}}\_{\text{/}}}{2} = \overline{\mathfrak{X}}\_{\text{(1-7)}}\tag{63}$$

then,

$$t\_{(1-8)} = \frac{\overline{\mathfrak{X}}\_{(1-7)} - \overline{\mathfrak{X}}\_{8}}{s\_{p(1-8)}\sqrt{\frac{1}{n\_1 + \ldots + n\_7 - 7} + \frac{1}{n\_8}}}\tag{64}$$

$$s\_{p(1-8)}^2 = \frac{(n\_1 + \dots + n\_7 - 7)s\_{p(1-7)}^2 + (n\_8 - 1)s\_8^2}{n\_1 + \dots + n\_8 - 8} \tag{65}$$

*s* 2 *<sup>p</sup>*ð Þ <sup>1</sup>�<sup>8</sup> is the pooled estimator of variance of SFs 1�8 with ð Þ *<sup>n</sup>*<sup>1</sup> <sup>þ</sup> *<sup>n</sup>*<sup>2</sup> � <sup>8</sup> degrees of freedom.

If *t*ð Þ <sup>1</sup>�<sup>8</sup> � � � �≻*t<sup>α</sup>=*2,*n*1<sup>þ</sup> … <sup>þ</sup>*n*8�<sup>8</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ … þ *n*<sup>8</sup> � 8 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 8 and doing some more adjustments on the frame and repeating this and the above tests until *H*<sup>0</sup> of yarn count is unable to be rejected. The operation steps can be summarized as below:

*(Continued)*

Step 34) Yarn count adjustment of SF 8.

Go to Step 34 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 35) Testing the variance of yarn count of SF 8 to be less than 1.

Go to Step 34 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 36) Testing the equality of variances of (SFs 1�7) and SF 8.

Go to Steps 34 and 35, and repeat until *H*<sup>0</sup> is unable to be rejected Step 37) Testing the yarn count of SF 8 with *μ*0.

Go to Steps 34, 35, and 36, and repeat until *H*<sup>0</sup> is unable to be rejected Step 38) Testing the equality of yarn counts of (SFs 1�7) and SF 8.

Go to Steps 34, 35, 36, and 37, and repeat until *H*<sup>0</sup> is unable to be rejected

The same will be done in a repeating manner for the rest of the spinning frames until SF 20.

Now the SFs 1�8 are producing the same yarn having the same yarn count and same variance of yarn count. The SFs (1�8) can be considered as a one machine producing the same product. A representation is given in **Figure 10**.

Suppose the same procedure is repeated for SFs 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, and 19, and now the last spinning frame is the 20th one, the procedure will continue with the statistics to make the SF 20 to produce what the customer ordered and also the same as (SFs 1�19). The necessary adjustments to produce *μ*<sup>0</sup> tex yarn is done on the SF 20 and *n*<sup>20</sup> bobbins from spindles are chosen randomly, having *s* 2 <sup>20</sup> variance of yarn count and *x*<sup>20</sup> average yarn count. To test if the variance of yarn count of SF 20 is less than 1, the one-sided hypothesis is:

$$\begin{aligned} H\_0: \sigma\_{20}^2 &= \mathbf{1} \\ H\_1: \sigma\_{20}^2 &< \mathbf{1} \end{aligned} \tag{66}$$

and the *χ*<sup>2</sup> test statistic is:

$$
\chi^2\_{20} = \frac{(n\_{20} - 1)s\_{20}^2}{1} \tag{67}
$$

The *H*<sup>0</sup> of variance of yarn count is rejected if *χ*<sup>2</sup> <sup>20</sup> ≺*χ*<sup>2</sup> <sup>1</sup>�*α*,*n*20�1. If it is unable to be rejected, then the procedure continues by going back to the SF 20 and doing some more adjustments on the frame and repeating this test until the *H*<sup>0</sup> of variance of yarn count is rejected.

**Figure 10.** *Representation of SFs 1*�*8 producing the same yarn.*

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

Both of the variances of yarn count of SFs (1�19) and SF 20 may be less than 1 but their equality has to be tested also because they will all mix into one lot. To test their equality, the hypothesis is:

$$\begin{aligned} H\_0: \sigma\_{(1-19)}^2 &= \sigma\_{20}^2\\ H\_1: \sigma\_{(1-19)}^2 &\neq \sigma\_{20}^2 \end{aligned} \tag{68}$$

and the F statistics is:

$$F\_{(1-20)} = \frac{s\_{p(1-19)}^2}{s\_{20}^2} \tag{69}$$

*H*<sup>0</sup> is rejected if *F*<sup>20</sup> ≻ *Fα=*2,*n*1<sup>þ</sup> … <sup>þ</sup>*n*19�19,*n*20�<sup>1</sup> or *F*<sup>20</sup> ≺*F*<sup>1</sup>�ð Þ *<sup>α</sup>=*<sup>2</sup> ,*n*1<sup>þ</sup> … <sup>þ</sup>*n*19�19,*n*20�<sup>1</sup> which denote the upper *α=*2 and lower 1 � ð Þ *α=*2 percentage points of the F distribution with degrees of freedom *n*<sup>1</sup> þ … þ *n*<sup>19</sup> � 19 and *n*<sup>20</sup> � 1, respectively. If the *H*<sup>0</sup> is rejected, then the procedure continues by going back to the SF 20 and doing some more adjustments on the frame and repeating these tests until the *H*<sup>0</sup> is unable to be rejected.

When it is guaranteed that both the variance of yarn count is less than 1 for SF 20 and it is equal to the other nineteen frames' pooled variance of yarn count, then come the yarn count statistics tests. The average of yarn count of SF 20 would be *μ*<sup>20</sup> and the two-sided hypothesis is:

$$\begin{aligned} H\_0: \mu\_{20} &= \mu\_0 \\ H\_1: \mu\_{20} &\neq \mu\_0 \end{aligned} \tag{70}$$

Variance of yarn count is estimated by *s* 2 20, *x*<sup>20</sup> is the average of the *n*<sup>20</sup> repeats of yarn count from SF 20, the *t*-test statistic is:

$$t\_{20} = \frac{\overline{\mathfrak{X}\_{20}} - \mu\_0}{s\_{20} / \sqrt{n\_{20}}} \tag{71}$$

where instead of a normal distribution it is a *t* distribution with *n*<sup>20</sup> � 1 degrees of freedom. If j j ð Þ *t*<sup>20</sup> ≻*t<sup>α</sup>=*2,*n*20�<sup>1</sup> then *H*<sup>0</sup> is rejected, *t<sup>α</sup>=*2,*n*20�<sup>1</sup> is the upper *α=*2 percentage point of the a *t* distribution with *n*<sup>20</sup> � 1 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> of yarn count is rejected, then the procedure continues by going back to the SF 20 and doing some more adjustments on the frame and repeating this test until the *H*<sup>0</sup> of yarn count is unable to be rejected.

Both of the yarn counts of (SFs 1�19) and SF 20 may be equal to what the customer ordered but their equality with each other also has to be tested because they will all mix into one lot. To test the yarn count equality of SFs (1�19) and SF 20, there is only one *μ*0, the hypothesis is:

$$\begin{aligned} H\_0: \mu\_{(1-19)} - \mu\_{20} &= \mathbf{0} \\ H\_1: \mu\_{(1-19)} - \mu\_{20} &\neq \mathbf{0} \end{aligned} \tag{72}$$

and the pooled *t*-test statistic is:

The average of yarn counts of (SFs 1�18) and SF 19 is:

$$\frac{\overline{\mathfrak{X}}\_{(1-18)} + \overline{\mathfrak{X}}\_{19}}{2} = \overline{\mathfrak{X}}\_{(1-19)}\tag{73}$$

then,

$$t\_{(1-20)} = \frac{\overline{\mathfrak{X}\_{(1-19)} - \overline{\mathfrak{X}}\_{20}}}{s\_{p(1-20)}\sqrt{\frac{1}{n\_1 + \ldots + n\_{19} - 19} + \frac{1}{n\_{20}}}}\tag{74}$$

$$s\_{p(1\to 20)}^2 = \frac{(n\_1 + \dots + n\_{19} - 19)s\_{p(1\to 19)}^2 + (n\_{20} - 1)s\_{20}^2}{n\_1 + \dots + n\_{20} - 20} \tag{75}$$

*s* 2 *<sup>p</sup>*ð Þ <sup>1</sup>�<sup>20</sup> is the pooled estimator of variance of SFs 1�20 with ð Þ *<sup>n</sup>*<sup>1</sup> <sup>þ</sup> *<sup>n</sup>*<sup>2</sup> � <sup>20</sup> degrees of freedom.

If *t*ð Þ <sup>1</sup>�<sup>20</sup> � � � �≻*tα=*2,*n*1<sup>þ</sup> … <sup>þ</sup>*n*20�<sup>20</sup> then *H*<sup>0</sup> is rejected, *tα=*<sup>2</sup> is the upper *α=*2 percentage point of the *t*-distribution with *n*<sup>1</sup> þ … þ *n*<sup>20</sup> � 20 degrees of freedom at a fixed significance level two-sided. If the *H*<sup>0</sup> is rejected, then the procedure continues by going back to the SF 20 and doing some more adjustments on the frame and repeating this and the above tests until the *H*<sup>0</sup> of yarn count is unable to be rejected. The operation steps can be summarized as below:

*(Continued)*

Step 94) Yarn count adjustment of SF20.

Go to Step 94 and repeat until *H*<sup>0</sup> is unable to be rejected Step 95) Testing the variance of yarn count of SF 20 to be less than 1. Go to Step 94 and repeat until *H*<sup>0</sup> is unable to be rejected

Step 96) Testing the equality of variances of SFs (1�19) and SF 20.

Go to Steps 94 and 95, and repeat until *H*<sup>0</sup> is unable to be rejected Step 97) Testing the yarn count of SF 20 with *μ*0.

Go to Steps 94, 95 and 96, and repeat until *H*<sup>0</sup> is unable to be rejected Step 98) Testing the equality of yarn counts of SFs (1�19) and SF 20.

Go to Steps 94, 95, 96 and 97, and repeat until *H*<sup>0</sup> is unable to be rejected. Now the SFs 1�20 are producing the same yarn having the same yarn count and same variance of yarn count. The SFs (1�20) can be considered as one machine producing the same product, no difference between the yarns of twenty different spinning frames. A representation is given in **Figure 11**.

The logic in this proposed statistical approach is in a spinning mill having twenty spinning frames to adjust the first spinning frame according to what the customer ordered and to the technology of the spinning frame; take samples, statistically test them and if rejected, correct the adjustments, do the statistic tests again, and if unable to be rejected, adjust the second spinning frame according to what the customer ordered and to the technology of the spinning frame, take samples, statistically test them and if rejected, correct the adjustments, do the statistic tests again, pool the output of the first and second frames, if rejected, repeat, and if unable to be rejected, go on to the third frame, and so on until the twentieth frame. This approach pools the output of all the spinning frames in multiple-stream process of ring spinning. This will guarantee that the production starts correct and is pooled, producing yarn as per customers' order by incorporating the necessary technology and reducing variability. The whole lot will have the same yarn property at the beginning of production. During production, control charts will be performed and assignable causes will be seen if they occur, and will be

*Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

**Figure 11.** *Representation of SFs 120 producing the same yarn.*

taken care of. Control charts will give much valuable information during production because it is assured that the production started correctly and all the frames are pooled. Additionally, instead of preparing separate control charts for each rational subgroup, even only one control chart for the whole lot would be enough, saving hence, time, cost, manpower, etc. This robust statistical approach can be incorporated in a statistics computer program, yielding a number of benefits for the enterprises.

On the other hand there is no restriction to employ boxplots, ANOVA, residual plots, etc. during production. These statistical methods will all add positive inferences on the data collected and support production and management. Claiming for better products and services alike will lead to new perspectives, ideas, point of views, etc.

Besides, spinning frames are not the only application area of this logic. Starting from the beginning of the stream, it can be applied to every machine in production, same two or more machines doing the same production, and so on. The first one will be adjusted at the beginning according to this logic, starting will be correct and will be pooled one by one, and continuing production will be controlled with the other statistical methods. Moreover, yarn count property is not the only application area falling under this logic. Yarn twist is also a property adjusted on the spinning frame. Other properties of textile materials adjusted on the machines can all be well worked with this proposed statistical approach.

A summary of the statistical procedures followed in this proposed statistical approach is given in **Table 2**.

#### **5. Conclusion**

This paper proposed a novel statistical approach for multiple-stream processes. Performed literature review suggests that control charts are used in multiple-stream processes but in this proposed statistical method, the expectations from the control


*Quality Control - An Anthology of Cases*

**Table 2.**

 *statistical approach.*

#### *Practicing Hypothesis Tests in Textile Engineering: Spinning Mill Exercise DOI: http://dx.doi.org/10.5772/intechopen.105643*

charts are divided into two: First adjust the machines correctly and pool production, then use control charts for assignable causes.

In this chapter, the proposed statistical approach is explained in detail being based on a spinning mill having twenty spinning frames. When the first spinning frame is adjusted according to what the customer ordered and to the technology of the spinning frame, the results of that adjustment is controlled statistically, by means of hypothesis testing. It is the yarn count property, being *μ*0, the examples are given. Yarn wrap on bobbins on the spindles, from rovings coming from the top, are drafted, and twisted to produce the yarn. *ni* samples are taken from independent, identical, and with equal probability of being chosen spindles, and yarn count property have a normal distribution, as the other properties of textile materials. The adjustments on the first spinning frame are done and the variance of yarn count is hypothesis tested with less than one because of the production year of the frame. The *χ*<sup>2</sup> test statistic is applied. If rejected, the adjustments are corrected, and the same test is repeated. If unable to be rejected, then yarn count is hypothesis tested with what the customer ordered *μ*0, the *t*-test statistic is applied; if rejected, the adjustments are corrected, and the same tests are repeated. If unable to be rejected, the second spinning frame is adjusted, the variance of yarn count is hypothesis tested with a *χ*<sup>2</sup> test statistic, and the equality of variances of yarn count of the two spinning frames is hypothesis tested with an F statistic. If rejected, the tests are repeated, if unable to be rejected, the yarn count hypothesis is tested with the *t*-test statistic. If rejected, adjustments on the frames are done and the tests are repeated, if both are unable to be rejected, then the yarn count of the two spinning machines are pooled. Now, the two frames are considered as one machine producing the same yarn, same variance of yarn count and same yarn count property, variability reduced the most. This statistical approach continues until the twentieth spinning frame and one by one, all the frames are considered as one machine producing the same yarn, same variance of yarn count and same yarn count property at the end.

This novel statistical approach guarantees that production starts with correct adjustments of the machines. In the performed literature review, this however has not been come across. By applying this statistical approach at the beginning of production, the correct starting will be assured and the machines will all be pooled one by one. On the other hand, during production, control charts will be applied to see the assignable causes and quick care ought to be taken. Additionally, instead of preparing separate control charts for each rational subgroup, even only one control chart for the whole lot would be enough, saving time, cost, manpower etc. This robust statistical approach can be incorporated in a statistics computer program, ending up with many benefits for the companies. Other statistical methods like boxplots, ANOVA, residual plots will all provide additional information about how the production proceeds. In addition to the above, this novel statistical approach can be applied to machines starting from the beginning of the multiple-stream like blowroom, carding, drawing, roving, examples for a spinning mill, more than one machine producing the same material. Besides, it can equally be applied to the other properties of textile materials, both adjusted directly on the machines or which result indirectly with machine settings like pressure, speed, etc. Raw materials, products, efficiency, yield, waste reduction, shift management of workers, faults, machine breakdowns, spare parts, electricity, economics, and much other application areas would emerge in due time.

*Quality Control - An Anthology of Cases*
