Question

Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments" (Textile Res. J., \(1997: 137 - 142\)) gave the accompanying data on extensibility\((\% )\)at\(100gm/cm\)for both high-quality (H) fabric and poor-quality (P) fabric specimens.

\(\begin{array}{*{20}{r}}H&{1.2}&{.9}&{.7}&{1.0}&{1.7}&{1.7}&{1.1}&{.9}&{1.7}\\{}&{1.9}&{1.3}&{2.1}&{1.6}&{1.8}&{1.4}&{1.3}&{1.9}&{1.6}\\{}&{.8}&{2.0}&{1.7}&{1.6}&{2.3}&{2.0}&{}&{}&{}\\P&{1.6}&{1.5}&{1.1}&{2.1}&{1.5}&{1.3}&{1.0}&{2.6}&{}\end{array}\)

a. Construct normal probability plots to verify the plausibility of both samples having been selected from normal population distributions.

b. Construct a comparative boxplot. Does it suggest that there is a difference between true average extensibility for high-quality fabric specimens and that for poor-quality specimens?

c. The sample mean and standard deviation for the highquality sample are\(1.508\)and\(.444\), respectively, and those for the poor-quality sample are\(1.588\)and\(.530.\)Use the two-sample\(t\)test to decide whether true average extensibility differs for the two types of fabric.

Short answer

(a) Plausible

(b) A small difference

(c) There is not sufficient evidence to support the claim that the true average extensibility differs for the two types of fabric.

Step-by-step solution

a)Step 1: Determine the normal probability plot H

Given:

H: \(\;1.2,{\rm{ }}0.9,{\rm{ }}0.7,{\rm{ }}1,{\rm{ }}1.7,{\rm{ }}1.7,{\rm{ }}1.1,{\rm{ }}0.9,{\rm{ }}1.7,{\rm{ }}1.9,{\rm{ }}1.3,{\rm{ }}2.1,{\rm{ }}1.6,{\rm{ }}1.8,{\rm{ }}1.4,{\rm{ }}1.3,{\rm{ }}1.9,{\rm{ }}1.6,{\rm{ }}0.8,{\rm{ }}2,{\rm{ }}1.7,{\rm{ }}1.6,{\rm{ }}2.3,{\rm{ }}2\)

P:\(1.6,1.5,1.1,2.1,1.5,1.3,1,2.6\)

(a) If we want to perform a two-sample\(t\)test, then we require that both sampling distributions of the sample mean are approximately normal.

We will create a normal probability plot.

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.

If the data contains\(n\)data values, then the standardized normal scores are the\(z\)-scores in the normal probability table of the appendix corresponding to an area of \(\frac{{j - 0.5}}{n}\)(or the closest area) with \(j \in \{ 1,2,3, \ldots .,n\} \).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

Determine the normal probability plot P

\(P\)

We will create a normal probability plot.

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.

If the data contains\(n\)data values, then the standardized normal scores are the\(z\)-scores in the normal probability table of the appendix corresponding to an area of \(\frac{{j - 0.5}}{n}\)(or the closest area) with\(j \in \{ 1,2,3, \ldots .,n\} \).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then the population distribution is approximately normal.

Both probability plots do not contain strong curvature and are roughly linear, thus both population distributions are approximately normal.

Since the population distributions are approximately normal, the sampling distribution of the sample mean(s) \(\bar x\)are also approximately normal thus it is appropriate to use the two-sample\(t\) test.

b)Step 3: Sort all data values from smallest to largest

\({\rm{H}}:0.7,0.8,0.9,0.9,1,1.1,1.2,1.3,1.3,1.4,1.6,1.6,1.6,1.7,1.7,1.7,1.7,1.8,1.9,1.9,2,2,2.1,2.3\)

P:\(1,1.1,1.3,1.5,1.5,1.6,2.1,2.6\)

The minimum is.

Since the number of data values is even, the median is the average of the two middle values of the sorted data set:

\(M = {Q_2} = \frac{{1.6 + 1.6}}{2} = 1.6\)

The first quartile is the median of the data values below the median (or at\(25\% \)of the data):

\({Q_1} = \frac{{1.1 + 1.2}}{2} = 1.15\)

The third quartile is the median of the data values above the median (or at\(75\% \)of the data):

\({Q_3} = \frac{{1.8 + 1.9}}{2} = 1.85\)

The maximum is \(2.3.\)

Find  the value for High Quartile

HIGH

The minimum is\(1.\)

Since the number of data values is even, the median is the average of the two middle values of the sorted data set:

\(M = {Q_2} = \frac{{1.5 + 1.5}}{2} = 1.5\)

The first quartile is the median of the data values below the median (or at\(25\% \)of the data):

\({Q_1} = \frac{{1.1 + 1.3}}{2} = 1.2\)

The third quartile is the median of the data values above the median (or at\(75\% \)of the data):

\({Q_3} = \frac{{1.6 + 2.1}}{2} = 1.85\)

The maximum is \(2.6.\)

Plot the graph

BOXPLOT

The whiskers of the boxplot are at the minimum and maximum value. The box starts at the first quartile, ends at the third quartile and has a vertical line at the median.

The first quartile is at\(25\% \)of the sorted data list, the median at \(50\% \) and the third quartile at\(75\% \)

There appears to be a small difference between the true average extensibility for high-quality fabric specimens and that for poor-quality specimens, because the vertical line in the box of the boxplot is not at roughly the same position in both boxplots.

C)Step 6: Determine the test statistic

(c)

\(\begin{array}{l}{{\bar x}_1} = 1.508\\{{\bar x}_2} = 1.588\\{s_1} = 0.444\\{s_2} = 0.530\end{array}\)

Let us assume:\(\alpha = 0.05\)

Given claim: differs

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _{\scriptstyle2\atop\scriptstyle}}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the test statistic:

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{1.508 - 1.588}}{{\sqrt {\frac{{{{0.444}^2}}}{{24}} + \frac{{{{0.530}^2}}}{8}} }} \approx - 0.384\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{0.444}^2}}}{{24}} + \frac{{{{0.530}^2}}}{8}} \right)}^2}}}{{\frac{{{{\left( {{{0.444}^2}/24} \right)}^2}}}{{24 - 1}} + \frac{{{{\left( {{{0.530}^2}/8} \right)}^2}}}{{8 - 1}}}} \approx 10\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's\(T\)distribution in the appendix containing the t-value in the row\(df = 10\):

\(P > 2 \times 0.10 = 0.20\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P > 0.05 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that the true average extensibility differs for the two types of fabric.