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The Paired \(t\) Test:

When \(D = X - Y\) (difference between observations within a pair), \({\mu _D} = \)\({\mu _1} - {\mu _2}\) and null hypothesis

\({H_0}:{\mu _D} = {\Delta _0},\)

the test statistic value for testing the hypotheses is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }},\)

where\(\bar d\) and \({s_D}\) are the sample mean and the sample standard deviation of differences\({d_i}\), respectively. In order to use this test, assume that the differences\({D_i}\) are from a normal population. Depending on alternative hypothesis, the\(P\) value can be determined as the corresponding area under the \({t_{n - 1}}\) curve.

The hypotheses of interest are\({H_0}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} > 0\). The following table represents the data for unabraded condition and abraded condition, as well as the difference between corresponding observed pairs: \(\begin{array}{*{20}{c}}i&{{\rm{ Unabraded, }}{x_i}}&{{\rm{ Abraded, }}{y_i}}&{{\rm{ Difference, }}{d_i} = {x_i} - {y_i}}\\1&{36.4}&{28.5}&{7.9}\\2&{55}&{20}&{35}\\3&{51.5}&{46}&{5.5}\\4&{38.7}&{34.5}&{4.2}\\5&{43.2}&{36.5}&{6.7}\\6&{48.8}&{52.5}&{ - 3.7}\\7&{25.6}&{26.5}&{ - 0.9}\\8&{49.8}&{46.5}&{3.3}\\{}&{}&{}&{}\end{array}\)

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean\(\bar d\)for the differences is

\(\bar d = \frac{1}{8} \cdot (7.9 + 35 + \ldots + 3.3) = 7.25\)

Reminder:\({\mu _D}\)is the true average of the difference of breaking load for fabric in unabraded and abraded condition and\({\Delta _0} = 0\). The only missing value to compute the test statistic value is the sample standard deviation of differences\({d_i}\).

The Sample Variance\({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

Where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{\infty } \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation\(s\)is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(s_D^2{\rm{ }} = \frac{1}{{8 - 1}} \cdot \left( {{{(7.9 - 7.25)}^2} + {{(35 - 7.25)}^2} + \ldots + {{(3.3 - 7.25)}^2}} \right) = 140.726\)

and the sample standard deviation\({s_D} = \sqrt {140.726} = 11.8628\)

The test statistic value is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{7.25 - 0}}{{11.8628/\sqrt 8 }} = 1.73\)

The test is one-sided where the alternative hypothesis is\({H_a}:{\mu _D} > 0\), therefore the\(P\)value is the area under the the\({t_{n - 1}} = {t_{8 - 1}} = {t_7}\)curve to the right of\(t\)value

\(P = P(T > 1.73) = 1 - P(T \le 1.73) = 1 - 0.936 = 0.064\)

Where\(T\)has student distribution with\(7\)degrees of freedom, and the probability as computed using software (you can use table in the appendix of the book). At significance level\(0.01\), because

\(P = 0.064 > 0.01\)

do not reject null hypothesis

at given significance level.