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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 goal here is to find data set for which t is infinite. From

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

t is infinite when \({s_D}\) is zero.

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

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

where

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

The Sample Standard Deviation s is

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

From the definition, \({s_D}\)is zero if and only if

\({x_i} = \bar x\)

for every\(i = 1,2, \ldots ,n\). This now indicates that all data should be equal.

NOTE: the data for the differences should be equal, not the paired data set.

From this conclusion, let's take that the difference is for example\(0.5\), the chosen data set is given in the table below (taking numbers for which \({x_i} - {y_i} = \)0.5)

Obviously, for the paired t test t=0, because the sample standard deviation for the differences is

\({s_D} = \sqrt {\frac{1}{{12}} \cdot \left( {{{(0.5 - 0.5)}^2} + {{(0.5 - 0.5)}^2} + \ldots + {{(0.5 - 0.5)}^2}} \right)} = 0\)

thus

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{0.5}}{{0/\sqrt {13} }} = \infty \)

Let's see what happens with t for the two sample t test.

The two-sample t test for testing \({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)uses the following value of test statistic

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

For the adequate alternative hypothesis the adequate area under the \({t_\nu }\)curve is the P value.

First data set:

the sample mean of the first data set is

\(\bar x = \frac{1}{{13}} \cdot (22.5 + 35.5 + \ldots + 178.5) = 100.5\)

and the sample standard deviation is

\(\begin{array}{l}{s_1} = \sqrt {\frac{1}{{12}} \cdot \left( {{{(22.5 - 100.5)}^2} + {{(35.5 - 100.5)}^2} + \ldots + {{(178.5 - 100.5)}^2}} \right)} \\ = 50.63\end{array}\)

and m=13.

Second data set:

there is only difference in the sample mean by 0.5 because the data set was chosen like that, however compute the sample mean of the second data set

\(\bar y = \frac{1}{{13}} \cdot (22 + 35 + \ldots + 178) = 100\)

and the sample standard deviation is

\(\begin{array}{l}{s_2} = \sqrt {\frac{1}{{12}} \cdot \left( {{{(22 - 100)}^2} + {{(35 - 100)}^2} + \ldots + {{(178 - 100)}^2}} \right)} \\ = 50.63\end{array}\)

and n=13.

The two sample t test statistic value is

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{100.5 - 100}}{{50.63 \cdot \sqrt {\frac{2}{{13}}} }} = 0.025\)

which is nearly zero.

The incorrect analysis yields an insignificant result

\({\rm{ The incorrect analysis yields an insignificant result }}\)