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The normal probability plot suggests that the differences are distributed normally - the plot of the differences is quite linear.

The given data is paired, this means that based on the sample standard deviation and the sample mean you can not perform a hypotheses test about the difference of the true averages. The problem is that the sample standard deviation of the differences can not be obtained from the given information and you need it to obtain the test statistic value for paired\(t\)test. If the samples were independent the given value would be useful. Thee answer is:

can not be carried out

The test that needs to be carried out to see if the lead hip peak IR velocity was significantly greater than the trail hip ER velocity is the upper tailed paired $t$ test.

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 t 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.

Based on the mentioned, the hypotheses of interest are \({H_0}:{\mu _D} = 0\) versus\({H_a}:{\mu _D} > 0\). The delta value\({\Delta _0} = 0\). The missing values are the sample mean and the sample standard deviation of the differences.

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

\(\bar d = \frac{1}{{15}} \cdot ( - 31.7 - 9.2 + \ldots + 146.4) = 82.5\)

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

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

where

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

The Sample Standard Deviation \(s\) is\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(\begin{aligned}{l}s_D^2 = \frac{1}{{15 - 1}} \cdot \left( {{{( - 31.7 - 82.5)}^2} + {{( - 9.2 - 82.5)}^2} + \ldots + {{(146.4 - 82.5)}^2}} \right)\\ = 7637.75\end{aligned}\)

and the sample standard deviation

\({s_D} = \sqrt {7637.75} = 87.4\)

The test statistic value is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{82.5 - 0}}{{87.4/\sqrt {15} }} = 3.66\)

The test is one-sided where the alternative hypothesis is\({H_a}:{\mu _D} > 00\), therefore the\(P\)value is the area under the the\({t_{n - 1}} = {t_{14 - 1}} = {t_1}3\)curve to the right of\(t\)value

\(P = P(T > 3.66) = 0.001\)

where\(T\)has student distribution with\(n - 1 = 15 - 1 = 14\)degrees of freedom, and the probability as computed using software (you can use table in the appendix of the book, or estimate the value using the mentioned table). At significance level\(0.05\)or\(0.01\), because

\(P = 0.001 < 0.01 < 0.05\)

reject null hypothesis

at given significance level. There is enough evidence to conclude that the true average difference in ER and IR velocity is positive. The conclusion is the same as in the article even thought the \(P\) value is incorrect.