Question

Torsion during hip external rotation (ER) and extension may be responsible for certain kinds of injuries in golfers and other athletes. The article "Hip Rotational Velocities During the Full Golf Swing" (J. of Sports Science and Medicine, 2009: 296-299) reported on a study in which peak ER velocity and peak IR (internal rotation) velocity (both in deg.sec\(^{ - 1}\)) were determined for a sample of 15 female collegiate golfers during their swings. The following data was supplied by the article's authors.

\(\begin{aligned}{*{20}{r}}{ Golfer }&{ \backslash multicolumn1c ER }&{ IR }&{ \backslash multicolumn1c diff }&{z perc }\\1&{ - 130.6}&{ - 98.9}&{ - 31.7}&{ - 1.28}\\2&{ - 125.1}&{ - 115.9}&{ - 9.2}&{ - 0.97}\\3&{ - 51.7}&{ - 161.6}&{109.9}&{0.34}\\4&{ - 179.7}&{ - 196.9}&{17.2}&{ - 0.73}\\5&{ - 130.5}&{ - 170.7}&{40.2}&{ - 0.34}\\6&{ - 101.0}&{ - 274.9}&{173.9}&{0.97}\\7&{ - 24.4}&{ - 275.0}&{250.6}&{1.83}\\8&{ - 231.1}&{ - 275.7}&{44.6}&{ - 0.17}\\9&{ - 186.8}&{ - 214.6}&{27.8}&{ - 0.52}\\{10}&{ - 58.5}&{ - 117.8}&{59.3}&{0.00}\\{11}&{ - 219.3}&{ - 326.7}&{107.4}&{0.17}\\{12}&{ - 113.1}&{ - 272.9}&{159.8}&{0.73}\\{13}&{ - 244.3}&{ - 429.1}&{184.8}&{1.28}\\{14}&{ - 184.4}&{ - 140.6}&{ - 43.8}&{ - 1.83}\\{15}&{ - 199.2}&{ - 345.6}&{146.4}&{0.52}\\{}&{}&{}&{}&{}\end{aligned}\)

a. Is it plausible that the differences came from a normally distributed population?

b. The article reported that mean\(( \pm SD) = - 145.3(68.0)\)for ER velocity and\( = - 227.8(96.6)\)for IR velocity. Based just on this information, could a test of hypotheses about the difference between true average IR velocity and true average ER velocity be carried out? Explain.

c. The article stated that " The lead hip peak IR velocity was significantly greater than the trail hip ER velocity\(\;\left( {p = 0.003, t value = 3.65} \right)\). "$ (The phrasing suggests that an upper-tailed test was used.) Is that in fact the case? (Note: "\(p = .033 \)in Table 2 of the article is erroneous.)

Short answer

a. Plausible;

b. Can not be carried out;

c. Reject null hypothesis. Same conclusion as in the article.

Step-by-step solution

A)Step 1: Plot normal probability

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

B)Step 2: Explain difference between true average IR velocity and true average ER velocity

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

C)Step 3: Determine Paired t Test:

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.

Find the Sample standard deviation

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

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

Find the value of P

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.