91Ó°ÊÓ

Let us assume

\(c = 95\% = 0.95\)

Other confidence levels can be determined similarly.

(a) Given:

\(n = 17\)

Determine the difference in value of each pair.

\(\begin{array}{*{20}{c}}{|{\rm{ sample }}1}&{{\rm{ sample 2}}}&{{\rm{ Difference D}}}\\{27.63}&{24.33}&{3.3}\\{30.57}&{26.36}&{4.21}\\{32.62}&{30.62}&2\\{39.79}&{33.74}&{6.05}\\{28.5}&{29.84}&{ - 1.34}\\{26.7}&{26.71}&{ - 0.01}\\{30.34}&{26.45}&{3.89}\\{28.69}&{21.49}&{7.2}\\{31.19}&{20.82}&{10.37}\\{36}&{21.75}&{14.25}\\{31.58}&{28.32}&{3.26}\\{32.55}&{27.22}&{5.33}\\{29.56}&{28.86}&{0.7}\\{28.64}&{28.58}&{0.06}\\{28.58}&{27.15}&{1.43}\\{31.99}&{29.46}&{2.53}\\{27.16}&{21.26}&{5.9}\\{{\rm{ Mean }}}&{}&{4.0665}\\{sd}&{}&{3.9549}\end{array}\)

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{3.3 + 4.21 + 2 + \ldots + 1.43 + 2.53 + 5.9}}{{17}} \approx 4.0665\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{(3.3 - 4.0665)}^2} + \ldots + {{(5.9 - 4.0665)}^2}}}{{17 - 1}}} \approx 3.9549\)

Determine the \({t_{\alpha /2}}\)using the Student's T distribution table in the appendix with $d f=n-1=17-1=16:

\({t_{0.025}} = 2.120\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.120 \cdot \frac{{3.9549}}{{\sqrt {17} }} \approx 2.0335\)

The endpoints of the confidence interval for \({\mu _d}\)are:

\(\begin{array}{l}\bar d - E = 4.0665 - 2.0335 = 2.0330\\\bar d + E = 4.0665 + 2.0335 = 6.1000\end{array}\)

We are 95 % confident that the true average difference in translation between dominant and nondominant arms for pitchers is between 2.0330 and 6.1000.

(b) Given:

\(n = 19\)

Determine the difference in value of each pair.

\(\begin{array}{*{20}{c}}{{\rm{ Sample 1 }}}&{{\rm{ Sample 2 }}}&{{\rm{ Difference D }}}\\{30.31}&{32.54}&{ - 2.23}\\{44.86}&{40.95}&{3.91}\\{22.09}&{23.48}&{ - 1.39}\\{31.26}&{31.11}&{0.15}\\{28.07}&{28.75}&{ - 0.68}\\{31.93}&{29.32}&{2.61}\\{34.68}&{34.79}&{ - 0.11}\\{29.1}&{28.87}&{0.23}\\{25.51}&{27.59}&{ - 2.08}\\{22.49}&{21.01}&{1.48}\\{28.74}&{30.31}&{ - 1.57}\\{27.89}&{27.92}&{ - 0.03}\\{28.48}&{27.85}&{0.63}\\{25.51}&{27.59}&{ - 2.08}\end{array}\)

\(\begin{array}{*{20}{c}}{22.49}&{21.01}&{1.48}\\{28.74}&{30.31}&{ - 1.57}\\{27.89}&{27.92}&{ - 0.03}\\{28.48}&{27.85}&{0.63}\\{25.6}&{24.95}&{0.65}\\{20.21}&{21.59}&{ - 1.38}\\{33.77}&{32.48}&{1.29}\\{32.59}&{32.48}&{0.11}\\{32.6}&{31.61}&{0.99}\\{29.3}&{27.46}&{1.84}\end{array}\)

Mean 0.2326

Sd 1.6034

\(\begin{array}{l}{\rm{Determine the sample mean of the differences}}{\rm{. The mean is the sum of all values }}\\{\rm{divided by the number of values}}{\rm{. }}\end{array}\)

\(\bar d = \frac{{ - 2.23 + 3.91 - 1.39 + \ldots + 0.11 + 0.99 + 1.84}}{{19}} \approx 0.2326\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 2.23 - 0.2326)}^2} + \ldots + {{(1.84 - 0.2326)}^2}}}{{19 - 1}}} \approx 1.6034\)

Determine the \({t_{\alpha /2}}\)using the Student's T distribution table in the appendix with \(df = n - 1\)=19-1=18:

\({t_{0.025}} = 2.101\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.101 \cdot \frac{{1.6034}}{{\sqrt {19} }} \approx 0.7728\)

The endpoints of the confidence interval for \({\mu _d}\) are:

\(\begin{array}{l}\bar d - E = 0.2326 - 0.7728 = - 0.5402\\\bar d + E = 0.2326 + 0.7728 = 1.0054\end{array}\)

We are 95 % confident that the true average difference in translation between dominant and nondominant arms for position players is between \( - 0.5402\) and\(1.0054\).

(c) Pitchers appear to have greater difference in side-to-side anteroposterior translation of their shoulders compared with position players, because the confidence interval in part (a) lies completely above the confidence interval in part (b).

Finally we get,

(a)\((2.0330,6.1000)\)

(b) \(( - 0.5402,1.0054)\)

(c) Pitchers appear to have greater difference in side-to-side anteroposterior translation of their shoulders compared with position players.