Question

How does energy intake compare to energy expenditure? One aspect of this issue was considered in the article “Measurement of Total Energy Expenditure by the Doubly Labelled Water Method in Professional Soccer Players” (J. of Sports Sciences, 2002: 391–397), which contained the accompanying data (MJ/day).

Test to see whether there is a significant difference between intake and expenditure. Does the conclusion depend on whether a significance level of .05, .01, or .001 is used?

Short answer

Reject null hypothesis at significance level 0.01 and 0.05;

Do not reject null hypothesis at significance level 0.001.

Step-by-step solution

To Find the

Given data suggest that the paired t test should be used.

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\({l_{n - 1}}\)curve.

The accompanying table gives the necessary values for the paired\(\iota \)test to be performed.

\( Player, i\)	\(Expenditure, {x_i}\)	\(Intake, {y_i}\)	\(Difference, {d_i} = {x_i} - {y_i}\)
1	14.4	14.6	-0.2
2	12.1	9.2	2.9
3	14.3	11.8	2.5
4	14.2	11.6	2.6
5	15.2	12.7	2.5
6	15.5	15	0.6
7	17.8	16.3	1.5

\(\)

Find the sample value

The sample mean and the sample standard deviation of the differences has to be computed.

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

\(\bar d = \frac{1}{7} \times ( - 0.2 + 2.9 \ldots 1.5) = 1.757\)

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

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

Where

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

The Sample Standard Deviation s is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)\(P = 0.008 > 0.001\)\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \times {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{7 - 1}} \times \left( {{{( - 0.2 - 1.757)}^2} + {{(2.9 - 1.757)}^2} + \ldots + {{(1.5 - 1.757)}^2}} \right)\\ = 1.433\end{array}\)

and the sample standard deviation

\({s_D} = \sqrt {1.433} = 1.197\)

To find P value

The hypotheses of interest are\({H_0}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} \ne 0\). The test statistic value is

\(t = \frac{{1.757 - 0}}{{1.197/\sqrt 7 }} = 3.88\)

The degrees of freedom are n-1=7-1=6. The P value is two times the area under\({t_6}\)curve to the right of |t|; thus

\(P = 2 \times P(T > 3.88)) = 2 \times 0.004 = 0.008\)

which was computed using software (you could use the table in the appendix).

Reject null hypothesis

at two of the given significance levels because

\(P = 0.008 < 0.01 < 0.05\)

At significance level 0.001 do

not reject null hypothesis

\(P = 0.008 > 0.001\)

Final proof 

Reject null hypothesis at significance level 0.01 and 0.05;

Do not reject null hypothesis at significance level 0.001.