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The body temperature of a set of five subjects is recorded, first at 8 AM and then at 12 AM. Therefore, the sample size is equal to \(n = 5\).

The following table shows the values of the differences for each pair by subtracting the temperature at 12 AM from the temperature at 8 AM.

Themean value of the differences between the body temperature at 8 AM and 12 AM is equal to:

\(\begin{array}{c}\bar d = \frac{{\sum d }}{n}\\ = \frac{{ - 1.4}}{5}\\ = - 0.28\end{array}\)

Therefore, the mean value \(\left( {\bar d} \right)\)is -0.28.

Thestandard deviation of the differences between the body temperatures at 8 AM and 12 AM is equal to:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 0.8 - \left( { - 0.28} \right)} \right)}^2} + {{\left( { - 0.5 - \left( { - 0.28} \right)} \right)}^2} + ...... + {{\left( { - 0.1 - \left( { - 0.28} \right)} \right)}^2}}}{{5 - 1}}} \\ = 0.36\end{array}\)

Therefore, the standard deviation \(\left( {{s_d}} \right)\)is equal to 0.36.

In general,\({\mu _d}\)represents the mean value of all the differences computed for the pairs of data when the population of all matched pairs of temperatures at 8 AM and 12 AM is taken into consideration.