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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 \(n = 5\).

It is known that the\({t_{\frac{\alpha }{2}}}\)follows \(\left( {n - 1} \right)\) degrees of freedom.

Thus,

\(\begin{array}{c}df = n - 1\\ = 5 - 1\\ = 4\end{array}\)

Therefore, the value of the degrees of freedom is equal to 4.

The confidence level is given to be equal to 99%. Hence, the level of significance becomes\(\alpha = 0.01\)

For the given level of significance\(\alpha = 0.01\)with 4 degrees of freedom, the criticalvalue is given in the t-distribution table.

Thus,

\(\begin{array}{c}{t_{crit}} = \left( {{t_{\frac{\alpha }{2}}},df} \right)\\ = \left( {{t_{\frac{{0.01}}{2}}},4} \right)\\ = 4.604\end{array}\)

Therefore, the critical value is 4.604.