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\(\begin{aligned}{l}{n_1} = 47\\{n_2} = 36\\{n_3} = 50\\\overline {{x_1}} = 174.7\\\overline {{x_2}} = 160.2\\\overline {{x_3}} = 153.5\\{s_1} = 50.9\\{s_2} = 37.2\\{s_3} = 45.9\\\alpha = 0.05\end{aligned}\)

The overall mean is the sum of all values divided by the number of values:

\(\overline x = \frac{{{n_1}\overline {{x_1}} + {n_2}\overline {{x_2}} + {n_3}\overline {{x_3}} }}{N} = \frac{{47(174.7) + 36(160.2) + 50(153.5)}}{{47 + 3 + 50}} \approx 162.8053\)

The mean square for treatment is:

\(\begin{aligned}{l}MSTr = \frac{{{n_1}{{(\overline {{x_1}} - \overline x )}^2} + {n_2}{{(\overline {{x_2}} - \overline x )}^2} + {n_2}{{(\overline {{x_3}} - \overline x )}^2}}}{{I - 1}}\\ = \frac{{47{{(147.7 - 162.8053)}^2} + 36{{(160.2 - 162.8053)}^2} + 50{{(153.5 - 162.8053)}^2}}}{{3 - 1}} = 5611.7632\end{aligned}\)

The mean square error is:

\(\begin{aligned}{l}MSE = \frac{{({n_1} - 1)s_1^2 + ({n_2} - 1)s_2^2 + ({n_3} - 1)s_3^2}}{{N - I}}\\ = \frac{{(47 - 1){{(50.9)}^2} + (36 - 1){{(37.2)}^2} + (50 - 1){{(45.9)}^2}}}{{(47 + 36 + 50) - 3}} = 2083.4258\end{aligned}\)

The ANOVA F statistic is the ratio of the MSTr and MSE:

\(F = \frac{{MSTr}}{{MSE}} = \frac{{5611.7632}}{{2083.4258}} \approx 2.694\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the F-distribution table in the appendix containing the F-value in the row\(dfn = I - 1 = 3 - 1 = 2\)and \(dfd = N - I = 47 + 36 + 50 - 3 = 130 > 100:\)

\(0.050 < P < 0.100\)

If the P-value is less than the significance level, then reject the null hypothesis.

\(P > 0.05 \Rightarrow \)Fail to reject\({H_0}\)

There is not sufficient evidence to support the claim that the true average cortisol concentration is different for the three groups.