91影视

\(\begin{aligned}{l}I = 5\\\alpha = 0.05\end{aligned}\)

The null hypothesis states that all population means are equal:

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3} = {\mu _4} = {\mu _5}\)

The alternative hypothesis states the opposite of the null hypothesis

\({H_1}:\)not all of \({\mu _1},{\mu _2},{\mu _3},{\mu _4},{\mu _5}\)are equal.

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

\(\overline x = \frac{{{n_1}\overline {{x_1}} + {n_2}\overline {{x_2}} \ldots + {n_k}\overline {{x_k}} }}{N} = \frac{{6(13.6833) + 6(15.95) + 6(13.6667) + 6(14.7333) + 6(13.0833)}}{{6 + 6 + 6 + 6 + 6}} \approx 14.2333\)

\(\begin{aligned}{l}MSTr = \frac{{{n_1}{{(\overline {{x_1}} - \overline x )}^2} + {n_2}{{(\overline {{x_2}} - \overline x )}^2} + \ldots + {n_k}{{(\overline {{x_k}} - \overline x )}^2}}}{{I - 1}}\\ = \frac{{6{{(13.6833 - 14.2233)}^2} + 6{{(15.95 - 14.2233)}^2} + 6{{(13.6667 - 14.2233)}^2} + 6{{(14.7333 - 14.2233)}^2} + 6{{(13.0833 - 14.2233)}^2}}}{{5 - 1}}\\ = 7.7138\end{aligned}\)The mean square for treatment is:

The mean square error is:

\(\begin{aligned}{l}MSE = \frac{{({n_1} - 1)s_1^2 + ({n_2} - 1)s_2^2 + \ldots + ({n_k} - 1)s_k^2}}{{N - I}}\\ = \frac{{(6 - 1){{(1.194)}^2} + (6 - 1){{(0.8165)}^2} + (6 - 1){{(0.9395)}^2} + (6 - 1){{(0.4729)}^2}}}{{(6 + 6 + 6 + 6 + 6) - 5}}\\ = 0.9135\end{aligned}\)

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

\(F = \frac{{MSTr}}{{MSE}} = \frac{{7.7138}}{{0.9135}} \approx 8.44\)

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 \(dfn = I - 1 = 5 - 1 = 4\)and \(dfd = N - I = 6 + 6 + 6 + 6 + 6 - 6 = 25\):

P < 0.001

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

P<0.05鈬扲eject \({H_0}\)

There is sufficient evidence to support the claim that the population means are different.