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Q3c) One Way ANOVA F Statistic

SOLUTION

Given

\bar{x}_{\text{ACD}} = 19.35, s_{\text{ACD}}^2 = 0.0058, n_{\text{ACD}} = 21

\bar{x}_{\text{MFM}} = 19.47, s_{\text{MFM}}^2 = 0.0562, n_{\text{MFM}} = 27

\bar{x}_{\text{GH}} = 19.16, s_{\text{GH}}^2 = 0.0330, n_{\text{GH}} = 19

Test Statistic

Overall Mean (Grand Mean)

\bar{x}_{\text{grand}} = \frac{(\bar{x}_{\text{ACD}} \times n_{\text{ACD}}) + (\bar{x}_{\text{MFM}} \times n_{\text{MFM}}) + (\bar{x}_{\text{GH}} \times n_{\text{GH}})}{n_{\text{ACD}} + n_{\text{MFM}} + n_{\text{GH}}}

\bar{x}_{\text{grand}} = \frac{(19.35 \times 21) + (19.47 \times 27) + (19.16 \times 19)}{21 + 27 + 19}

\bar{x}_{\text{grand}} = 19.3444

Sum of Squares for Treatments Groups (SST)

SST = n_{\text{ACD}}(\bar{x}_{\text{ACD}} - \bar{x}_{\text{grand}})^2 + n_{\text{MFM}}(\bar{x}_{\text{MFM}} - \bar{x}_{\text{grand}})^2 + n_{\text{GH}}(\bar{x}_{\text{GH}} - \bar{x}_{\text{grand}})^2

SST = 21(19.35 - 19.3444)^2 + 27(19.47 - 19.3444)^2 + 19(19.16 - 19.3444)^2

SST = 1.0726

Sum of Squares for Error (SSE)

SSE = (n_{\text{ACD}} - 1)s_{\text{ACD}}^2+ (n_{\text{MFM}} - 1)s_{\text{MFM}}^2 + (n_{\text{GH}} - 1)s_{\text{GH}}^2

SSE = (21 - 1)0.0058 + (27 - 1)0.0562 + (19 - 1)0.0330

SSE = 2.0917

Mean Squares

MST (Mean Square Treatment): MST = \frac{SST}{k - 1} = \frac{1.0726}{3 - 1}

MST = 0.5363

MSE(Mean Square Error): MSE = \frac{SSW}{n - k} = \frac{2.1712}{67 - 3}

MSE = 0.0339

F-Statistic

F = \frac{MST}{MSE} = \frac{0.5363}{0.0339}

F = 15.8200

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