Q3c) One Way ANOVA F Statistic

ADMS 2320: FINAL EXAM PREP · Winter 2024 2022 Summer Final 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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