# Two Way ANOVA Table for N = 12, k = 3, h = 4 & α = 5%

One way ANOVA table or summary to estimate the test of hypothesis (H_{0}) at 5% or 0.05 significance level for sample *means* by using *F-distribution table*

ANOVA Table for Two-Way Classification | ||||
---|---|---|---|---|

Sources of Variation | df | SS | MSS | F-Ratio |

between Treatment | 2 | 6.5 | 6.5/2 = 3.25 | 3.25/3.1389 = 1.0354 |

between Varieties | 3 | 5.6667 | 5.6667/3 = 1.8889 | 1.8889/3.1389 = 0.6018 |

Error | 6 | 18.8333 | 18.8333/6 = 3.1389 | |

Total | 11 |

## Work with Steps for N = 12, k = 3, h = 4 & α = 5%

The below is the example work with steps shows how to generate the two way ANOVA table for the different means of treatments (rows) & subjects (columns) to find the F test statistic for test hypothesis (H_{0}) at a 5% or 0.05 significance level for three sample means x_{1}, x_{2} & x_{3}.

__Workout :__

step 1 Address the formula input parameters and values

x

_{1}= 8, 7, 5 & 5

x

_{2}= 7, 4, 4 & 8

x

_{3}= 3, 4, 5 & 6

step 2 Form the below table to carry out the analysis of

*variance*

Data sets | 1 | 2 | 3 | 4 | Total | Square |

x_{1} | 8 | 7 | 5 | 5 | ∑x_{1} = 25 | (∑x_{1})^{2} = 625 |

x_{2} | 7 | 4 | 4 | 8 | ∑x_{2} = 23 | (∑x_{2})^{2} = 529 |

x_{3} | 3 | 4 | 5 | 6 | ∑x_{3} = 18 | (∑x_{3})^{2} = 324 |

Total | 18 | 15 | 14 | 19 | G = 66 | ∑T_{i}^{2} = 1478 |

Square | 324 | 225 | 196 | 361 | ∑T_{j}^{2} = 1106 |

Data sets | 1 | 2 | 3 | 4 | Total |

x_{1}^{2} | 64 | 49 | 25 | 25 | 163 |

x_{2}^{2} | 49 | 16 | 16 | 64 | 145 |

x_{3}^{2} | 9 | 16 | 25 | 36 | 86 |

Total | 122 | 81 | 66 | 125 | ∑x_{ij}^{2} = 394 |

N = sum of Values of all items = 4 + 4 + 4

N = 12

Correction factor (CF) = G

^{2}N

= 66

^{2}12

= 435612

CF = 363

step 4 Find the sum of squares of all individual items (∑x

_{ij}) and then total sum of squares (TSS)

Total sum of squares (TSS) =k∑i = 1 k∑j = 1 x

_{ij}

^{2}- CF

= 394 - 363

= 31

step 5 Find sum of squares between rows (SSR)

SSR = ∑T

_{i}

^{2}h - CF

where h is the number of observation in each row

= 14784 - 363

= 369.5 - 363

= 6.5

step 6 Find sum of squares between columns (SSC)

SSC = ∑T

_{j}

^{2}k - CF

where k is the number of observation in each column

= 11063 - 363

= 368.6667 - 363

= 5.6667

step 7 Find Sum of square due to error (SSE)

SSE = TSS - SSR - SSC

= 31 - 6.5 - 5.6667

= 18.8333

step 8 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

Sources of variation | df | SS | MSS | F-Ratio |

between Treatments | k - 1 | SSR | SSR/k-1 = MST | MST/MSE = F _{R} |

between Varieties | h - 1 | SSC | SSC/h-1 = MSV | MSV/MSE = F _{C} |

Error | (h - 1)(k - 1) | SSE | SSE/(k-1)(h-1) = MSE | |

Total | N - 1 |

Sources of variation | d.f | SS | MSS | F-Ratio |

between Treatments | 3 - 1 = 2 | 6.5 | 6.5/2 = 3.25 | 3.25/3.1389 = 1.0354 |

between Varieties | 4 - 1 = 3 | 5.6667 | 5.6667/3 = 1.8889 | 1.8889/3.1389 = 0.6018 |

Error | (3 - 1)(4 - 1) = 6 | 18.8333 | 18.8333/6 = 3.1389 | |

Total | 12 - 1 = 11 |

step 9 Find Critical values F

_{e}or Table values of F

- The critical value variance between treatments is obtained from F table for

[(k –1), (k – 1) (h – 1)] degrees of freedom at 5% level of significance. - The critical value variance between varieties is obtained from F table for [(h – 1), (k –1) (h – 1)] degrees of freedom at 5% level of significance.

__Critical values of F from F-distribution table__

- For treatments, the critical value F
_{e}(2,6) from F-distribution table at 5% of significance level is 5.1433 - For varieties, the critical value F
_{e}(3,6) from F-distribution table at 5% of significance level is 4.7571

__Inference__- There is no significance difference between treatments, since the calculated value of F
_{0}= 1.0354 is smaller than the table value of F_{e}= 5.1433. Therefore the null hypothesis H_{0}is accepted. - There is no significance difference between varieties or subjects, since the calculated value of F
_{0}= 0.6018 is smaller than the table value of F_{e}= 4.7571. Therefore the null hypothesis H_{0}is accepted.