A (10) | B (15) | C (20) |

B (25) | C (10) | A (15) |

C (25) | A (20) | B (15) |

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Login NowWhen the experimental material is not homogeneous the LSD is better than RBD. In RBD local control is used according to one way grouping. i.e according to blocks but in LSD local control is used according to two way grouping i.e. rows and columns. Hence it is used when two sources of errors are to be controlled simultaneously. In this design number of treatments are equal to the number of replication and the treatment occurs once and only once in each row and column. In this design, Latin alphabet are used to denote the treatments, and shape is square due to equal number of treatments and replication so called Latin Square design. It is based upon the all principles of design namely replication, randomization and local control.

Let us consider m treatments with m replication each so that there are N = m^{2} experiments unit.

Let us divide the experimental material into m^{2} experimental units arranged in square so that each row as well as column contains m units. In this design none of treatments are replicated along row wise or column wise. In this case we study the variation between treatments, the variation between rows and variations between columns. It has only m^{2} experimental treatments. Hence it is the case of incomplete three ways ANOVA. for complete three way ANOVA, we need m^{3} experimental unit.

Let us consider t = 4(A, B, C, D) then 4 x 4 LSD is as shown below.

A | D | B | C |

B | C | D | A |

C | B | A | D |

D | A | C | B |

**Merits of LSD**

- Due to the use of two way grouping of controls more variation than CD and RBD
- It is incomplete three way layout. It’s advantage over complete three way layout is that instead of m
^{3}experimental units only m^{3}units are needed. - The statistical analysis remains simple if some observations are missing.

**Demerits:**

- The assumptions of factors are independent is not always true.
- It is suitable for treatments 5 to 10.
- It is not easy in the field layout.

**Problem Part:**

Now,

Problem to test

H_{0R} = Rows are insignificant

H_{1R} = Rows are significant

H_{0c} = Columns are insignificant

H_{1c} = Columns are significant

H_{0T} = Treatments are insignificant

H_{1T} = Treatments are significant

T_{i} |
T_{i}^{2} |
||||

A 10 | B 15 | C 20 | 45 | 2025 | |

B 25 | C 10 | A 15 | 50 | 2500 | |

C 25 | A 20 | B 15 | 60 | 3600 | |

T_{j} |
60 | 45 | 50 | G=155 | ΣT_{i}^{2} = 8125 |

T_{j}^{2} |
3600 | 2025 | 2500 | ΣT_{j}^{2} = 8125 |

T_{. . A} = 10 + 20 + 15 = 45

T_{. . B} = 25 + 15 + 15 = 55

T_{. . C} = 25 + 10 + 20 = 55

ΣT^{2}_{. . k} = 45^{2} + 55^{2} + 55^{2} = 8075

K = A, B, C

m = 3

N = m^{2} = 3^{2} = 9

CG = \(\frac{G^2}{N}\) = \(\frac{155^2}{9}\) = 2669.444

\(\sum_{(i, j, k)} y^{2}_{ijk}\) = 10^{2} + 15^{2} + 25^{2} + 10^{2} + 15^{2} + 25^{2} + 20^{2} + 15^{2} = 2925

SSR = \(\frac{\sum_{i} T_{i . . }^{2}}{m}\) – CF = \(\frac{1}{3}\) x 8125 – 2669.444 = 38.889

SSC = \(\frac{\sum_{i} T_{j . . }^{2}}{m}\) – CF = \(\frac{1}{3}\) x 8125 – 2669.444 = 38.889

SST = \(\frac{\sum_{i} T_{k . . }^{2}}{m}\) – CF = \(\frac{1}{3}\) x 8075 – 2669.444 = 22.222

SSE = TSS – SSR – SSC – SST

= 255.556 – 38.889 – 38.889 – 22.222 = 155.556

**ANOVA Table**

S.V. | d.f. | S.S. | M.S. | F_{cat} |
F_{Tab} |

Row | 2 | 38.889 | 19.444 | 0.249 | F_{0.05 (2, 2)} = 19 |

Column | 2 | 38.889 | 19.444 | 0.249 | F_{0.05 (2, 2)} = 19 |

Treatment | 2 | 22.222 | 11.111 | 0.142 | F_{0.05 (2, 2)} = 19 |

Error | 2 | 155.556 | 77.778 | ||

Total | 8 | 255.556 |

**Decision:**

F_{R} = 0.249 <> F_{0.05(2,2)} = 19, Accept H_{0.05} at 5% level of significance.

F_{c} = 0.249 <> F_{0.05(2,2)} = 19, Accept H_{0.05} at 5% level of significance.

F_{T} = 0.142 <> F_{0.05(2,2)} = 19, Accept H_{0.05} at 5% level of significance.

**Conclusion:**

Row are insignificant.

Columns are insignificant.

Treatments are insignificant.

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