_{1}

For the analysis of square contingency tables with the same row and column ordinal classifications, this article proposes new models which indicate the structures of symmetry with respect to the anti-diagonal of the table. Also, this article gives a simple decomposition in 3 ′ 3 contingency table using the proposed models. The proposed models are applied to grip strength data.

Consider the data in

For the analysis of an

where

Right hand grip strength level | Left hand grip strength level | Total | ||
---|---|---|---|---|

Excellent (1) | Good (2) | Poor (3) | ||

Excellent (1) | 74 | 89 | 3 | 166 |

(77.00) | (85.90) | (3.00) | (165.90) | |

Good (2) | 10 | 444 | 93 | 547 |

(10.81) | (444.00) | (96.48) | (551.29) | |

Poor (3) | 0 | 23 | 69 | 92 |

(0.00) | (21.39) | (66.41) | (87.81) | |

Total | 84 | 556 | 165 | 805 |

(87.81) | (551.29) | (165.90) | (805.00) |

model states that the probability that an observation will fall in the (i,j)th cell of the table is equal to the probability that it falls in the (j,i)th cell. Namely, this model describes a structure of symmetry with respect to the main diagonal of the table. Stuart [

where

column marginal distribution. Read [

This model states that the probability that an observation will fall in one of the upper-right triangle cells above the main diagonal of the table is equal to the probability that it falls in one of the lower-left triangle cells below the main diagonal.

For the data in

The present paper proposes three models and gives a simple decomposition using the proposed models in

Firstly, we propose a model defined by

where

Secondly, we propose a model defined by

Let X and Y denote the row and column variables, respectively. Then, this model is also expressed as

We shall refer to this model as the anti-diagonal global symmetry (AGS) model.

Finally, we propose a model defined by

This model states that the row marginal distribution is identical to the column marginal distribution in reverse order. We shall refer to this model as the anti-diagonal marginal homogeneity (AMH) model.

We obtain the following theorem.

Theorem 1. When

Proof. If the AS model holds, then the AGS and AMH models hold. Assuming that both the AGS and AMH models hold, then we shall show that the AS model holds. If the AMH and AGS models hold, then we have

Note that this theorem does not hold when

the table (

Assume that

1) The MLE of

2) The MLE of

where

The MLEs of

for anti-diagonal cells (since

of freedom (df) for testing goodness-of-fit of the AS model is

We shall consider the comparison between two nested models. Suppose that model

Consider the data in

Under the AMH model, the probability that an examinee’s right hand grip strength level is “Excellent (1)”, is estimated to be equal to the probability that an another examinee’s left hand grip strength level is “Poor (3)”. Also, the probability that an examinee’s right hand grip strength level is “Poor (3)”, is estimated to be equal to the probability that an examinee’s left hand grip strength level is “Excellent (1)”.

The decomposition of the AS model into the AGS and AMH models, given by Theorem 1, would be useful for seeing the reason for its poor fit when the AS model fits the

The authors would like to thank the referee for their helpful comments.

Kiyotaka Iki, (2016) Analysis of the Grip Strength Data Using Anti-Diagonal Symmetry Models. Open Journal of Statistics,06,590-593. doi: 10.4236/ojs.2016.64049