Applications of Chi-Square Tests

川长思鸟来 2022-08-10 00:38 11阅读 0赞

November 3, 2015

By  [arthur charpentier][]

(This article was first published on  **[Freakonometrics » R-english][Freakonometrics _ R-english]**, and kindly contributed to  [R-bloggers)][R-bloggers]

This morning, in our mathematical statistical class, we’ve seen the use of the chi-square test. The first one was related to some goodness of fit of a multinomial distribution. Assume that ![gif.latex_5Cboldsymbol_7BN_7D_28N_1_5Ccdots_N_k_29_5Csim_5Cmathcal_7BM_7D_28n_5Cboldsymbol_7Bp_7D_29][]. In order to test ![gif.latex_H_0_5Cboldsymbol_7Bp_7D_5Cboldsymbol_7Bp_7D_0][] against ![gif.latex_H_1_5Cboldsymbol_7Bp_7D_5Cneq_5Cboldsymbol_7Bp_7D_0][], use the statistic

![gif.latex_Q_5Csum_7Bj_1_7D_5Ek_20_5Cfrac_7B_5BN_j-np_7B0_j_7D_5D_5E2_7D_7Bnp_7B0_j_7D_7D][]Under ![gif.latex_H_0][], ![gif.latex_Q_5Csim_5Cchi_5E2_28k-1_29][]. For instance, we have the number of weddings, in a large city, per season,

> n=c(301,356,413,262)

We want to test if weddings are celebrated uniformely over the year, i.e. ![4][].

> np=rep(sum(n)/4,4)
    > cbind(n,np)
           n  np
    [1,] 301 333
    [2,] 356 333
    [3,] 413 333
    [4,] 262 333
    > Q=sum( (n-np)^2/np  )
    > Q
    [1] 39.02102

This quantity should be compared with the quantile of the chi-square distribution

> qchisq(.95,df=4-1)
    [1] 7.814728

but it is also possible to compute the *p*\-value,

> 1-pchisq(Q,df=4-1)
    [1] 1.717959e-08

Here, we reject the assumption that weddings are celebrated uniformly over the year.

A second application is a goodness-of-fit test, for some parametric distribution. Assume that ![gif.latex_X_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_7D][] takes discrete data, say ![gif.latex_5C_7B1_2_5Ccdots_k_5C_7D][]. Here ![gif.latex_5Cmathbb_7BP_7D_5BX_j_5D_p_j_28_5Cboldsymbol_7B_5Ctheta_7D_29][]. In order to test ![gif.latex_H_0_X_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_0_7D][] against ![gif.latex_H_1_X_5Cnot_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_0_7D][], use

![gif.latex_Q_5Csum_7Bj_1_7D_5Ek_20_5Cfrac_7B_5BN_j-np_7Bj_7D_28_5Cboldsymbol_7B_5Ctheta_7D_0_29_5D_5E2_7D_7Bnp_7Bj_7D_28_5Cboldsymbol_7B_5Ctheta_7D_0_29_7D][]

One can prove that ![gif.latex_Q_5Csim_5Cchi_5E2_28k-1-_5Ctext_7Bdim_7D_28_5Cboldsymbol_7B_5Ctheta_7D_29_29][] under ![gif.latex_H_0][]. For instance, consider the popular example of [von Bortkiewicz’s horsekicks][von Bortkiewicz_s horsekicks] data.

> n=c(109,65,22,3,1)
    > sum(n*0:4)/sum(n) 
    [1] 0.61

The  first thing is that we should regroup 3 and 4+, in order to have enough observation in each cell of the table

> n_correc=c(109,65,22,4)

Now we can try a ![gif.latex_5Cmathcal_7BP_7D_280.6_29][] distribution

> np=200*c(dpois(0:2,lambda=.6),
    +    1-ppois(2,lambda=.6))
    > n_correc=c(109,65,22,4)
    > cbind(n_correc,np)
         n_correc         np
    [1,]      109 109.762327
    [2,]       65  65.857396
    [3,]       22  19.757219
    [4,]        4   4.623058
    > Q=sum( (n_correc-np)^2/np  )
    > Q
    [1] 0.3550214

The quantile of the chi-square distribution is

> qchisq(.95,df=4-1-1)
    [1] 5.991465

and the *p*\-value is

> 1-pchisq(Q,df=4-1-1)
    [1] 0.837352

Finally, it is possible to use the chi-square test in order to test for independence. Consider here two categorical variables![gif.latex_X][]and ![gif.latex_Y][], e.g. the color of the hair, and the color of the eyes, and summarize the information in a contingency table

> n=HairEyeColor[,,1]+HairEyeColor[,,2]
    > n
           Eye
    Hair    Brown Blue Hazel Green
      Black    68   20    15     5
      Brown   119   84    54    29
      Red      26   17    14    14
      Blond     7   94    10    16

In that case, use

![gif.latex_Q_5Csum_7Bi_j_7D_5Cfrac_7B_5BN_7Bi_j_7D-N_7Bi_j_7D_5E_5Cperp_5D_5E2_7D_7BN_7Bi_j_7D_5E_5Cperp_7D][]

with

![gif.latex_N_7Bi_j_7D_5E_5Cperp_5Cfrac_7BN_7Bi_5Ccdot_7D_20N_7B_5Ccdot_20j_7D_7D_7Bn_7D][]

where ![gif.latex_N_7Bi_5Ccdot_7D][] and ![gif.latex_N_7B_5Ccdot_j_7D][] denote respectively the number of observation per row and per column.

> ni=apply(n,1,sum)         # sum per row [hair]
    > nj=apply(n,2,sum)         # sum per colum [eye]
    > n_ind= ni %*% t(nj)/sum(n)
    > rownames(n_ind)=rownames(n)
    > n_ind
              Brown      Blue    Hazel     Green
    Black  40.13514  39.22297 16.96622 11.675676
    Brown 106.28378 103.86824 44.92905 30.918919
    Red    26.38514  25.78547 11.15372  7.675676
    Blond  47.19595  46.12331 19.95101 13.729730

Under ![gif.latex_H_0_X_5Cperp_20Y][],

![gif.latex_Q_5Csim_5Cchi_5E2_28_28I-1_29_28J-1_29_29][]

> Q= sum( (n-n_ind)^2/n_ind )
    > Q
    [1] 138.2898

The quantile is here

> qchisq(.95,df=(4-1)*(4-1))
    [1] 16.91898

and the *p-*value is way below 5%,

> 1-pchisq(Q,df=(4-1)*(4-1))
    [1] 0

[arthur charpentier]: http://www.r-bloggers.com/author/arthur-charpentier/
[Freakonometrics _ R-english]: http://freakonometrics.hypotheses.org/20531
[R-bloggers]: http://www.r-bloggers.com/
[gif.latex_5Cboldsymbol_7BN_7D_28N_1_5Ccdots_N_k_29_5Csim_5Cmathcal_7BM_7D_28n_5Cboldsymbol_7Bp_7D_29]: http://latex.codecogs.com/gif.latex?%5Cboldsymbol%7BN%7D=%28N_1,%5Ccdots,N_k%29%5Csim%5Cmathcal%7BM%7D%28n,%5Cboldsymbol%7Bp%7D%29
[gif.latex_H_0_5Cboldsymbol_7Bp_7D_5Cboldsymbol_7Bp_7D_0]: /images/20220731/c1e5b051d84c488fa8fc8bd5ef535d88.png:%5Cboldsymbol%7Bp%7D=%5Cboldsymbol%7Bp%7D_0
[gif.latex_H_1_5Cboldsymbol_7Bp_7D_5Cneq_5Cboldsymbol_7Bp_7D_0]: /images/20220731/da26117f6da24a78b7f164626163bf3b.png
[gif.latex_Q_5Csum_7Bj_1_7D_5Ek_20_5Cfrac_7B_5BN_j-np_7B0_j_7D_5D_5E2_7D_7Bnp_7B0_j_7D_7D]: http://latex.codecogs.com/gif.latex?Q=%5Csum_%7Bj=1%7D%5Ek%20%5Cfrac%7B%5BN_j-np_%7B0,j%7D%5D%5E2%7D%7Bnp_%7B0,j%7D%7D
[gif.latex_H_0]: /images/20220731/c1e5b051d84c488fa8fc8bd5ef535d88.png
[gif.latex_Q_5Csim_5Cchi_5E2_28k-1_29]: /images/20220731/33e136dfc8ff4c568f09b837b7492a4a.png
[4]: /images/20220731/473a8cfda9ff401a8bc3279630d4a828.png
[gif.latex_X_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_7D]: http://latex.codecogs.com/gif.latex?X%5Csim%20F_%7B%5Cboldsymbol%7B%5Ctheta%7D%7D
[gif.latex_5C_7B1_2_5Ccdots_k_5C_7D]: http://latex.codecogs.com/gif.latex?%5C%7B1,2,%5Ccdots,k%5C%7D
[gif.latex_5Cmathbb_7BP_7D_5BX_j_5D_p_j_28_5Cboldsymbol_7B_5Ctheta_7D_29]: http://latex.codecogs.com/gif.latex?%5Cmathbb%7BP%7D%5BX=j%5D=p_j%28%5Cboldsymbol%7B%5Ctheta%7D%29
[gif.latex_H_0_X_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_0_7D]: /images/20220731/c1e5b051d84c488fa8fc8bd5ef535d88.png:X%5Csim%20F_%7B%5Cboldsymbol%7B%5Ctheta%7D_0%7D
[gif.latex_H_1_X_5Cnot_5Csim_20F_7B_5Cboldsymbol_7B_5Ctheta_7D_0_7D]: /images/20220731/ad59b5abb0d24a11927ff8a58ceab762.png
[gif.latex_Q_5Csum_7Bj_1_7D_5Ek_20_5Cfrac_7B_5BN_j-np_7Bj_7D_28_5Cboldsymbol_7B_5Ctheta_7D_0_29_5D_5E2_7D_7Bnp_7Bj_7D_28_5Cboldsymbol_7B_5Ctheta_7D_0_29_7D]: /images/20220731/68cb51e031a943739b404c50f3aabd2e.png
[gif.latex_Q_5Csim_5Cchi_5E2_28k-1-_5Ctext_7Bdim_7D_28_5Cboldsymbol_7B_5Ctheta_7D_29_29]: http://latex.codecogs.com/gif.latex?Q%5Csim%5Cchi%5E2%28k-1-%5Ctext%7Bdim%7D%28%5Cboldsymbol%7B%5Ctheta%7D%29%29
[von Bortkiewicz_s horsekicks]: http://www.datavis.ca/courses/grcat/grc1.html
[gif.latex_5Cmathcal_7BP_7D_280.6_29]: /images/20220731/360f1104898a45d4bbf9f14a1cbe6a11.png
[gif.latex_X]: http://latex.codecogs.com/gif.latex?X
[gif.latex_Y]: /images/20220731/3d8d4b50e3b246f69b236967b6f64b27.png
[gif.latex_Q_5Csum_7Bi_j_7D_5Cfrac_7B_5BN_7Bi_j_7D-N_7Bi_j_7D_5E_5Cperp_5D_5E2_7D_7BN_7Bi_j_7D_5E_5Cperp_7D]: /images/20220731/af5d6d6991dc46af88079fc727eca0e4.png
[gif.latex_N_7Bi_j_7D_5E_5Cperp_5Cfrac_7BN_7Bi_5Ccdot_7D_20N_7B_5Ccdot_20j_7D_7D_7Bn_7D]: /images/20220731/63afb1c878744c83949a5ef7b982580c.png
[gif.latex_N_7Bi_5Ccdot_7D]: http://latex.codecogs.com/gif.latex?N_%7Bi,%5Ccdot%7D
[gif.latex_N_7B_5Ccdot_j_7D]: /images/20220731/b632353349d144009b4047e30a1434f7.png
[gif.latex_H_0_X_5Cperp_20Y]: /images/20220731/c1e5b051d84c488fa8fc8bd5ef535d88.png:X%5Cperp%20Y
[gif.latex_Q_5Csim_5Cchi_5E2_28_28I-1_29_28J-1_29_29]: /images/20220731/ef5cedc473a64389a78f76dc2b985ade.png