Create two sample data vectors, containing three and four distinct values, respectively.

x = [1 1 2 3 1]; y = [1 2 5 3 1];

Cross-tabulatexandy.

table = crosstab(x,y)

table = 2 1 0 0 0 0 0 1 0 0 1 0

The rows intablecorrespond to the three distinct values inx, and the columns correspond to the four distinct values iny.

Generate two independent vectors,x1andx2, each containing 50 discrete uniform random numbers in the range1:3.

rngdefault;% for reproducibilityx1 = unidrnd(3,50,1); x2 = unidrnd(3,50,1);

Cross-tabulatex1andx2.

[table,chi2,p] = crosstab(x1,x2)

table = 1 6 7 5 5 2 11 7 6 chi2 = 7.5449 p = 0.1097

The returnedpvalue of0.1097indicates that, at the 5% significance level,crosstabfails to reject the null hypothesis thattableis independent in each dimension.

加载示例数据,contains measurements of large model cars during the years 1970-1982.

loadcarbig

Cross-tabulate the data of four-cylinder cars (cyl4) based on model year (when) and country of origin (org).

[table,chi2,p,labels] = crosstab(cyl4,when,org);

Uselabelsto determine the index location intablefor the number of four-cylinder cars made in the USA during the late period of the data.

labels

labels = 3×3 cell array 'Other' 'Early' 'USA' 'Four' 'Mid' 'Europe' [] 'Late' 'Japan'

The first column oflabelscorresponds to the data incyl4, and indicates that row2oftablecontains data on cars with four cylinders. The second column oflabelscorresponds to the data inwhen, and indicates that column3oftablecontains data on cars made during the late period. The third column oflabelscorresponds to the data inorg, and indicates that location1of the third dimension oftablecontains data on cars made in the USA.

Therefore,table(2,3,1)contains the number of four-cylinder cars made in the USA during the late period.

table(2,3,1)

ans = 38

The data contains 38 four-cylinder cars made in the USA during the late period.

Load the hospital data.

loadhospital

Thehospitaldataset array contains data on 100 hospital patients, including last name, gender, age, weight, smoking status, and systolic and diastolic blood pressure measurements.

To determine whether smoking status is independent of gender, usecrosstab创建一个2×2列联表的吸烟者d nonsmokers, grouped by gender.

[tbl,chi2,p,labels] = crosstab(hospital.Sex, hospital.Smoker)

tbl = 40 13 26 21 chi2 = 4.5083 p = 0.0337 labels = 2×2 cell array 'Female' '0' 'Male' '1'

The rows of the resulting contingency tabletblcorrespond to the patient's gender, with row 1 containing data for females and row 2 containing data for males. The columns correspond to the patient's smoking status, with column 1 containing data for nonsmokers and column 2 containing data for smokers. The returned resultchi2 = 4.5083is the value of the chi-squared test statistic for a Pearson's chi-squared test of independence. The returned valuep = 0.0337is an approximate-value based on the chi-squared distribution.