Create sample data matrixywith columns that are constants, plus random normal disturbances with mean 0 and standard deviation 1.

y = meshgrid(1:5); rngdefault;% For reproducibilityy = y + normrnd(0,1,5,5)

y =5×51.5377 0.6923 1.6501 3.7950 5.6715 2.8339 1.5664 6.0349 3.8759 3.7925 -1.2588 2.3426 3.7254 5.4897 5.7172 1.8622 5.5784 2.9369 5.4090 6.6302 1.3188 4.7694 3.7147 5.4172 5.4889

Perform one-way ANOVA.

p = anova1(y)

p = 0.0023

The ANOVA table shows the between-groups variation (Columns) and within-groups variation (Error).SSis the sum of squares, anddf的自由度。总度freedom is total number of observations minus one, which is 25 - 1 = 24. The between-groups degrees of freedom is number of groups minus one, which is 5 - 1 = 4. The within-groups degrees of freedom is total degrees of freedom minus the between groups degrees of freedom, which is 24 - 4 = 20.

MSis the mean squared error, which isSS/dffor each source of variation. TheF-statistic is the ratio of the mean squared errors (13.4309/2.2204). Thep-value is the probability that the test statistic can take a value greater than the value of the computed test statistic, i.e., P(F > 6.05). The smallp-value of 0.0023 indicates that differences between column means are significant.

Input the sample data.

strength = [82 86 79 83 84 85 86 87 74 82...78 75 76 77 79 79 77 78 82 79]; alloy = {'st','st','st','st','st','st','st','st',...'al1','al1','al1','al1','al1','al1',...'al2','al2','al2','al2','al2','al2'};

The data are from a study of the strength of structural beams in Hogg (1987). The vector strength measures deflections of beams in thousandths of an inch under 3000 pounds of force. The vector alloy identifies each beam as steel ('st'), alloy 1 ('al1'), or alloy 2 ('al2'). Although alloy is sorted in this example, grouping variables do not need to be sorted.

Test the null hypothesis that the steel beams are equal in strength to the beams made of the two more expensive alloys. Turn the figure display off and return the ANOVA results in a cell array.

[p,tbl] = anova1(strength,alloy,'off')

p = 1.5264e-04

tbl=4×6 cell array列1到5{‘源’}{“党卫军”}{“df”}{女士' } {'F' } {'Groups'} {[184.8000]} {[ 2]} {[ 92.4000]} {[ 15.4000]} {'Error' } {[102.0000]} {[17]} {[ 6.0000]} {0x0 double} {'Total' } {[286.8000]} {[19]} {0x0 double} {0x0 double} Column 6 {'Prob>F' } {[1.5264e-04]} {0x0 double } {0x0 double }

总度freedom is total number of observations minus one, which is $20-1=19$. The between-groups degrees of freedom is number of groups minus one, which is $3-1=2$. The within-groups degrees of freedom is total degrees of freedom minus the between groups degrees of freedom, which is $19-2=17$.

MSis the mean squared error, which isSS/dffor each source of variation. TheF-statistic is the ratio of the mean squared errors. Thep-value is the probability that the test statistic can take a value greater than or equal to the value of the test statistic. Thep-value of 1.5264e-04 suggests rejection of the null hypothesis.

You can retrieve the values in the ANOVA table by indexing into the cell array. Save theF-statistic value and thep-value in the new variablesFstatandpvalue.

Fstat = tbl{2,5}

Fstat = 15.4000

pvalue = tbl{2,6}

pvalue = 1.5264e-04

Input the sample data.

strength = [82 86 79 83 84 85 86 87 74 82...78 75 76 77 79 79 77 78 82 79]; alloy = {'st','st','st','st','st','st','st','st',...'al1','al1','al1','al1','al1','al1',...'al2','al2','al2','al2','al2','al2'};

The data are from a study of the strength of structural beams in Hogg (1987). The vector strength measures deflections of beams in thousandths of an inch under 3000 pounds of force. The vector alloy identifies each beam as steel (st), alloy 1 (al1), or alloy 2 (al2). Although alloy is sorted in this example, grouping variables do not need to be sorted.

Perform one-way ANOVA usinganova1. Return the structurestats, which contains the statisticsmultcompareneeds for performingMultiple Comparisons.

[~,~,stats] = anova1(strength,alloy);

The smallp-value of 0.0002 suggests that the strength of the beams is not the same.

Perform a multiple comparison of the mean strength of the beams.

[c,~,~,gnames] = multcompare(stats);

In the figure, the blue bar represents the comparison interval for mean material strength for steel. The red bars represent the comparison intervals for the mean material strength for alloy 1 and alloy 2. Neither of the red bars overlaps with the blue bar, which indicates that the mean material strength for steel is significantly different from that of alloy 1 and alloy 2. You can confirm the significant difference by clicking the bars that represent alloy 1 and 2.

Display the multiple comparison results and the corresponding group names in a table.

tbl = array2table(c,"VariableNames",...["Group A","Group B","Lower Limit","A-B","Upper Limit","P-value"]); tbl.("Group A") = gnames(tbl.("Group A")); tbl.("Group B") = gnames(tbl.("Group B"))

tbl=3×6 tableGroup A Group B Lower Limit A-B Upper Limit P-value _______ _______ ___________ ___ ___________ __________ {'st' } {'al1'} 3.6064 7 10.394 0.00016831 {'st' } {'al2'} 1.6064 5 8.3936 0.0040464 {'al1'} {'al2'} -5.628 -2 1.628 0.35601

The first two columns show the pair of groups that are compared. The fourth column shows the difference between the estimated group means. The third and fifth columns show the lower and upper limits for the 95% confidence intervals of the true difference of means. The sixth column shows thep-value for a hypothesis that the true difference of means for the corresponding groups is equal to zero.

The first two rows show that both comparisons involving the first group (steel) have confidence intervals that do not include zero. Because the correspondingp-values (1.6831e-04 and 0.0040, respectively) are small, those differences are significant.

The third row shows that the differences in strength between the two alloys is not significant. A 95% confidence interval for the difference is [-5.6,1.6], so you cannot reject the hypothesis that the true difference is zero. The correspondingp-value of 0.3560 in the sixth column confirms this result.