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13.2.1 GO Tutorial X Note that we reject the null hypothesis of equal treatment means when the test statistic is large enough

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using excel>data>data analysis>ANOVA without replication

we have

Anova: Two-Factor Without Replication
SUMMARY Count Sum Average Variance
0.37 4 329 82.25 4.916667
0.51 4 308 77 5.333333
0.71 4 300 75 3.333333
1.02 4 287 71.75 10.91667
1.4 4 259 64.75 11.58333
1.99 4 251 62.75 7.583333
RR 1 6 418 69.66667 62.66667
RR 2 6 424 70.66667 60.26667
RR 3 6 442 73.66667 54.66667
RR 4 6 450 75 47.6
ANOVA
Source of Variation SS df MS F P-value F crit
Office diameter 1107.5 5 221.5 179.5946 7.88E-13 2.901295
Radon Released 112.5 3 37.5 30.40541 1.28E-06 3.287382
Error 18.5 15 1.233333
Total 1238.5 23

a ) Since F0 = 179.5946

is in the rejection regionthere is sufficient evidence to conclude that mean percetage varies according to office diameter

b ) p valuje <0.01

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