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SSTR = 6,750 H 0 : 1= 2= 3= 4 SSE = 8,000 H a : at least one mean is different n T =

SSTR = 6,750

H0: 1=2=3=4

SSE = 8,000

Ha: at least one mean is different

nT = 20

The null hypothesis is to be tested at the 5% level of significance. The p-value is

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