Jon S. answered 12/31/20
Patient and Knowledgeable Math and English Tutor
this is a left-sided test at alpha = 0.05 and n=7, so the critical t value is for 6 degrees of freedom and the negative of the 0.05 tail value for that DOF = -1.943
the test statistic is tcalc = DBAR/(STD DEV/SQRT(n) = -7,1 / (5.9/sqrt(7)) = -3.18
since tcalc < tcrit, you would reject the null hypothesis at alpha = 0.05