Answer)
As the population variance is unknown here we will use t distribution to estimate the interval
N = 13
Mean = 11.95
S.d = √153.76
Degrees of freedom is = n-1 = 12
For 12 dof and 95% confidence level
Critical.value t from t table is 2.18
Margin of error (MOE) = t*s.d/√n = 2.18*√153.76/√13 = 7.4932
Lower limit = mean - moe = 11.95 - 7.49
So answer is
u > 11.95 - 7.49
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