**8. Incorrect. The answer is
false, not true **

The test can be formulated as follows

H_{0}: μ=0.100

H_{a}: μ> 0.100

Under the H_{0}, t_{0}=(X_{bar}-μ_{0})(n)^{0.5}/s=(0.106-0.100)*16^{0.5}/0.012=2.0

At 0.05 level of significance the decision rule is to reject H_{0}
if t_{0.05,15} = 1.753 < t_{0},
where a=0.05 and there
are 16-1=15 degrees of freedom. Since t_{0}=2, we reject H_{0}
at the 5% level.

At 0.01 level of significance the decision rule is to reject H_{0}
if t_{0.01,15} = 2.602 < t_{0},
where a=0.05 and there are
16-1=15 degrees of freedom. Since t_{0}=2, we fail to reject H_{0}
at the 1% level.