Answer
With a significance level of .05, there is insufficient evidence to reject the null hypothesis.
Work Step by Step
The z value is the sample mean is decreased by mean and divided by the deviation.
Mean = $\frac{10.08+9.89+10.05+10.16+10.21+10.11}{6}$ = 10.0833
z = $\frac{x-u}{o/\sqrt n}$ = $\frac{10.0833-10.1}{0.1/\sqrt 6}$ = -0.41
P = .6818