Answer
There is not sufficient evidence to support that the mean lead concentration is less than 14.
Work Step by Step
$H_{0}:\mu=14$. $H_{a}:\mu <14.$ Hence the value of the test statistic: $\frac{\overline{x}-\mu}{s/\sqrt n}=\frac{11.05-14}{6.4612/\sqrt{10}}=-1.444.$ The P-value is the interval of probabilities between which the value of the test-statistic lies in the table with degree of freedom=sample size-1=10-1=9, hence P is between 0.05 and 0.1. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.05$, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to support that the mean lead concentration is less than 14.
