Answer
There is not sufficient evidence to reject that the weights are consistent.
Work Step by Step
$H_{0}:\mu=0.8535.$ $H_{a}:\mu \ne0.8535.$ Hence the value of the test statistic: $\frac{\overline{x}-\mu}{s/\sqrt n}=\frac{0.8635-0.8535}{0.057/\sqrt{19}}=0.765.$ 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=19-1=18, hence P is more than 0.2. 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 $\alpha=0.05$, because it is more than 0.2, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to reject that the weights are consistent.
