Answer
There is sufficient evidence to support that the mean wight of discarded plastic is more than 1.8.
Work Step by Step
$H_{0}:\mu=1.8$. $H_{a}:\mu>1.8.$ Hence the value of the test statistic: $\frac{\overline{x}-\mu}{s/\sqrt n}=\frac{1.911-1.8}{1.065/\sqrt{62}}=0.82.$ The P-value is the probability of z being bigger than 0.82 is 1 minus the probability of the z-score being less than 0.82, hence:P=1-0.7939=0.2061. 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 0.2061, hence we fail to reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the mean wight of discarded plastic is more than 1.8.
