Consider an ordinary binary max-heap data structure with $n$ elements that supports insert and extract-max in $O(\log n)$ worst-case time.
Question: If extract max is $O(1)$ amortized does that mean worst case is $O(n)$?
My answer: No, A sequence of $n$ EXTRACT_MAX will cost $O(n\log n)$. This is because in a Heap the leaves are almost on the same level and the number of leaves in a Heap is $O(n/2) = O(n)$. Calling EXTRACT_MAX each time removes the maximum element from the root and replace it with a new second maximum element and decrement a leaf. Thus, decrementing all the leaves (which are of $O(n)$) will take $O(\log n)$ time each as all leaves are almost in the same level totally giving a time complexity of $O(n\log n)^{1/4}$.
Does this make sense?