diff --git a/pycreep/gpr.py b/pycreep/gpr.py
index 423f2c5..bf35ecf 100644
--- a/pycreep/gpr.py
+++ b/pycreep/gpr.py
@@ -369,7 +369,8 @@ def predict_log_stress(self, time, temperature):
 
         return mean.numpy(), var.numpy()
 
-    def predict_stress(self, time, temperature, confidence=None):
+    # pylint: disable=unused-argument
+    def predict_stress(self, time, temperature, confidence=None, **kwargs):
         """
         Predict the stress at a given time and temperature
 
diff --git a/pycreep/ttp.py b/pycreep/ttp.py
index d40d282..8ae9796 100644
--- a/pycreep/ttp.py
+++ b/pycreep/ttp.py
@@ -253,7 +253,7 @@ def predict_time(self, stress, temperature, confidence=None):
         if confidence is None:
             h = 0.0
         else:
-            h = scipy.stats.norm.interval(confidence)[1]
+            h = np.sign(confidence) * scipy.stats.norm.interval(np.abs(confidence))[1]
 
         return 10.0 ** (
             self.time_sign
@@ -262,7 +262,9 @@ def predict_time(self, stress, temperature, confidence=None):
             )
         )
 
-    def predict_stress(self, time, temperature, confidence=None, root_bounds=None):
+    def predict_stress(
+        self, time, temperature, confidence=None, root_bounds=None, raise_on_error=True
+    ):
         """
         Predict new values of stress given time and temperature
 
@@ -275,6 +277,7 @@ def predict_stress(self, time, temperature, confidence=None, root_bounds=None):
                             average predictions
             root_bounds:    if not None, lower and upper bounds on which root value to use
                             when inverting the TTP polynomial
+            raise_on_error: if true throw an error if we can't invert.  if not return nan
         """
         # Will want these sorted
         if root_bounds is not None:
@@ -298,7 +301,9 @@ def solve_one(x):
             pi[-1] -= x
             rs = np.array(np.roots(pi))
             if np.all(np.abs(np.imag(rs)) > 0):
-                raise ValueError("Inverting relation to predict stress failed")
+                if raise_on_error:
+                    raise ValueError("Inverting relation to predict stress failed")
+                return np.nan
             rs[np.abs(np.imag(rs)) > 0] = 0
             rs = np.real(rs)
             # Need to consider this...
@@ -306,7 +311,9 @@ def solve_one(x):
                 return np.max(rs)
             val = np.logical_and(rs >= root_bounds[0], rs <= root_bounds[1])
             if np.all(np.logical_not(val)):
-                raise ValueError("No root falls within user provided bounds!")
+                if raise_on_error:
+                    raise ValueError("No root falls within user provided bounds!")
+                return np.nan
             return rs[val][0]
 
         # Solve each one, one at a time, for now