When testing vs. , the observed value of the z-score was found to be -2.15. Then, the p-value for this test would be
A) .0158
B) .0316
C) .9842
D) .9684

Question

When testing  vs.  , the observed value of the z-score was found to be -2.15. Then, the p-value for this test would be
A) .0158 
B) .0316 
C) .9842 
D) .9684

Quizplus · Accepted Answer

The difference between the two strings (11eaa8e1_5384_eec2_93a6_2b181f5a2311_TB4382_11 and 11eaa8e1_5384_eec3_93a6_01376e915e61_TB4382_11) is in the third to the last position, where one has "c2" and the other has "c3". The observed z-score is 2.15, indicating that the two strings are significantly different. Since this is a two-tailed test, we need to find the area under the normal distribution curve in both tails. Using a standard normal distribution table, we find that the area to the right of z = 2.15 is 0.0158. Therefore, the p-value is 2*0.0158 = 0.0316.

When Testing Vs