Better Students Ask More Questions.
The test statistic in a left tailed z-score is -1.25. Use the given information to find...
1 Answer | add yours
When the z-score is given, the p-value can be estimated using the formula `int_(-oo)^z int 1/sqrt 2*pi)*e^(u^2/2) du`
For a z-score of -1.25, the corresponding p-value is 0.1056
This is a significantly high p-value and the null hypothesis can be accepted based on this value.
Posted by justaguide on June 5, 2012 at 5:39 AM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.