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When the z-score is given, the p-value can be estimated using the formula `int_(-oo)^z 1/sqrt (2*pi)*e^(-u^2/2) du`
For a z-score of 2.5, the corresponding p-value is 0.0062
This is a very small value less than 0.01 or even 0.05 and it usually leads to the rejection of the null hypothesis.
Due to the extremely small p-value the null hypothesis is rejected.
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