The test statistic in a left tailed z-score is -1.25. Use the given information to find the p-value. State the conclusion about the null hypothesis

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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.

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