Question

Nicolas was testing H_0: \mu=24H 0 ​ :μ=24H, start subscript, 0, end subscript, colon, mu, equals, 24 versus H_\text{a}: \mu\neq24H a ​ :μ  ​ =24H, start subscript, start text, a, end text, end subscript, colon, mu, does not equal, 24 with a sample of 121212 observations. His test statistic was t=-1.79t=−1.79t, equals, minus, 1, point, 79. Assume that the conditions for inference were met. What is the approximate P-value for Nicolas' test?

Answer 1

Considering the hypotheses tested, it is found that the p-value for Nicolas's test is of 0.101.

What are the hypotheses tested?

At the null hypotheses, it is tested if the mean is of 24, that is:

$H_0: \mu = 24$

At the alternative hypotheses, it is tested if it is different, hence:

$H_1: \mu \neq 24$

As we are testing if the mean is different of a value, we have a two-tailed test, with t = -1.79 and 12 - 1 = 11 df. Hence, using a t-distribution calculator, the p-value is of 0.101.

More can be learned about p-values at brainly.com/question/26454209

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