Question

The time needed for college students to complete a certain puzzle is modelled using a normal distribution with a mean of 30 seconds and a standard deviation of 3 seconds. You wish to see if the
population mean time µ is changed by vigorous exercise, so you have a group of nine college students
exercise vigorously for 30 minutes and then complete the puzzle. What is the rejection region at the
significance level 0.01?
What is the power of the test at µ = 28 and σ = 1 in the above question?

Answer 1

As we are trying to test here whether the mean is changed from mean value of 30 seconds, therefore this is a two tailed test, and for 0.01 level of significance, we have from the standard normal tables that:

P( -2.576 < Z < 2.576) = 0.99

Therefore the rejection region here is given as: Z < -2.576 or Z > 2.576

For this the raw scores are computed as:

The standard error is computed as:

Mean - 2.576*Std Dev = 30 - 2.576 = 27.424
Mean + 2.576*Std Dev = 30 + 2.576 = 32.256

Therefore the rejection region here is given as:
X < 27.424 or X > 32.256

b) Power of the test is defined as the probability of rejecting a false null hypothesis.

Given that the true mean is 28 and standard deviation is 1, the power is the probability that the value lies in the above computed rejection region for this distribution. Therefore it is computed here as:

Converting it to a standard normal variable, we get here:

Getting these from standard normal tables, we get here:

Therefore 0.0420 is the required probability here.