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# Two-Sided Test

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15.54 Power of a two-sided test. Power calculations for two-sided tests follow the same outline as for one-sided tests. Look at the example with a test of H0: u = 128 versus Ha: u /= 128 at the 5% level of significance. The company medical director failed to find significant evdence that the mean blood pressure of a population of executives differed from the national mean u = 128. The data used were an SRS of size 72 from a population with standard deviation sigma = 15. The medical director now wonders if the test used would detect an important difference if one were present. What would be the power of this test against the alternative u = 134?
(a) The test in the example rejects H0 when |z| > 1.96. The test statistic is z = (xbar - 128)/(15/ sqrt72). Write the rule for rejecting H0 in terms of the values of xbar. (Because the test is two-sided, it rejects H0 when xbar is either too large or too small.)
(b) Now find the probability that xbar takes values that lead to rejecting H0 if the true mean is u = 134. This probability is the power.
(c) What is the probability that this test makes a Type II error when u = 134?

15.55 Power of a two-sided test, continued. Let's now use technology to perform the power calculations for the two-sided test of the previous exercise.
(a) Use the Power of a Test applet to find the power of the two-sided test against the alternative u = 134. Make sure that it is similar to what you calculated in the previous exercise.
(b) Use the applet to calculate the power of the test against the alternative u = 122. Can the test be relied on to detect a mean that differs from 128 by 6?
(c) If the alternative were further from H0, say u = 136, would the power be higher or lower than the values calculated in (a) and (b)?

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The expert examines a two-sided test for power calculations. Examples are determined for significant evidence.

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15.54.
(a) (x-128)/(15/sqrt(72))=±1.96
X=131.5 or 124.5
When x bar is smaller than 124.5 or bigger than 131.5, we could reject the null hypothesis.

(b)z=(134-128)/(15/sqrt(72))=3.39
From the z table, the probability is 0.000362

(c) we ...

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