Whenever a hypothesis test is conducted, we need to ascertain that test is of high qualitity. One way to check the power or sensitivity of a test is to compute the probability of test that it can reject the null hypothesis correctly when an alternate hypothesis is correct. In other words, power of a test is the probability of accepting the alternate hypothesis when it is true, where alternative hypothesis detects an effect in the statistical test.

$ {Power = \ P(\ reject\ H_0 | H_1 \ is \ true) } $

Power of a test is also test by checking the probability of Type I error($ { \alpha } $) and of Type II error($ { \beta } $) where Type I error represents the incorrect rejection of a valid null hypothesis whereas Type II error represents the incorrect retention of an invalid null hypothesis.

Lesser the chances of Type I or Type II error, more is the power of statistical test.

## Example

A survey has been conducted on students to check their IQ level. Suppose a random sample of 16 students is tested. The surveyor tests the null hypothesis that the IQ of student is 100 against the alternative hypothesis that the IQ of student is not 100, using a 0.05 level of significance and standard deviation of 16. What is the power of the hypothesis test if the true population mean were 116?

**Solution:**

As distribution of the test statistic under the null hypothesis follows a Student t-distribution. Here n is large, we can approximate the t-distribution by a normal distribution. As probability of committing Type I error($ { \alpha } $) is 0.05 , we can reject the null hypothesis ${H_0}$ when the test statistic $ { T \ge 1.645 } $. Let’s compute the value of sample mean using test statistics by following formula.

$ {T = \frac{ \bar X – \mu}{ \frac{\sigma}{\sqrt \mu}} \\[7pt]

\implies \bar X = \mu + T(\frac{\sigma}{\sqrt \mu}) \\[7pt]

\, = 100 + 1.645(\frac{16}{\sqrt {16}})\\[7pt]

\, = 106.58 } $

Let’s compute the power of statistical test by following formula.

$ {Power = P(\bar X \ge 106.58 \ where\ \mu = 116 ) \\[7pt]

\, = P( T \ge -2.36) \\[7pt]

\, = 1- P( T \lt -2.36 ) \\[7pt]

\, = 1 – 0.0091 \\[7pt]

\, = 0.9909

} $

So we have a 99.09% chance of rejecting the null hypothesis ${H_0: \mu = 100 } $ in favor of the alternative hypothesis $ {H_1: \mu \gt 100 } $ where unknown population mean is $ {\mu = 116 } $.

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