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Statistics – Power Calculator

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.

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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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