Statistics Power Calculator – lesscss

By | September 3, 2019

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.

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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