Statistical Power
How to detect a false $H_0$?
- '’The power of a test’’ is the probability of making a correct decision (by rejecting the $H_0$) when the $H_0$ is false.
- The higher the power, the more sensitive the test in detecting the false hypothesis.
How to have higher power?
- the further the alternative value is away from the $H_0$, the higher the power
- A higher level of significance $\alpha$ gives higher power
- less variability - less power
- the larger the sample size - the greater the power
To determine the sample size needed for a study set $\alpha$ and the desired power, decide of the $H_A$, estimate $\sigma$ and calculate the sample size
Power Of Test
Consider this test:
- $H_0$: average blood pressure of employers is the same as national average,
- i.e. $H_0: \mu = 130$
- $H_A$: it’s different
- $H_A: \mu \ne 130$
Suppose that $H_A$ is actually true
- what is our chance to make Type II Errors? - i.e. fail to reject $H_0$ when we should reject it
Suppose that the actual average is 132: i.e. $\mu = 132$ we sample 100 individuals then the true sampling distribution of $\bar{x}$ is $N(132, 2.5)$ since $\text{SE} = \cfrac{25}{\sqrt{100}}$ what is the probability of successfully rejecting $H_0$?
We can divide it onto two probability questions:
-
what are possible values of $\bar{x}$ sufficient to reject $H_0$? (under $H_0$ ) - use this hypothetical Sampling Distribution to find the probability of observing such values of $\bar{x}$ (from the 1st step)
Step 1 The null distribution is $N(130, 2.5)$ the 2.5% tails are those with $Z = \pm 1.96$
$-1.96 = z_1 = \cfrac{x_1 - 130}{2.5} $x_1 = 125.1$
$+1.96 = z_2 = \cfrac{x_2 - 130}{2.5} $x_2 = 134.9$
Step 2
Now we compute the probability of rejecting $H_0$ if $\bar{x}$ actually came from $N(132, 2.5)$
$z_\text{left} = \cfrac{125.1 - 132}{2.5} = -2.76$ area: 0.003 $z_\text{right} = \cfrac{134.9 - 132}{2.5} = 1.16$ area: 0.123
so the probability of rejecting $H_0$ if the true mean is 132 is 0.004 + 0.123 = 0.126
This is the power of a test the probability of rejecting the $H_0$
The power varies depending on what we suppose the truth is
If the power of a test is 0.979, what’s the type 2 error rate? Type 2 error rate is the probability of failing to reject $H_0$ so type 2 error rate is 1 - 0.979 = 0.021
x = seq(120, 140, 0.1)
null.mu = 130; se = 2.5
null.y = dnorm(x, mean=null.mu, sd=se)
plot(x, null.y, type='l', lty=2, bty='n',
ylab='Probability')
x1 = 125.1; x2 = 134.9
abline(v=c(null.mu, x1, x2), lty=2)
real.mu = 132
real.y = dnorm(x, mean=real.mu, sd=se)
lines(x, real.y, col='red', lwd=2)
abline(v=real.mu, col='red', lwd=2)
x1.left = max(which(x <= x1))
polygon(x=x[c(1, 1:x1.left, x1.left)],
y=c(0, real.y[1:x1.left], 0),
col=adjustcolor('red', 0.5), border=NA)
x2.left = min(which(x >= x2))
x2.right = length(x)
polygon(x=x[c(x2.left, x2.left:x2.right, x2.right)],
y=c(0, real.y[x2.left:x2.right], 0),
col=adjustcolor('red', 0.5), border=NA)
Power Analysis
|- e.g. see here [http://stats.stackexchange.com/questions/108186]
- http://www.statmethods.net/stats/power.html - in R
- http://www.marketingdistillery.com/2014/08/10/multiple-abn-tests-in-marketing-with-anova-and-r/ - sample size
Visualization
- http://homepage.stat.uiowa.edu/~mbognar/applets/power.html