library(knitr) # creating slides

library(ggplot2) # making plots

library(reshape2) # handling data frames

Where (X’-30)/(\(s\)/\(\sqrt(n)\)) measures the number of standard

errors the sample mean is from the mean hypothesized by \(H_0\)

and the denominator, (\(s\)/\(\sqrt(n)\)) is the standard error of

the sample mean.

:: flip the following reasoning

:: look at the right tail

Power is the probability of rejecting the NULL HYPOTHESIS

\(H_0\), when it is false.

:: Used to determine if your sample size was big enough

to yield a meaningful, rather than random result

:: Detect if your ALTERNATIVE HYPOTHESIS, \(H_a\) is true,

to lower the risk of a Type II errors.

As beta, \(\beta\), is the probability of a Type II error,

for accepting a false null hypothesis, then the complement

of this the power is (1-\(\beta\)).

\[ P = (1-\beta) \]

:: As \(\mu_a\) gets bigger, the test gets more powerful

:: As \(n\) gets bigger, the test gets more powerful

:: Power decreases with increases in variation

:: As \(\alpha\) increases, power increases

:: In a one sided test the power is greater as \(\alpha\)

is bigger than \(\alpha\)/2

Power is the probability that the true mean \(\mu\) is greater

than the (1-\(\alpha\)) quantile, in our sleep example:

:: if \(\alpha\) = .05 and as \(H_a\) \(\mu_a\) > 30

:: for normal distributions of which we know the variances

:: we make p = qnorm(.95) our reference

If a test statistic fell in the shaded portion, 5% of the area under the curve, we would reject \(H_0\) in favor \(H_a\).

The two hypotheses, \(H_0\) and \(H_a\), represent two distributions

:: since they’re talking about means or

:: centers of distributions.

For a random variable X which is distributed as Normal

with a mean mu, \(\mu\) and variance sigma squared, \(\sigma^2\):

:: under \(H_0\), X’ is N(\(\mu_0\) , \(\sigma^2\)/n)

:: under \(H_a\), X’ is N(\(\mu_a\) , \(\sigma^2\)/n)

The distribution represented by \(H_a\) will moved to the right,

most of the blue curve is to the right of the vertical line,

indicating that with \(\mu_a\) = 34, the test is more powerful,

so it is

:: correct to reject the \(H_0\) as it appears to be false.

The distribution represented by \(H_a\), in the above graph

moved under the blue curve, indicating that with

:: \(\mu_a\) = 30 = \(\mu_0\) , the test,

:: \(H_a\) is almost as powerful as \(H_0\), so it is

:: incorrect to reject the null hypothesis since it does

not appear to be false.

The distribution represented by \(H_a\) will move to the left of

\(\mu_0\) = 30, the area under the blue curve is less than the

5% our \(\alpha\), so the test is not only less powerful, it even

contradicts \(H_a\)

:: it is therefore not worth looking into.

Power is a function that depends on a specific value of an

alternative mean, \(\mu_a\), which is any value greater than

\(\mu_0\), the mean hypothesized by \(H_0\)

:: Recall that H_a specified mu > 30, in the sleep case.

If \(\mu_a\) is much bigger than \(\mu_0\) = 30 then the power

a probability, is bigger than if \(\mu_a\) is close to 30.

:: As \(\mu_a\) approaches 30, the mean under \(H_0\), the power

approaches \(\alpha\).

# mean=30

z <- qnorm(.95)

pnorm(30+z,mean=30,lower.tail=FALSE) 

## [1] 0.05

With the mean set to \(\mu_0\) the two distributions,

null and alternative, are the same and power = \(\alpha\) = 5%.

#mean=32

z <- qnorm(.95)

pnorm(30+z, mean=32,lower.tail=FALSE) 

## [1] 0.63876

With \(\mu_a\) > \(\mu_0\)mu_a the power is greater than \(\alpha\), at 64%.

:: When the sample mean is many standard errors greater than

the mean hypothesized by the null hypothesis,

:: the probability of rejecting \(H_0\) is false is much higher.

# sd=1

z <- qnorm(.95)

pnorm(30+z, mean=32, sd= 1, lower.tail=FALSE) 

## [1] 0.63876

# sd=2

z <- qnorm(.95)

pnorm(30+z, mean=32, sd= 2, lower.tail=FALSE) 

## [1] 0.5704709

:: Power decreases with increases in variation

:: As \(\alpha\) increases, power increases

:: In a one sided test the power is greater as \(\alpha\) is bigger than \(\alpha\)/2

Assumed that X’~ N(\(\mu_a\), \(\sigma^2/n\))

Supposed that \(H_a\) says that \(\mu\) > \(\mu_0\)

:: As power = 1 - beta

= Prob ( X’ > \(\mu_0\) + z_(1-\(\alpha\)) * \(\sigma\)/\(\sqrt(n)\))

:: the population mean equals \(\mu_0\)

:: the level size of the test is \(\alpha\)

:: But are given \(\mu_a\) > \(\mu_0\), \(\mu_0\) and \(\alpha\)

:: and X’ depends on the data.

So we can solve for the quantities \(\beta\), \(\sigma\), \(n\),and \(\mu_a\)

:: As power = 1 - beta

= Prob ( X’ > \(\mu_0\) + z_(1-\(\alpha\)) * \(\sigma\)/\(\sqrt(n)\))

:: only \(\sqrt(n)\) * \(\frac{\mu_a - \mu_0}{\sigma}\) is needed

:: this is the effect size

:: t quantile instead of the z

Power is still a probability, namely

:: \(\frac{P(X' - \mu_0)}{(S /\sqrt(n)}\) > t_(1-\(\alpha\), \(n\)-1)

for \(H_a\) that \(\mu\) > \(mu_a\)

:: the proposed distribution is not centered at \(mu_0\),

:: so we have to use the non-central t distribution.

:: omit one of the arguments and power.t.test()

solves for it

power.t.test(n = 16,delta = 2 / 4, sd=1, type = "one.sample", 
             alt = "one.sided")$power

## [1] 0.6040329

power.t.test(n = 16, delta = 2 , sd=4, type = "one.sample", 
             alt ="one.sided")$power

## [1] 0.6040329

power.t.test(n = 16, delta = 100 , sd=200, type = "one.sample", 
             alt ="one.sided")$power

## [1] 0.6040329

:: keeping the effect size (the ratio delta/sd) constant preserved the power

:: get a constant effect size with

with the same values for n (16) & alpha (.05), but

different delta and standard deviation values

:: specify a power we want and solve for the sample size \(n\).

power.t.test(power = .8, delta = 2 / 4, sd=1, type = "one.sample", 
             alt = "one.sided")$n

## [1] 26.13751

power.t.test(power = .8, delta = 2, sd=4, type = "one.sample", 
             alt = "one.sided")$n

## [1] 26.13751

power.t.test(power = .8, delta = 100, sd=200, type = "one.sample", 
             alt = "one.sided")$n

## [1] 26.13751

# use power.t.test to find delta for a power=.8 and n=26 and sd=1

power.t.test(power = .8, n=26, sd=1, type = "one.sample",  
             alt = "one.sided")$delta

## [1] 0.5013986

#use power.t.test to find delta for a power=.8 and n=27 and sd=1

power.t.test(power = .8, n=27, sd=1, type = "one.sample",  
             alt = "one.sided")$delta

## [1] 0.4914855

#use power.t.test to find delta for a power=.8 and n=26 and sd=2

power.t.test(power = .8, n=26, sd=2, type = "one.sample",  
             alt = "one.sided")$delta

## [1] 1.002797