A human liver has an average mass of 1.5kg with a standard deviation of 0.2kg.
How this works in R
dnorm(x =1.0, mean =1.5, sd =0.2)
[1] 0.0876415
dnorm(x =1.2, mean =1.5, sd =0.2)
[1] 0.647588
dnorm(x =1.4, mean =1.5, sd =0.2)
[1] 1.760327
dnorm(x =1.6, mean =1.5, sd =0.2)
[1] 1.760327
dnorm(x =1.8, mean =1.5, sd =0.2)
[1] 0.647588
dnorm(x =2.0, mean =1.5, sd =0.2)
[1] 0.0876415
dnorm(), pnorm(), qnorm(), and rnorm()
dnorm() calculates the relative probability of a specific outcome value
pnorm() calculates the sum of probabilities less than (or greater than), an outcome value
this is the cumulative probability
adds up a whole bunch of dnorm()’s
qnorm() calculates the value below which there is a certain probability
like the inverse function of pnorm()
give it a probability and it works out the value
rnorm() generates random value from the normal distribution
pnorm()
pnorm(q =seq(0, 3, by =0.01), mean =1.5, sd =0.2, lower.tail =TRUE)
pnorm(q =seq(0, 3, by =0.01), mean =1.5, sd =0.2, lower.tail =FALSE)
lower.tail = TRUE means values less than
lower.tail = FALSE means values more than
because this is a continuous distribution we never have to worry about the pesky -1
Examples
The concentration of \(PM_{2.5}\) in air has an average value of 12 \(\mu g/m^{3}\) with a standard deviation of 5 \(\mu g/m^{3}\). What is the probability that a given day will exceed the recommended 25\(\mu g/m^{3}\)?
pnorm(q = 25, mean =12, sd =5, lower.tail = FALSE)
gives 0.47%
Using the values above, what is the highest value of \(PM_{2.5}\) we can expect to see once a year?
qnorm(p = 1/365.25, mean =12, sd =5, lower.tail = FALSE)
gives 25.9\(\mu g/m^{3}\)
rnorm()
this gives a random number from the normal distribution
a random sample of one person’s liver mass
command is rnorm(n = 1, mean = 1.5, sd = 0.2)
n = 1 means just one sample
Confidence Intervals
we rarely know population statistics about our data
more common to have to infer them from sample statistics
if we don’t know the real standard deviation
estimate it from the sample
replace the normal distribution by its little cousin, the t-distribution
t-distribution is more spread out (more uncertain) than the normal
as sample size gets bigger the t converges to the normal
Confidence Intervals
imagine taking a bunch of different samples from a population
each one will have a slightly different mean
for one sample, its mean is our best guess as to the true population mean
but also need to express our degree of certainty (or otherwise) in this guess
this gives the confidence interval
Confidence Interval Imagined
Confidence Intervals in R
function t.test() is handy for confidence interval calculation
in fact it’s a bit of a Swiss army knife of conf. int., hypothesis testing….
give it a set of values and it’ll output the confidence interval for their mean
t.test(airquality$Wind)
t.test(airquality$Wind)$conf.int
One Sample t-test
data: airquality$Wind
t = 34.961, df = 152, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
9.394804 10.520229
sample estimates:
mean of x
9.957516
Hypothesis Testing
do two samples have difference means?
consider gapminder dataset
was life expectancy in 2007 higher in Southern Europe than Northern Europe?
z <- dslabs::gapminder |>filter(year ==2007, region %in%c("Northern Europe", "Southern Europe"))t.test(life_expectancy ~ region, data = z)
Welch Two Sample t-test
data: life_expectancy by region
t = -0.10871, df = 14.573, p-value = 0.9149
alternative hypothesis: true difference in means between group Northern Europe and group Southern Europe is not equal to 0
95 percent confidence interval:
-3.374106 3.047439
sample estimates:
mean in group Northern Europe mean in group Southern Europe
77.67000 77.83333
Paired T-tests
Ten mice were placed on a high fat diet. Their weights were recorded before and afterwards. Is there a significant weight gain?
mouse
before
after
1
265.6
274.6
2
84.4
100.2
3
196.7
206.0
4
187.7
194.0
5
199.0
203.9
6
235.7
247.9
7
259.2
262.9
8
172.3
172.8
9
188.2
198.1
10
241.1
248.0
Graphically
With Paired t-test
t.test(x = before, y = after)
Welch Two Sample t-test
data: before and after
t = -0.33694, df = 17.986, p-value = 0.7401
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-56.79972 41.09972
sample estimates:
mean of x mean of y
202.99 210.84
t.test(x = before, y = after, paired =TRUE)
Paired t-test
data: before and after
t = -5.6619, df = 9, p-value = 0.0003089
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
-10.986396 -4.713604
sample estimates:
mean difference
-7.85
Human gestation has an average period of 266 days with a standard deviation of 16 days. What is the probability that a baby will arrive exactly on their due date? Assume normal distribution. [2.5%]
Make a 95% confidence interval for animal weight for the bisons recorded in the knz_bison dataset. [742.16 - 756.74]
Perform a hypothesis test that female and male bisons are the same weight. [ \(p_{value} = 7 \times 10^{-5} << 0.05\)]
Assignments - Week Two
Complete week two moodle quiz
Complete swirl() exercises
install.packages("swirl")
library(swirl)
install_course("Statistical Inference")
swirl()
choose course Statistical Inference
do the three exercises 1 (Introduction), 9 (T Confidence Intervals), and 10 (Hypothesis Testing)
email a screen shot of the end of each lesson to eugene.hickey@associate.atu.ie
