Concept of hypothesis

A hypothesis is a statement about a given population. Usually it is stated as a population parameter having a given value or being in a certain interval.
Examples:
- Quality control of products: The hypothesis is that the products e.g. have a certain weight, a given power consumption or a minimal durability.
- Scientific hypothesis: There is no dependence between a company’s age and level of return.

Significance test

A significance test is used to investigate, whether data is contradicting the hypothesis or not.
If the hypothesis says that a parameter has a certain value, then the test should tell whether the sample estimate is “far” away from this value.
For example:
- Waiting times in a queue. We sample $n$ customers and count how many that have been waiting more than 5 minutes. The company policy is that at most $10\%$ of the customers should wait more than 5 minutes. In a sample of size $n=32$ we observe 4 with waiting time above 5 minutes, i.e. the estimated proportion is $\hat{\pi} = \frac{4}{32} = 12.5\%$. Is this “much more” than (i.e. significantly different from) $10\%$?
- The blood alcohol level of a student is measured 4 times with the values $0.504,0.500,0.512,0.524$, i.e. the estimated mean value is $\bar{y}=0.51$. Is this “much different” than a limit of $0.5$?

Null and alternative hypothesis

The null hypothesis - denoted $H_0$ - usually specifies that a population parameter has some given value. E.g. if $\mu$ is the mean blood alcohol level we can state the null hypothesis
- $H_0 : \mu = 0.5$.
The alternative hypothesis - denoted $H_a$ - usually specifies that the population parameter is contained in a given set of values different than the null hypothesis. E.g. if $\mu$ again is the population mean of a blood alcohol level measurement, then
- the null hypothesis is $H_0 : \mu = 0.5$
- the alternative hypothesis is $H_a : \mu \neq 0.5$.

Test statistic

We consider a population parameter $\mu$ and write the null hypothesis \[ H_0:\mu = \mu_0, \] where $\mu_0$ is a known number, e.g. $\mu_0 = 0.5$.
Based on a sample we have an estimate $\hat{\mu}$.
A test statistic $T$ will typically depend on $\hat{\mu}$ and $\mu_0$ (we may write this as $T(\hat{\mu}, \mu_0)$) and measure “how far from $\mu_0$ is $\hat{\mu}$?”
Often we use $T(\hat{\mu},\mu_0)$ = “the number of standard deviations from $\hat{\mu}$ to $\mu_0$”.
For example it would be very unlikely to be more than 3 standard deviations from $\mu_0$, i.e. in that case $\mu_0$ is probably not the correct value of the population parameter.

$P$-value

We consider
- $H_0$: a null hypothesis.
- $H_a$: an alternative hypothesis.
- $T$: a test statistic, where the value calculated based on the current sample is denoted $t_{obs}$.
To investigate the plausibility of $H_0$, we measure the evidence against $H_0$ by the so-called $p$-value:
- The $p$-value is the probability of observing a more extreme value of $T$ (if we were to repeat the experiment) than $t_{obs}$ under the assumption that $H_0$ is true.
- “Extremity” is measured relative to the alternative hypothesis; a value is considered extreme if it is “far from” $H_0$ and “closer to” $H_a$.
- If the $p$-value is small then there is a small probability of observing $t_{obs}$ if $H_0$ is true, and thus $H_0$ is not very probable for our sample and we put more support in $H_a$, so:
\[ \textbf{The smaller the $p$-value, the less we trust $H_0$.} \]
What is a small $p$-value? If it is below $5\%$ we say it is significant at the $5\%$ level.

Significance level

We consider
- $H_0$: a null hypothesis.
- $H_a$: an alternative hypothesis.
- $T$: a test statistic, where the value calculated based on the current sample is denoted $t_{obs}$ and the corresponding $p$-value is $p_{obs}$.
Small values of $p_{obs}$ are critical for $H_0$.
In practice it can be necessary to decide whether or not we are going to reject $H_0$.
The decision can be made if we previously have decided on a so-called $\alpha$-level, where
- $\alpha$ is a given percentage
- we reject $H_0$, if $p_\text{obs}$ is less than or equal to $\alpha$
- $\alpha$ is called the significance level of the test
- typical choices of $\alpha$ are $5\%$ or $1\%$.

Significance test for mean

Two-sided $t$-test for mean:

We assume that data is a sample from $\texttt{norm}(\mu,\sigma)$.
The estimates of the population parameters are $\hat{\mu}=\bar{y}$ and $\hat{\sigma}=s$ based on $n$ observations.
Null hypothesis: $H_0:\ \mu = \mu_0$, where $\mu_0$ is a known value.
Two-sided alternative hypothesis: $H_a:\ \mu \neq \mu_0$.
Observed test statistic: $t_{obs} = \frac{\bar{y} - \mu_0}{se}$, where $se = \frac{s}{\sqrt{n}}$.
I.e. $t_{obs}$ measures, how many standard deviations (with $\pm$ sign) the empirical mean lies away from $\mu_0$.
If $H_0$ is true, then $t_{obs}$ is an observation from the $t$-distribution with $df = n - 1$.
$P$-value: Values bigger than $|t_{obs}|$ or less than $-|t_{obs}|$ puts more support in $H_a$ than $H_0$.
The $p$-value = 2 x “upper tail probability of $|t_{obs}|$”. The probability is calculated in the $t$-distribution with $df$ degrees of freedom.

Example: Two-sided $t$-test

Blood alcohol level measurements: $0.504, 0.500, 0.512, 0.524$.
These are assumed to be a sample from a normal distribution.
We calculate
- $\bar{y} = 0.51$ and $s = 0.0106$
- $se = \frac{s}{\sqrt{n}} = \frac{0.0106}{\sqrt{4}} = 0.0053$.
- $H_0: \mu = 0.5$, i.e. $\mu_0 = 0.5$.
- $t_{obs} = \frac{\bar{y}-\mu_0}{se} = \frac{0.51-0.5}{0.0053} = 1.89$.
So we are almost 2 standard deviations from $0.5$. Is this extreme in a $t$-distribution with 3 degrees of freedom?

library(mosaic)
1 - pdist("t", q = 1.89, df = 3)

## [1] 0.07757725

The $p$-value is 2$\cdot$ 0.078, i.e. more than 15%. On the basis of this we do not reject $H_0$.

One-sided $t$-test for mean

The book also discusses one-sided $t$-tests for the mean, but we will not use those in the course.

Agresti: Overview of $t$-test

Significance test for proportion

Consider a sample of size $n$, where we observe whether a given property is present or not.
The relative frequency of the property in the population is $\pi$, which is estimated by $\hat{\pi}$.
Null hypothesis: $H_0: \pi = \pi_0$, where $\pi_0$ is a known number.
Two-sided alternative hypothesis: $H_a: \pi\neq\pi_0$.
If $H_0$ is true the standard error for $\hat{\pi}$ is given by $se_0 = \sqrt{\frac{\pi_0(1-\pi_0)}{n}}$.
Observed test statistic: $z_{obs} = \frac{\hat{\pi}-\pi_0}{se_0}$
I.e. $z_{obs}$ measures, how many standard deviations (with $\pm$ sign) there is from $\hat{\pi}$ to $\pi_0$.

Approximate test

If both $n\hat{\pi}$ and $n(1 - \hat{\pi})$ are larger than 15 we know from previously that $\hat{\pi}$ follows a normal distribution (approximately), i.e.
- If $H_0$ is true, then $z_{obs}$ is an observation from the standard normal distribution.
$P$-value for two-sided test: Values greater than $|z_{obs}|$ or less than $-|z_{obs}|$ point more towards $H_a$ than $H_0$.
The $p$-value=2 x “upper tail probability for $|z_{obs}|$”. The probability is calculated in the standard normal distribution.

Example: Approximate test

We consider a study from Florida Poll 2006:
- In connection with problems financing public service a random sample of 1200 individuals were asked whether they preferred less service or tax increases.
- 52% preferred tax increases. Is this enough to say that the proportion is significantly different from fifty-fifty?
Sample with $n = 1200$ observations and estimated proportion $\hat{\pi} = 0.52$.
Null hypothesis $H_0: \pi = 0.5$.
Alternative hypothesis $H_a: \pi\neq 0.5$.
Standard error $se_0 = \sqrt{\frac{\pi_0(1-\pi_0)}{n}} = \sqrt{\frac{0.5\times0.5}{1200}} = 0.0144$
Observed test statistic $z_{obs} = \frac{\hat{\pi}-\pi_0}{se_0}=\frac{0.52-0.5}{0.0144}=1.39$
“upper tail probability for 1.39” in the standard normal distribution is 0.0823, i.e. we have a $p$-value of 2$\cdot$ 0.0823$\approx$ 16%.
Conclusion: There is not sufficient evidence to reject $H_0$, i.e. we do not reject that the preference in the population is fifty-fifty.
Note, the above calculations can also be performed automatically in R by (a little different results due to rounding errors in the manual calculation):

count <- 1200 * 0.52 # number of individuals preferring tax increase
prop.test(x = count, n = 1200, correct = F)

## 
##  1-sample proportions test without continuity correction
## 
## data:  count out of 1200
## X-squared = 1.92, df = 1, p-value = 0.1659
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.4917142 0.5481581
## sample estimates:
##    p 
## 0.52

Binomial (exact) test

Consider again a sample of size $n$, where we observe whether a given property is present or not.
The relative frequency of the property in the population is $\pi$, which is estimated by $\hat{\pi}$.
Let $y_+=n\hat{\pi}$ be the frequency (total count) of the property in the sample.
It can be shown that $y_+$ follows the binomial distribution with size parameter $n$ and success probability $\pi$. We use $Bin(n,\pi)$ to denote this distribution.
Null hypothesis: $H_0: \pi=\pi_0$, where $\pi_0$ is a known number.
Alternative hypothesis: $H_a: \pi \neq \pi_0$, where $\pi_0$ is a known number.
$P$-value for two-sided binomial test:
- If $y_+\geq n\pi_0$: 2 x “upper tail probability for $y_+$” in the $Bin(n,\pi_0)$ distribution.
- If $y_+< n\pi_0$: 2 x “lower tail probability for $y_+$” in the $Bin(n,\pi_0)$ distribution.

Example: Binomial test

Experiment with $n=30$, where we have $y_+=14$ successes.
We want to test $H_0:\pi=0.3$ vs. $H_a:\pi\not=0.3$.
Since $y_+>n\pi_0=9$ we use the upper tail probability corresponding to the sum of the height of the red lines to the right of 14 in the graph below. (Notice, the graph continues on the right hand side to $n=30$, but it has been cut off for illustrative purposes.)
The upper tail probability from 14 and up (i.e. greater than 13) is:

lower_tail <- pdist("binom", q = 13, size = 30, prob = 0.3)

1 - lower_tail

## [1] 0.04005255

The two-sided $p$-value is then 2 x 0.04 = 0.08.

Binomial test in R

We return to the Chile data, where we again look at the variable sex.
Let us test whether the proportion of females is different from 50 %, i.e., we look at $H_0:\ \pi=0.5$ and $H_a:\ \pi \neq 0.5$, where $\pi$ is the unknown population proportion of females.

Chile <- read.delim("https://asta.math.aau.dk/datasets?file=Chile.txt")
binom.test( ~ sex, data = Chile, p = 0.5, conf.level = 0.95)

## 
## 
## 
## data:  Chile$sex  [with success = F]
## number of successes = 1379, number of trials = 2700, p-value =
## 0.2727
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
##  0.4916971 0.5297610
## sample estimates:
## probability of success 
##              0.5107407

The $p$-value for the binomial exact test is $27\%$, so there is no significant difference between the proportion of males and females.
The approximate test has a $p$-value of $26\%$, which can be calculated by the command

prop.test( ~ sex, data = Chile, p = 0.5, conf.level = 0.95, correct = FALSE)

(note the additional argument correct = FALSE).

Hypothesis test

Statistical inference: Hypothesis and test

Concept of hypothesis

Significance test

Null and alternative hypothesis

Test statistic

\(P\)-value

Significance level

Significance test for mean

Two-sided \(t\)-test for mean:

Example: Two-sided \(t\)-test

One-sided \(t\)-test for mean

Agresti: Overview of \(t\)-test

Significance test for proportion

Approximate test

Example: Approximate test

Binomial (exact) test

Example: Binomial test

Binomial test in R

Agresti: Overview of tests for mean and proportion