How it started

How it ended

Our research question:

• Is there an association between the social class of passengers and their survival?

• Independent variable: social class (ordinal: 1st / 2nd / 3rd class)

• Dependent variable: survival (nominal: survived / died)

Loading our Titanic data

load("C:/Users/vxs15qru/OneDrive - University of East Anglia/PPLX7012A 19-20/labs/data/titanic.RData")
names(titanic)

##  [1] "row.names" "pclass"    "survived"  "name"      "age"       "embarked" 
##  [7] "home.dest" "room"      "ticket"    "boat"      "sex"

glimpse(titanic)

## Rows: 1,313
## Columns: 11
## $ row.names <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 1~
## $ pclass    <fct> 1st, 1st, 1st, 1st, 1st, 1st, 1st, 1st, 1st, 1st, 1st, 1st, ~
## $ survived  <int> 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, ~
## $ name      <fct> "Allen, Miss Elisabeth Walton", "Allison, Miss Helen Loraine~
## $ age       <dbl> 29.0000, 2.0000, 30.0000, 25.0000, 0.9167, 47.0000, 63.0000,~
## $ embarked  <fct> Southampton, Southampton, Southampton, Southampton, Southamp~
## $ home.dest <fct> "St Louis, MO", "Montreal, PQ / Chesterville, ON", "Montreal~
## $ room      <fct> B-5, C26, C26, C26, C22, E-12, D-7, A-36, C-101, , , , B-35,~
## $ ticket    <fct> 24160 L221, , , , , , 13502 L77, , , , 17754 L224 10s 6d, 17~
## $ boat      <fct> 2, , (135), , 11, 3, 10, , 2, (22), (124), 4, 9, B, , 6, , ,~
## $ sex       <fct> female, female, male, female, male, male, female, male, fema~

Our variables

titanic%>%
  count(pclass,survived)

##   pclass survived   n
## 1    1st        0 129
## 2    1st        1 193
## 3    2nd        0 161
## 4    2nd        1 119
## 5    3rd        0 574
## 6    3rd        1 137

Or, slightly nicer, using the tabyl() function from the janitor library

titanic %>%
  tabyl(pclass, survived)

##  pclass   0   1
##     1st 129 193
##     2nd 161 119
##     3rd 574 137

As you would report it:

The null hypothesis:

• Null hypothesis: there is no association between class and survival

• To reject the null: we have to show the numbers we have look very different than the no association scenario

• Two approaches:

  • show proportions (intuitive)

  • Calculate the expected value what we would observe if that would there was no association, and then:

  • Compare that value to observed data (less intuitive, but we do it all the time)

Proportions: chances of survival by class

titanic%>%
  group_by(pclass)%>%
  summarise(mean(survived))

## # A tibble: 3 x 2
##   pclass `mean(survived)`
##   <fct>             <dbl>
## 1 1st               0.599
## 2 2nd               0.425
## 3 3rd               0.193

This seems to provide evidence against the null. Make sure you understand why!

The independence / null world

• Remember that we always contrast the null with our research hypothesis

• But what exactly is the null?

• The null, in this case at least, implies Independence between the two variable

• This means that we can use the laws of probability to calculate the null world

• \(P(A \cap B) = P(A) \times P(B)\) or: the joint probability of independent events is calculated as the probability of event A multiplied by the probability of event B.

The expected value under the null

• Suppose you just had information on the marginal totals.

The expected value

\[\texttt{Expected value for a cell}= \frac{\texttt{Row total}\times \texttt{Column total}}{\texttt{Sample Size}}\]

• Or, in general:

\(\texttt{E}=\frac{\texttt{R}\times\texttt{C}}{\texttt{N}}\)

In our case (using the proportions we specified above):

The statistical test: the idea

• Chi-square (pron. “kai”, written \(\chi^2\))

• Proposed by Karl Pearson, who noticed that relying on normal distributions can lead to errors

• Allows us to summarize the differences between observed and expected

• We also know how the chi-square statistics are distributed (along the \(\chi^2\) distribution)

• so we can work out the p-value!

Building the formula

• Call O the observed number

• Call E the expected number

• We subtract E from O, square the results, divide by E and sum

\(\chi^2=\sum{\frac{(\texttt{O}-\texttt{E})^2}{\texttt{E}}}\)

Calculating: step 1 (0-E)

Calculating: step 2 (Squaring)

Calculating: step 3 (Dividing by E and summing)

• The \(\chi^2\) statistic is 173.8

• Is that statistically significant?

We check against the distribution:

The \(\chi^2\) distribution with \(df\) degrees of freedom is the distribution of the sums of the squares of \(df\) independent standard normal random variables…

So, we need to know the degrees of freedom for each test:

Degrees of freedom

• equals number of rows, minus one \(\times\) number of columns, minus one

• For a 3 by 2 table: 2 \(\times\) 1 equals 2

• Hence, two degrees of freedom

And the distribution for 2 degrees of freedom is:

Using Base R

chisq.test(with(titanic, table(pclass, survived)))

## 
##  Pearson's Chi-squared test
## 
## data:  with(titanic, table(pclass, survived))
## X-squared = 173.81, df = 2, p-value < 2.2e-16

Using tabyl()

my.table<-titanic %>%
    tabyl(pclass, survived)
chisq.test(my.table)

## 
##  Pearson's Chi-squared test
## 
## data:  my.table
## X-squared = 173.81, df = 2, p-value < 2.2e-16

Writing this up:

``To determine the degree of association between passenger class and survival, I carried out a chi-square test on the cross-tabulation of class and survival. The value of the chi-square statistic was 173.81 (d.f. = 2, N = 1313), and the corresponding p-value was less than 0.05, allowing us to reject the null hypothesis that there is no association.’’