W3: Tests of Association

Question 1

Another possibility is that response we are interested in is in the form of a count, examples of count variables are (2)

Answer 1

• number of students with graduate jobs
• number of individuals passing a test

Question 2

One question of interest is if the counts change (Y) based on

Answer 2

one or more binary covariates (X)

Question 3

We are interested whether there is an association between

Answer 3

a binary factor and their response –> not regression

Question 4

Suppose we are interested in association of binary factor and response, summarise in table below:

If there is no association between binary factor (e.g., degree = psych, enginerring) and response (yes/no to having a grad job) then expect

Answer 4

proportion of a/g to be similar to the proportion b/h

Question 5

Formally test the contigency table of whether proportion are similar or not using

Answer 5

Chi-squared tets for independence

Question 6

The null hypothesis for this test is that

Answer 6

There is no association between the factor and response

Question 7

Alternative hypothesis is that

Answer 7

There is an association between factor and response

Question 8

Example of null and alternate hypothesis (2) using this table:

Chi squared

Answer 8

HO: No association between region and access to services
H1: Association between region and access to services

Question 9

Calculate expected frequencies for each cell of the contigency table assuming no association (H0 is true) using formula:

Answer 9

Question 10

Whats column and whats row?

Chi squared

Answer 10

Question 11

How to calculate overall total of a table? - for example in this table?

Answer 11

10 + 5 + 2 +27 + 25 + 8 = 77

Question 12

How to note down E while doing it for each cell?

Answer 12

ETL
EBL
EBR
ETR
EMR
T = top
B = Bottom
L = Left
R = Right
M = Middle

Question 13

After hypothesis and expected frequencies, we calculate the test statistic
Formula is:

Answer 13

We do it for each cell in table for observed (O) then expected frequencies we just calculated and add it together

Question 14

Chi-squared test formula still holds the same when response and/or factorrs have

Answer 14

more than two levels

Question 15

After calculating x^2 chi-squared statistic, we calculate degrees of freedoom by:

Answer 15

R = no. of rows in contigency actual table
C = no. of columns in contigency actual table.

Question 16

Calculate DF for this table

Chi squared

Answer 16

r = 3
c = 2
(3 - 1) * (2 - 1) = 2 * 1 = 2

Question 17

For tables with 2 rows and 2 columns we use the

Chi squared

Answer 17

x2/1 disturbition

Question 18

The DF calculated corresponds to what p value you look at:
If x2/1 disturbition then look at

Chi squared

Answer 18

Question 19

What do we say before calculating expected frequencies?

Chi squared

Answer 19

We calculate the expected frequencies

Question 20

What do we say before stating hypotheses

Chi squared

Answer 20

We test:
List HO/H1

Question 21

What do we say before calculating test statistic

Answer 21

We calculate the test statistic as

Question 22

After calculating test statistic and then calculating DF we say:

Chi squared

Answer 22

We compare this x^2 with the x2/1 disturbition

Question 23

What table of critical sig levels do we list to compare x^2 value?

Chi square

Answer 23

10%
5%
1%
0.1%

Question 24

If x^2 larger than 0.1% value then

Answer 24

p <0.001

Question 25

Example of a conclusion of chi-squared test being significant at 0.1% level

Answer 25

We see p < 0.001 and so we reject the H0 at the 0.1% level. We have very strong evidence of an association between personality and degree.

Question 26

List of concluding statements - p-value greater than 10%

Answer 26

The p-value is greater than 10% so there is no evidence against the null hypothesis H0 and we do no reject it.

Question 27

List of concluding statements - p-value less than 10%

Answer 27

The p-value is less than 10% but greater than 5% so there may be slight evidence against the null hypothesis H0 but we do no reject it.

Question 28

List of concluding statements - p-value less 5%

Answer 28

he p-value is less than 5% but greater than 1% so there is moderate evidence against the null hypothesis H0 and we reject it and accept the alternate hypothesis H1.

Question 29

List of concluding statements - p-value less 1%

Answer 29

The p-value is less than 1% but greater than 0.1% so there is strong evidence against the null hypothesis H0 and we reject it and accept the alternate hypothesis H1.

Question 30

List of concluding statements - p-value less 0.1% level

Answer 30

The p-value is less than 0.1% so there is very strong evidence against the null hypothesis H0 and we reject it and accept the alternate hypothesis H1.

Question 31

Assumptions of Chi-squared test - (2)

Answer 31

• We require independent observations
• For 2x2 table, all expected frequencies must be larger than 5.
  For a larger table, no more than 20% of all cells may have an expected frequency less than 5 and all expected frequencies must be larger than 1

Question 32

In assumptions of chi-squared , we should always check the

Answer 32

second assumption

Question 33

An alternative test to Chi-squared test is

Answer 33

Fisher’s exact test

Question 34

Why is Fisher’s exact test a suitable alternative test to chi squared

Answer 34

This is based on an exact p-value rather than asymptotic properties
of the test statistic, and so is approriate even when there are small
counts in some cells of the contingency table.

Question 35

When is Fisher’s exact test utilised? - (2)

Answer 35

The Chi-squared test is exact asymptotically, which means that it is more
and more accurate as the sample size gets bigger.
This is why, if any of the expected frequencies are less than 5, we should
not use it, as the sample size is not large enough for the test to be
accurate.

Question 36

Chi-squared 2x2 contiegency table in SPSS

Answer 36

Interpret same as before - seeing difference in proportion between the rows

Question 37

Chi-squared test statistics, DF and p-value

Answer 37

Pearson Chi-squared = 0.024
DF = 1
P -value = 0.877

Question 38

Checking chi-squared test if all expected frequencies are above five in SPSS

Answer 38

Little note that 0 cells have expected counts less than 5
So all assumptin are satisfied so use results of Chi-squared test

Question 39

Fisher’s exact test gives same 2x2 contiegency table as well as data on DF and p-value:

Answer 39

DF 1 since exact-sig (2-sided)
P-value is 0.534

Question 40

2X2 contigency table in R

Answer 40

Question 41

Chi-squared test statistic, DF, p-value in R

Answer 41

Question 42

Chi-squared test expected frequencies output

Answer 42

Question 43

Fisher’s exact test results in R

Answer 43

P value = 1 (not sig)