Lecture 7 Two-Way Tables Slides available from Statistics

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Lecture 7Two-Way TablesSlides available from
Statistics SPSS page of www.gpryce.com

• Social Science Statistics Module I
• Gwilym Pryce

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Notices

• Register

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Aims and Objectives

• Aim
• This session introduces methods of examining
  relationships between categorical variables
• Objectives
• By the end of this session you should be able to
  
• Understand how to examine relationships between
  categorical variables using
• 2 way tables
• Chi square test for independence.

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Plan

• 1. Independent events
• 2. Contingent events
• 3. Chi square test for independence
• 4. Further Study

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1. Probability of two Independent events occurring

• If knowing that one event occurs does not affect
  the outcome of another event, we say those two
  outcomes are independent.
• And if A and B are independent, and we know the
  probability of each of them occurring, we can
  calculate the probability of them both occurring

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Example You have a two sided die and a coin,
find Pr(1 and H).

• Answer ½ x ½ ¼
• Rule P(A ? B) P(A) x P(B)

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e.g. You have one fair coin which you toss twice
whats the probability of getting two heads?

• Suppose
• A 1st toss is a head
• B 2nd toss is a head
• what is the probability of A ? B?
• Answer A and B are independent and are not
  disjoint (i.e. not mutually exclusive). P(A)
  0.5 and P(B) 0.5.
• P (A ? B) 0.5 x 0.5 0.25.

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2. Probability of two contingent events occurring

• If knowing that one event occurs does change the
  probability that the other occurs, then two
  events are not independent and are said to be
  contingent upon each other
• If events are contingent then we can say that
  there is some kind of relationship between them
• So testing for contingency is one way of testing
  for a relationship

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Example of contingent events

• There is a 70 chance that a child will go to
  university if his/her parents are middle class,
  but only a 10 chance if his/her parents are
  working class. Given that there is a 60 chance
  of a childs parents being working class
• What are the chances that a child will be born
  working class and go to University?
• What proportion of people at university will be
  from working working class backgrounds?

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A tricky one...
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This diagram illustrates graphically how the
probability of going to university is contingent
upon the social class of your parents.
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6 of all children are both working class and end
up going to University
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as percent of all children
Working class Middle class
Go to University 6 28
Do not go to University 54 12
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at Uni from WC parents?

• Of all children, only 34 end up at university
  (6 WC 28 MC)
• i.e. 6 out of every 34 University students are
  from WC parents
• 6/34 17.6 of University students are WC

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• Probability theory states that
• if x and y are independent, then the probability
  of events x and y simultaneously occurring is
  simply equal to the product of the two events
  occurring
• But, if x and y are not independent, then

Prob(x ? y) Prob(x) ? Prob(y given that x has occurred)
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Test for independence

• We can use these two rules to test whether events
  are independent
• Does the distribution of observations across
  possible outcomes resemble
• the random distribution we would get if events
  were independent?
• I.e. if we assume independence and calculate the
  expected number of of cases in each category, do
  these figures correspond fairly closely to the
  actual distribution of outcomes found in our
  data?
• Or a distribution of outcomes more akin to
  contingency
• i.e. one event contingent on the other

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Example 1 Is there a relationship between social
class and education? We might test this by
looking at categories in our data of WC, MC,
University, no University.
Suppose we have 300 observations distributed as
follows
Working class Middle class
Go to University 18 84
Do not go to University 162 36
Given this distribution, would you say these two
variables are independent?
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• To do the test for independence we need to
  compare expected with observed.
• But how do we calculate ei, the expected number
  of observations in category i?
• ei number of cases expected in cell i assuming
  that the two categorical variables are
  independent
• ei is calculated simply as the probability of an
  observation falling into category i under the
  independence assumption, multiplied by the total
  number of observations.
• I.e. No contingency

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• So, if UNIY or UNIN and WC or MC are independent
  (i.e. assuming H0) then

Prob(UNIY ? WC) Prob(UNIY)?Prob(WC) so the
expected number of cases for each of the four
mutually exclusive categories are as follows
Working class Middle class
Go to University P(UNIY) x P(WC) x n P(UNIY) x P(MC) x n
Do not go to University P(UNIN) x P(WC) x n P(UNIN) x P(MC) x n
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• But how do we work out
• Prob(UNIY) and Prob(WC)
• which are needed to calcluate Prob(UNIY ? WC)
• Prob(UNIY ? WC) Prob(UNIY)?Prob(WC)
• Answer we assume independence and so estimate
  them from the data by simply dividing the total
  observations by the total number in the given
  category
• E.g. Prob(UNIY) Total no. cases UNIY ? All
  observations
• (18 84) / 300 0.34
• Prob(WC) is calculated the same way
• E.g. Prob(WC) Total no. cases WC ? All
  observations
• (18 162)/300 0.6
• Prob(UNIY ? WC) .34 x .6 x 300 61.2

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Working class Middle class
Go to University P(UNIY) x P(WC) x n (no. at Uni / n) x (no. WC/n) x n P(UNIY) x P(MC) x n (no. at Uni / n) x (no. MC/n) x n
Do not go to University P(UNIN) x P(WC) x n (no.not Uni / n) x (no. WC/n) x n P(UNIN) x P(MC) x n (no. not Uni / n) x (no. MC/n) x n
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Working class Middle class
Go to University 18 84 102
Do not go to University 162 36 198
180 120 300
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Working class Middle class
Go to University P(UNIY) x P(WC) x n (102 / 300) x (180 /300) x 300 P(UNIY) x P(MC) x n (102 / 300) x (120 /300) x 300
Do not go to University P(UNIN) x P(WC) x n (198 / 300) x (180 /300) x 300 P(UNIN) x P(MC) x n (198 / 300) x (120 /300) x 300
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Expected count in each category
Working class Middle class
Go to University (102 / 300) x (180 /300) x 300 .34 x .6 x 300 61.2 (102 / 300) x (120 /300) x 300 .34 x .4 x 300 40.8
Do not go to University (198 / 300) x (180 /300) x 300 .66 x .6 x 300 118.8 (198 / 300) x (120 /300) x 300 .66 x .4 x 300 79.2
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We have the actual count (I.e. from our data set)
Working class Middle class
Go to University 18 84
Do not go to University 162 36
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And the expected count
(I.e. the numbers wed expect if we assume class
education to be independent of each other)
Working class Middle class
Go to University 61.2 40.8
Do not go to University 118.8 79.2
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What does this table tell you?
Working class Middle class
Go to University Actual count 18 84
Expected count 61.2 40.8
Do not go to University Actual count 162 36
Expected count 118.8 79.2
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• It tells you that if class and education were
  indeed independent of each other
• I.e. the outcome of one does not affect the
  chances of outcome of the other
• Then youd expect a lot more working class people
  in the data to have gone to university than
  actually recorded (61 people, rather than 18)
• Conversely, youd expect far fewer middle class
  people to have gone to university (half the
  number actually recorded 41 people rather than
  80).

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But remember, all this is based on a sample, not
the entire population

• Q/ Is this discrepancy due to sampling variation
  alone or does it indicate that we must reject the
  assumption of independence?
• To answer this within the standardised hypothesis
  testing framework we need to know the chances of
  false rejection

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3. Chi-square test for independence
(non-parametric -- I.e. no presuppositions re
distribution of variables sample size not
relevant)

• (1) H0 expected actual ? x y are
  independent
• I.e. Prob(x) is not affected by whether or not y
  occurs
• H1 expected ? actual ? there is some
  relationship
• I.e. Prob(x) is affected by y occurring.
• (2) a 0.05

k no. of categories ei expected (given H0)
no. of sample observations in the ith category oi
actual no. of sample observations in the ith
category d no. of parameters that have to be
estimated from the sample data.
r no. of rows in table c no. of colums
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Chi-square distribution changes shape for
different df
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• (3) Reject H0 iff P lt a
• (4) Calculate P
• P Prob(c2 gt c2c)
• N.B. Chi-square tests are always an upper tail
  test
• c2 Tables are usually set up like a t-table with
  df down the side, and the probabilities listed
  along the top row, with values of c2c actually
  in the body of the table. So look up c2c in the
  body of the table for the relevant df and then
  find the upper tail probability that heads that
  column.
• SPSS - CDF.CHISQ(c2c,df) calculates Prob(c2 lt
  c2c), so use the following syntax
• COMPUTE chi_prob 1 - CDF.CHISQ(c2c,df).
• EXECUTE.

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Do a chi-square test on the following table
Working class Middle class
Go to University Actual count 18 84
Expected count 61.2 40.8
Do not go to University Actual count 162 36
Expected count 118.8 79.2
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• H0 expected actual
• ? class and Higher Education are independent
• H1 expected ? actual
• ? there is some relationship between class and
  Higher Education

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(2) State the formula calc c2

• c2 ( (18 - 61.2)2 / 61.2
• (84 - 40.8)2/ 40.8
• (162-118.8)2 / 118.8
• (36 - 79.2)2/ 79.2 )

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• c2 ((18 - 61.2)2 / 61.2 (84 -
  40.8)2/ 40.8
• (162-118.8)2 /118.8 (36 - 79.2)2/
  79.2 )
• 30.49 45.74 15.71 23.56
• 115.51
• df (r-1)(c-1) 1
• Sig P(c2 gt 115.51) 0

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• (3) Reject H0 iff P lt a
• (4) Calculate P
• COMPUTE chi_prob 1 - CDF.CHISQ(115.51,1).
• EXECUTE.
• Sig P(c2 gt 115.51) 0
• ? Reject H0

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Caveat

• As with the 2 proportions tests, the chi-square
  test is,
• an approximate method that becomes more
  accurate as the counts in the cells of the table
  get larger (Moore, Basic Practice of Statistics,
  2000, p. 485)
• Cell counts required for the Chi-square test
• You can safely use the chi-square test with
  critical values from the chi-square distribution
  when no more than 20 of the expected counts are
  less than 5 and all individual expected counts
  are 1 or greater. In particular, all four
  expected counts in a 2x2 table should be 5 or
  greater (Moore, Basic Practice of Statistics,
  2000, p. 485)

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Example 2 Is there a relationship between
whether a borrower is a first time buyer and
whether they live in Durham or Cumberland?

• Only real problem is how do we calculate ei the
  expected number of observations in category i?
• (I.e. number of cases expected in i assuming that
  the variables are independent)
• the formula for ei is the probability of an
  observation falling into category i multiplied by
  the total number of observations.

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As noted earlier

• Probability theory states that
• if x and y are independent, then the probability
  of events x and y simultaneously occurring is
  simply equal to the product of the two events
  occurring
• But, if x and y are not independent, then

Prob(x ? y) Prob(x) ? Prob(y given that x has occurred)
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• So, if FTBY or N and CountyD or C are independent
  (i.e. assuming H0) then
• Prob(FTBY ? CountyD) Prob(FTBY)?Prob(CountyD)
• so the expected number of cases for each of the
  four mutually exclusive categories are as
  follows

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Prob(FTBN) Total no. cases FTBN ? All
observations
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This gives us the expected count
To obtain this table in SPSS, go to Analyse,
Descriptive Statistics, Crosstabs, Cells, and
choose expected count rather than observed
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• What does this table tell you?
• Does it suggest that the probability of being an
  FTB independent of location?
• Or does it suggest that the two factors are
  contingent on each other in some way?
• Can it tell you anything about the direction of
  causation?
• What about sampling variation?

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Summary of Hypothesis test

• (1) H0 FTB and County are independent
• H1 there is some relationship
• (2) a 0.05
• (3) Reject H0 iff P lt a
• (4) Calculate P
• P Prob(c2 gt c2c) 0.29557 Do not reject H0
• I.e. if we were to reject H0, there would be a 1
  in 3 chance of us rejecting it incorrectly, and
  so we cannot do so. In other words, FTB and
  County are independent.

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Contingency Tables in SPSS
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• Click Cells button to select counts s
• If you select all three (row, column and total),
  you will end up with

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• Click the Statistics button to choose which stats
  you want.
• If you click Chi-square, the results of a range
  of tests will be listed

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We have been calculating the Pearson Chi-square
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4. For further study

• The Pearson Chi square test only tests for the
  existence of a relationship
• It tells you little about the strength of the
  relationship
• SPSS includes a raft of measures that try to
  measure the level of association between
  categorical variables.
• Click on the name of one of the statistics and
  SPSS will give you a brief definition (see below)
• In the lab exercises, take a look at these
  statistics and copy and paste the definitions
  along side your answers
• Right click on the definition and select Copy.
  Then open up a Word document and paste along with
  your output.

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Nominal variables

• Contingency coefficient
• A measure of association based on chi-square.
  The value ranges between zero and 1, with zero
  indicating no association between the row and
  column variables and values close to 1 indicating
  a high degree of association between the
  variables. The maximum value possible depends on
  the number of rows and columns in a table.
• Phi and Cramers V
• Phi is a chi-square based measure of association
  that involves dividing the chi-square statistic
  by the sample size and taking the square root of
  the result. Cramer's V is a measure of
  association based on chi-square.
• Lambda
• A measure of association which reflects the
  proportional reduction in error when values of
  the independent variable are used to predict
  values of the dependent variable. A value of 1
  means that the independent variable perfectly
  predicts the dependent variable. A value of 0
  means that the independent variable is no help in
  predicting the dependent variable.
• Uncertainty coefficient
• A measure of association that indicates the
  proportional reduction in error when values of
  one variable are used to predict values of the
  other variable. For example, a value of 0.83
  indicates that knowledge of one variable reduces
  error in predicting values of the other variable
  by 83. The program calculates both symmetric and
  asymmetric versions of the uncertainty
  coefficient.

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Ordinal Variables

• Gamma
• A symmetric measure of association between two
  ordinal variables that ranges between negative 1
  and 1. Values close to an absolute value of 1
  indicate a strong relationship between the two
  variables. Values close to zero indicate little
  or no relationship. For 2-way tables, zero-order
  gammas are displayed. For 3-way to n-way tables,
  conditional gammas are displayed.
• Somers d
• A measure of association between two ordinal
  variables that ranges from -1 to 1. Values close
  to an absolute value of 1 indicate a strong
  relationship between the two variables, and
  values close to 0 indicate little or no
  relationship between the variables. Somers' d is
  an asymmetric extension of gamma that differs
  only in the inclusion of the number of pairs not
  tied on the independent variable. A symmetric
  version of this statistic is also calculated.
• Kendalls tau-b
• A nonparametric measure of association for
  ordinal or ranked variables that take ties into
  account. The sign of the coefficient indicates
  the direction of the relationship, and its
  absolute value indicates the strength, with
  larger absolute values indicating stronger
  relationships. Possible values range from -1 to
  1, but a value of -1 or 1 can only be obtained
  from square tables.
• Kendalls tau-c
• A nonparametric measure of association for
  ordinal variables that ignores ties. The sign of
  the coefficient indicates the direction of the
  relationship, and its absolute value indicates
  the strength, with larger absolute values
  indicating stronger relationships. Possible
  values range from -1 to 1, but a value of -1 or
  1 can only be obtained from square tables.

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Correlations

• Pearson correlation coefficient r
• a measure of linear association between two
  variables
• Spearman correlation coefficient
• a measure of association between rank orders.
  Values of both range between -1 (a perfect
  negative relationship) and 1 (a perfect positive
  relationship). A value of 0 indicates no linear
  relationship.

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When you have a dependent variable measured on an
interval scale an independent variable with a
limited number of categories

• Eta
• A measure of association that ranges from 0 to
  1, with 0 indicating no association between the
  row and column variables and values close to 1
  indicating a high degree of association. Eta is
  appropriate for a dependent variable measured on
  an interval scale (e.g., income) and an
  independent variable with a limited number of
  categories (e.g., gender). Two eta values are
  computed one treats the row variable as the
  interval variable the other treats the column
  variable as the interval variable.