STAT131 W6La Association from Contingency Tables

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STAT131W6La Association from Contingency Tables

• by
• Anne Porter
• alp_at_uow.edu.au

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Null and Alternative hypothesesActivity

• Card game

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Activity Outcomes

• We draw a card from a pack until such time as
  there is a protest.
• The cards have been stacked such that all red
  come first (or all black)
• The draw is meant be be random ie a mix of red
  and black.
• At some point students reject the idea of
  fairness
• The proportion of reds is higher than expected by
  chance
• (or blacks depending on which was drawn first)
• Students are in fact rejecting the null
  hypothesis that the proportion
• of red cards is 0.5.
• (or that the proportion of red and black cards
  is equal)
• They are accepting the hypothesis that the
  proportion is not equal 0.5

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Null and Alternative hypotheses

• Null hypothesis is that the proportion of red
  cards (females) is 0.5 (or that the proportion of
  red and black cards is equal)
• Alternative hypothesis is that the proportion of
  red cards (females) is not equal 0.5

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Null and Alternative hypothesesformal
Tests of proportions

• H0 p 0.5 and
• HA p ? 0.5
• The p we refer to is the population proportion
• We do not hypothesise about a sample proportion
• We make inference about a population parameter p

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Lecture Outline

• Test hypotheses about association between
  categorical variables
• Testing Hypotheses (5 steps)
• Null and alternative hypotheses
• a level of significance
• Select test and state decision rule
• Perform experiment
• Draw conclusions
• test of association AND
• model fit
• p values

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Contingency tables

• For this contingency table what is
• P(Male)
• P(Support)

20/70
40/70
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Contingency tables

• If event male is independent of event support
  then
• P(Male and Support)

P(Male)xP(Support)
20/70 x 40/70 0.1632
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Contingency tables

• Given 70 observed people, if P(Male
  Support)0.1632
• How many are expected to be male and support
  given independence?

11.43
0.1632 x70 11.43 if events Males and Support
are independent
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Contingency tables

• Knowing the expected frequency for (male and
  support) we have no more degrees of freedom, the
  remaining values are fixed.

11.43
20-11.438.57
30-8.5721.43
40-11.4338.57
Note We had 1 degree of freedom
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Contingency tables

• If we observe a sample of data we may ask if the
  variables sex and level of support are
  associated? To test this we formally test the
  hypotheses

E11.43
E8.57
E38.57
E21.43
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Hypotheses no association

• Ho Under model of independence, E distributed
• (Row total column total)/grand total
• Ha E not distributed
• (row totalcolumn total)/grand total

E8.57
E11.43
E38.57
E21.43
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2. Assign a

• a is determined such that we have a desired
  level of confidence in our procedures (ie in our
  results).
• For the chi-square test for association we will
  use a0.05
• We will examine choosing alpha (a) later

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Degrees of freedom

• Knowing the expected frequency for (male and
  support) we have no more degrees of freedom, the
  remaining values are fixed.

11.43
20-11.438.57
21.43
38.57
Note We had 1 degree of freedom
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Degrees of freedom

• The degrees of freedom for a rows x column matrix
  may be calculated as (r-1)x(c-1)(2-1)x(2-1)1
• r is the number or rows and c is the number of
  columns

11.43
8.57
21.43
38.57
Note We had 1 degree of freedom
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Hypotheses no association

• Ho Under model of independence, E distributed
• (Row total column total)/grand total
• Ha E not distributed
• (row totalcolumn total)/grand total

E8.57
E11.43
E38.57
E21.43
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3. Select a test statistic and... determine the
rejection region

• To test about association in contingency tables
  we calculate
• And determine the region of rejection ie how big
  chi-square has to be before we conclude that the
  observed are sufficiently different to the
  expected to reject the null hypothesis
• eij expected count for the ith row and jth column
  of the table

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3... determine the rejection region

• For our contingency table
• df1,

a0.05
Then reject Ho there is evidence that the
variables are not independent
If the calculated gt
3.841
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3... determine the rejection region

• For our contingency table
• df1,

a0.05
Then reject Ho there is evidence that the
variables are not independent
If the calculated gt
3.841
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4. Calculate
E11.43
E8.57
E28.57
E21.43
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Decision

• As calculated value of 0.70 lt 3.841 (tabulated
  value) there insufficient evidence to reject the
  model that sex and level of support are
  independent. That is there is no evidence of an
  association between sex and level of support. The
  profile of support by males is similar to the
  profile of support for females. 13/40
  (32.5)males support, 7/30 (23.3) females support

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SPSS data entry looks like

• Data, weight cases by freq has been selected
• Analyse, Descriptives, Crosstabs and options have
  been selected

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SPSS output contingency table
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SPSS output Pearson Chi-Square
Value of chi-square
Assumption of expected frequencies gt 5 hold
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SPSS output Pearson Chi-Square
Probability of getting a statistic as high or
greater than 0.706 is 0.401. This is high gt0.05
therefore retain Ho, we can get this chi value by
chance under independence
Value of chi-square
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Example from Utts p. 528SPSS data

• Yes / No Ear infection
• P Placebo gum
• X xylitol gum
• L xylitol lozenge
• Is there an association between ear infection and
  gum used?

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Under Independence Expected frequency
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Under Independence Expected
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Under Independence Expected
Degrees of freedom
2
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Hypotheses

• Ho Under model of independence, E distributed
• - (Row total column total)/grand total
• Ha E not distributed
• - (row totalcolumn total)/grand total
• If
• p1proportion who get an infection in population
  given placebo
• p2proportion who get an infection given Xylitol
  gum
• P3proportion who get an infection in a
  population given Xylitol lozenges
• Ho
• Ha

p1p2p3

p1, p2 and p3 are not all the same
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5 step hypothesis test

• Ho Under model of independence, E distributed
• (Row total column total)/grand total
• Ha E not distributed in this manner
• a
• df
• Statistic and Region of rejection

0.05
(3-1)x(2-1)2
If calculated chi-square gt5.991 reject Ho there
is evidence that the variables are not independent
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Conclusion using decision rule SPSS

• Chi-square 6.690

gt5.991 therefore there is evidence that the data
do not fit the model of independence
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P values (sig)

• For chi-square test (one tailed) the p value is
• the probability of getting this statistic or
  greater

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Conclusion using p value from SPSS
The probability of getting a chi-square as high
as this or higher is 0.035. This is a small
probability (lt0.05) if the H0 were true. There is
evidence of an association between infection and
gum used

• Chi-square 6.690

Assumptions re expected frequencygt5 OK
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Significance Tests - Formal

• 1. Null and alternative hypotheses
• 2. Assign a
• 3. Select a statistic and determine the rejection
  region
• 4. Perform the experiment and calculate the
  observed value of c2 or T or Z orother statistic
  
• 5. Draw conclusions in context of problem

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Previous hypothesis testing situations

• Model fit
• Ho Expected distributed Binomial(2,0.5)
• Ha Expected not distributed Binomial (2,0.5)
• 2. Ho Expected distributed Poisson (0.4)
• Ha Expected not Poisson (0.4)
• 3. Ho Expected distributed as per the random
  stopping model
• Ha Expected not distributed as per random
  stopping model

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Future hypothesis testing situations
Tests of proportions

• the null hypothesis may be proportion 0.5 and
• alternative hypothesis proportion ? 0.5
• the null hypothesis may be m 0 and
• alternative hypothesis m ? 0.

Tests of means