Week 6: Chi-squared tests

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Week 6 Chi-squared tests

• Learning outcomes
• use a chi-squared test to determine whether two
  traits are independent

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Frequency table
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Contingency table 2 way
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Contingency table 2 way
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Question

• Are the proportions of males having each of the
  qualifications the same as the corresponding
  proportions for females?
• Males 15 A-level, 61 higher, 24 other
• Females 17, 62, 21
• Are qualifications independent of gender?
• For this example the answer is yes.

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Mendels peas experiment
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Chi-squared test for independence

• Null hypothesis the attribute smooth or
  wrinkled is independent of whether the pea is
  green or yellow.
• Alternative hypothesis the proportion which are
  smooth depends on the colour of the pea.

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Expected frequencies

• If the null hypothesis is true then the
  proportion of smooth peas is
• 423/556
• regardless of the colour of the pea

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Expected number green and smooth
Therefore the expected number green and smooth is
this proportion of the number of green peas
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In the same way
Yellow and smooth
Green and wrinkled
Green and smooth
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Observed/Expected frequencies
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Observed/Expected frequencies
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Formula

• The expected values in each row and column add up
  to the corresponding marginal total.
• Therefore, in a 2x2 table, once one expected
  value is calculated the rest can be obtained by
  subtraction.
• A 2x2 contingency table as 1 degree of freedom.

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Chi-squared test for a 2x2 table
To compare the observed and expected frequencies

• follows a chi-squared distribution with 1 degree
  of freedom.

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Peas example
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conclusion

• Since the value of chi-squared is low
• observed frequencies are close to expected
  frequencies
• the null hypothesis can be accepted
• there is no evidence of an association between
  colour and texture

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Using tables

• Using a 5 significance level with 1 degree of
  freedom the null hypothesis would be rejected if

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Note

• The chi-squared test is only valid if the
  expected frequencies are greater than 5.
• The degrees of freedom for a table with r rows
  and c columns are
• (r - 1) x ( c - 1)

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Age and drinking categories of a sample of males
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Expected frequencies
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Chi-squared statistic

• Degrees of freedom (3 - 1) x (5 - 1) 8
• Value from tables for a 5 significance level is
  15.51.

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Conclusions

• The value of chi-squared is large therefore
• There is very convincing evidence from this
  survey that drinking habits are related to the
  age of the drinker.
• A comparison of the observed values with the
  expected values in the table shows that there are
  very few heavy drinkers in the oldest age group
  and fewer than expected in the 45 to 54 years age
  group.

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SPSS

• The original form of the data will be a row for
  each individual e.g.
• Units Age
• 1 4
• 2 3
• 1 1
• SPSS does two things
• Creates the contingency table
• Carries out a chi-squared test on the table.

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AnalysegtDescriptive StatisticsgtCrosstabs
Select Statistics and Chi-square Select Cells
and Expected
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Contingency Table
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Chi-squared test
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Goodness of fit tests
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Expected frequencies
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Chi-squared test

• follows a chi-squared distribution with d.f.
• number of categories - 1

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Result

• Since chi-squared is small (lt the appropriate
  value from the tables with 3 degrees of freedom)
  the theoretical model is a good fit to the data.