Quantitative Methods Partly based on materials by Sherry O

1
Quantitative MethodsPartly based on materials
by Sherry OSullivan

• Part 3
• Chi - Squared Statistic

2
Recap on T-Statistic

• It used the mean and standard error of a
  population sample
• The data is on an interval or scale
• Mean and standard error are the parameters
• This approach is known as parametric
• Another approach is non-parametric testing

3
Introduction to Chi-Squared

• It does not use the mean and standard error of a
  population sample
• Each respondent can only choose one category
  (unlike scale in t-Statistic)
• The expected frequency must be greater than 5 in
  each category for the test to succeed.
• If any of the categories have less than 5 for the
  expected frequency, then you need to increase
  your sample size
• Or merge categories

4
Example using Chi-Squared

• Is there a preference amongst the UW student
  population for a particular web browser? (Dr C
  Prices Data)
• They could only indicate one choice
• These are the observed frequencies responses from
  the sample
• This is called a contingency table

Firefox IExplorer Safari Chrome Opera
Observed frequencies 30 6 4 8 2
5
Was it just chance?

• How confident am I?
• Was the sample representative of all UW students?
• Was the variation in the measurements just
  chance?
• Chi-Squared test for significance
• Several ways to use the test
• Simplest is Null Hypothesis
• H0 The students show no preference for a
  particular browser

6
Chi-Squared Goodness of fit (No
preference)

• H0 The students show no preference for a
  particular browser
• This leads to Hypothetical or Expected
  distribution of frequency
• We would expect an equal number of respondents
  per category
• We had 50 respondents and 5 categories

Firefox IExplorer Safari Chrome Opera
Expected frequencies 10 10 10 10 10
Expected frequency table
7
Stage1 Formulation of Hypothesis

• H0 There is no preference in the underlying
  population for the factor suggested.
• H1 There is a preference in the underlying
  population for the factors suggested.
• The basis of the chi-squared test is to compare
  the observed frequencies against the expected
  frequencies

8
Stage 2 Expected Distribution

• As our null- hypothesis is no preference, we
  need to work out the expected frequency
• You would expect each category to have the same
  amount of respondents
• Show this in Expected frequency table
• Each expected frequency must be more than 5 to be
  valid

Firefox IExplorer Safari Chrome Opera
Expected frequencies 10 10 10 10 10
9
Stage 3a Level of confidence

• Choose the level of confidence (often 0.05
  sometimes 0.01)
• 0.05 means that there is 5 chance that
  conclusion is chance
• 95 chance that our conclusions are accurate

Stage 3b Degree of freedom

• We need to find the degree of freedom
• This is calculated with the number of categories
• We had 5 categories, df 5-1 (4)

10
Stage 3b Critical value of Chi-Squared

• In order to compare our calculated chi-square
  value with the critical value in the
  chi-squared table we need
• Level of confidence (0.05)
• Degree of freedom (4)
• Our critical value from the table 9.49

11
Chi-Squared Table from http//ourwayit.com/CA517/L
earningActivities.htm
12
Stage 4 Calculate statistics

• We find the differences between the observed and
  the expected values for each category
• We square each difference, and divide the answer
  by its expected frequency
• We add all of them up

Firefox IExplorer Safari Chrome Opera
Observed 30 6 4 8 2
Expected 10 10 10 10 10
52
13
Stage 5 Decision

• Can we reject the H0 that students show no
  preference for a particular browser?
• Our value of 52 is way beyond 9.49. We are (at
  least) 95 confident the value did not occur by
  chance
• And probably much more confident than that
• So yes we can safely reject the null hypothesis
• Which browser do they prefer?
• Firefox as it is way above expected frequency of
  10

14
Alternative Method

• Outline Calculate chi-squared, and use the table
  to find the confidence
• In this case, calculated ?2 52
• Go to the appropriate row of the table, and look
  across for the highest value that is LOWER than
  the measured value
• The top of that column gives our confidence that
  the effect is real

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Chi-Squared Table from http//ourwayit.com/CA517/L
earningActivities.htm

• The probability of this result happening by
  chance is less than 0.001
• We can be at least 99.9 confident of our result

16
Chi-Squared No Difference from a Comparison
Population.

• RQ Are drivers of high performance cars more
  likely to be involved in accidents?
• Sample n 50 and Market Research data of
  proportion of people driving these categories

High Performance Compact Midsize Full size
FO observed accident frequency 20 14 9 9
Ownership () 10 40 30 20
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Contingency Table

• Null hypothesis H0 type of car has no effect on
  accident frequency
• Once the expected frequencies (under the null
  hypothesis) have been calculated, the analysis is
  the same as the no preference calculation

High Performance Compact Midsize Full size
FO observed accident frequency 20 14 9 9
Ownership () 10 40 30 20
FE expected accident frequency 5 (10 of 50) 20 15 10
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Chi-Squared test for Independence.

• What makes computer games fun?
• Review found the following
• Factors (Mastery, Challenge and Fantasy)
• Is there a different opinion depending on gender?
• Research sample of 50 males and 50 females

Mastery Challenge Fantasy
Male 10 32 8
Female 24 8 18
Observed frequency table
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What is the research question?

• A single sample with individuals measured on 2
  variables
• RQ Is there a relationship between fun factor
  and gender?
• HO There is no such relationship
• Two separate samples representing 2 populations
  (male and female)
• RQ Do male and female players have different
  preferences for fun factors?
• HO Male and female players do not have
  different preferences

20
Chi-Squared analysis for Independence.

• Establish the null hypothesis (previous slide)
• Determine the critical value of chi-squared
  dependent on the confidence limit (0.05) and the
  degrees of freedom.
• df (Rows 1)(Columns 1) 1 2 2 (R2,
  C3)
• Look up in chi-squared table
• Critical chi-squared value 5.99

Mastery Challenge Fantasy
Male 10 32 8
Female 24 8 18
21
Chi-Squared Table from http//ourwayit.com/CA517/L
earningActivities.htm
22
Chi-Squared analysis for Independence.

• Calculate the expected frequencies
• Add each column and divide by types (in this case
  2)
• Easier if you have equal number for each gender
  (if not come and see me)

Mastery Challenge Fantasy Respondents
Male (FObs) 10 32 8 50
Female (FObs) 24 8 18 50
Cat total 34 40 26

Male (FExp) 17 20 13
Female (FExp) 17 20 13
23
Chi-Squared analysis for Independence.

• Calculate the statistics using the chi-squared
  formula
• Ensure you include both male and female data

Mastery Challenge Fantasy
Male (FObs) 10 32 8
Female (FObs) 24 8 18

Male (FExp) 17 20 13
Female (FExp) 17 20 13
24
Stage 5 Decision

• Can we reject the null hypothesis?
• Our value of 24.01 is way beyond 5.99. We are 95
  confident the value did not occur by chance
• Conclusion We are 95 confident that there is a
  relationship between gender and fun factor
• But else can we get from this?
• Significant fun factor for males Challenge
• Significant fun factor for females Mastery and
  Fantasy

Mastery Challenge Fantasy
Male (FObs) 10 32 8
Female (FObs) 24 8 18

Male (FExp) 17 20 13
Female (FExp) 17 20 13
25
Alternative Method

• Outline Calculate chi-squared, and use the table
  to find the confidence
• In this case, calculated ?2 24.01
• Go to the appropriate row of the table, and look
  across for the highest value that is LOWER than
  the measured value
• The top of that column gives our confidence that
  the effect is real

26
Chi-Squared Table from http//ourwayit.com/CA517/L
earningActivities.htm

• The probability of this result happening by
  chance is less than 0.001
• We can be at least 99.9 confident of our result

27
Computers

• A computer can be used to calculate the expected
  values but you have to tell it how
• Use formulae in Excel
• Then the computer will calculate the p value for
  you
• p probability that the observed difference is
  due to chance
• There is a nice command in Excel that will do
  this

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End