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Quantitative MethodsQuantitative Methods
Part 3Chi - Squared Statistic
Recap on T-StatisticRecap on T-StatisticIt used the mean and standard error
of a population sampleThe data is on an “interval” or scaleMean and standard error are the
parametersThis approach is known as
parametric Another approach is non-parametric
testing
Introduction to Chi-Introduction to Chi-SquaredSquaredIt does not use the mean and standard
error of a population sampleEach respondent can only choose one
category (unlike scale in T-Statistic)The expected frequency must be
greater than 5 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
Example using Chi-Example using Chi-SquaredSquared“Is there a preference amongst
the UW student population for a particular web browser? “ (Dr C Price’s Data)◦They could only indicate one choice◦These are the observed frequencies
responses from the sampleFirefox IExplore
rSafari Chrome Opera
Observed frequencies
30 6 4 8 2
Was it just chance?Was it just chance?How confident am I?
◦Was the sample representative of all UW students?
◦Was it just chance?Chi-Squared test for significance
◦Some variations on test◦Simplest is Null Hypothesis
:The students show “no preference” for a particular browser
Chi-Squared: “Goodness of Chi-Squared: “Goodness of fit” (No preference)fit” (No preference)
: 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
categoriesFirefox IExplore
rSafari Chrome Opera
Expected frequencies
10 10 10 10 10
Expected frequency table
Stage1: Formulation of Stage1: Formulation of HypothesisHypothesis
: There is no preference in the underlying population for the factor suggested.
: 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
Stage 2: Expected Stage 2: Expected DistributionDistribution
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
◦Has to have more than 5 to be validFirefox IExplore
rSafari Chrome Opera
Expected frequencies
10 10 10 10 10
Stage 3a: Level of confidenceStage 3a: Level of confidence
Choose the level of confidence (often 0.05)◦0.05 means that there is 5% chance that
conclusion is chance◦95% chance that our conclusions are certain
Stage 3b: Degree of freedomStage 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)
Stage 3: Critical value of Chi-Stage 3: Critical value of Chi-SquaredSquared
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
Stage 4: Calculate Stage 4: Calculate statisticsstatisticsWe compare the observed
against the expected for each category
We square each oneWe add all of them up
Firefox IExplorer
Safari Chrome Opera
Observed
30 6 4 8 2
Expected 10 10 10 10 10
= 52
Stage 5: DecisionStage 5: DecisionCan we reject the That students
show no preference for a particular browser?◦Our value of 52 is way beyond 9.49. We
are 95% confident the value did not occur by chance
So yes we can safely reject the null hypothesis
Which browser do they prefer?◦Firefox as it is way above expected
frequency of 10
Chi-Squared: Chi-Squared: “No Difference “No Difference from a Comparison Population”.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
◦Once null hypothesis of expected frequency has been done, the analysis is the same as no preference calculation
High Performance
Compact Midsize Full size
FO20 14 9 9
MR%10% 40% 30% 20%
FE5 (10% of 50) 20 15 10
Chi-Squared test for Chi-Squared test for “Independence”.“Independence”.
What makes computer games fun?
Review found the following◦Factors (Mastery, Challenge and
Fantasy)◦Different opinion depending on
genderResearch sample of 50 males and 50
females
Mastery Challenge Fantasy
Male10 32 8
Female24 8 18
Observed frequency table
What is the research What is the research question?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”
Chi-Squared analysis for Chi-Squared analysis for “Independence”.“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 = (R – 1)*(C – 1) = 1 * 2 = 2 (R=2, C=3)
Look up in chi-squared table◦ Chi-squared value = 5.99
Mastery Challenge
Fantasy
Male10 32 8
Female24 8 18
Chi-Squared analysis for Chi-Squared analysis for “Independence”.“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 (FO)10 32 8 50
Female (FO)24 8 18 50
Cat total34 40 26
Male (FE)17 20 13
Female (FE)17 20 13
Chi-Squared analysis for Chi-Squared analysis for “Independence”.“Independence”.
Calculate the statistics using the chi-squared formula◦ Ensure you include both male and female
data
Mastery Challenge Fantasy
Male (FO)10 32 8
Female (FO)24 8 18
Male (FE)17 20 13
Female (FE)17 20 13
2 2 2 22 (10 17) (32 20) (24 17) (8 20)
...17 20 17 20
24.01
Stage 5: DecisionStage 5: DecisionCan 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 (FO)10 32 8
Female (FO)24 8 18
Male (FE)17 20 13
Female (FE)17 20 13
WorkshopWorkshopWork on Workshop 7 activitiesYour journal (Homework)Your Literature Review
(Complete/update)
ReferencesReferences Dr C. Price’s notes 2010 Gravetter, F. and Wallnau, L. (2003) Statistics for the
Behavioral Sciences, New York: West Publishing Company