Post on 02-Jan-2016
Quantitative Methods
Partly based on materials by Sherry O’Sullivan
Part 3Chi - Squared Statistic
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
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
Example using Chi-Squared
• “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 sample– This is called a ‘contingency table’
Firefox IExplorer Safari Chrome Opera
Observed frequencies
30 6 4 8 2
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
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
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
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 validFirefox IExplorer Safari Chrome Opera
Expected frequencies
10 10 10 10 10
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 freedomStage 3b: Degree of freedom
We need to find the degree of freedomThis is calculated with the number of
categories◦We had 5 categories, df = 5-1 (4)
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
Chi-Squared Table from http://ourwayit.com/CA517/LearningActivities.htm
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
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
Alternative Method
• Outline: Calculate chi-squared, and use the table to find the confidence
• In this case, calculated Χ2 = 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
Chi-Squared Table from http://ourwayit.com/CA517/LearningActivities.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
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%
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
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
What is the research question?1. 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”
2. 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 “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 (R=2, C=3)
• Look up in chi-squared table– Critical chi-squared value = 5.99
Mastery Challenge Fantasy
Male 10 32 8
Female 24 8 18
Chi-Squared Table from http://ourwayit.com/CA517/LearningActivities.htm
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
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
2 2 2 22 (10 17) (32 20) (24 17) (8 20)
...17 20 17 20
24.01
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
Alternative Method:
• Outline: Calculate chi-squared, and use the table to find the confidence
• In this case, calculated Χ2 = 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
Chi-Squared Table from http://ourwayit.com/CA517/LearningActivities.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
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
End