# Quantitative Methods - PowerPoint PPT Presentation

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Quantitative Methods. Part 3 Chi - 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

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Quantitative Methods

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## Quantitative Methods

Part 3

Chi - 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 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-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

### 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 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 categories

Expected frequency table

### Stage1: Formulation of Hypothesis

• : 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 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

• Has to have more than 5 to be valid

### Stage 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 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-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

### Stage 4: Calculate statistics

• We compare the observed against the expected for each category

• We square each one

• We add all of them up

= 52

### Stage 5: Decision

• Can 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: “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

• Once null hypothesis of expected frequency has been done, the analysis is the same as no preference calculation

### Chi-Squared test for “Independence”.

• What makes computer games fun?

• Review found the following

• Factors (Mastery, Challenge and Fantasy)

• Different opinion depending on gender

• Research sample of 50 males and 50 females

Observed frequency table

### 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”

### 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 = (R – 1)*(C – 1) = 1 * 2 = 2 (R=2, C=3)

• Look up in chi-squared table

• Chi-squared value = 5.99

### 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)

### Chi-Squared analysis for “Independence”.

• Calculate the statistics using the chi-squared formula

• Ensure you include both male and female data

### 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

### Workshop

• Work on Workshop 7 activities