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Quantitative Methods

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Quantitative Methods

Part 3

Chi - Squared 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

- 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

- “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

- 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

: 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

- : 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

- 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

- 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)

- 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

- We compare the observed against the expected for each category
- We square each one
- We add all of them up

= 52

- 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

- 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

- 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

- 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”

- 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

- 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)

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

- 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

- Work on Workshop 7 activities
- Your journal (Homework)
- Your Literature Review (Complete/update)

References

- Dr C. Price’s notes 2010
- Gravetter, F. and Wallnau, L. (2003) Statistics for the Behavioral Sciences, New York: West Publishing Company