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Quantitative Methods

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Quantitative Methods Part 3 Chi - Squared Statistic T-test used the mean and standard error of a population sample. To calculate the mean and standard error means ... – PowerPoint PPT presentation

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Title: Quantitative Methods


1
Quantitative Methods
  • 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 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

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

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 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

6
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

Firefox IExplorer Safari Chrome Opera
Expected frequencies 10 10 10 10 10
Expected frequency table
7
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

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
  • Has to have 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)
  • 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)

10
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

11
Stage 4 Calculate statistics
  • We compare the observed against the expected for
    each category
  • We square each one
  • We add all of them up

Firefox IExplorer Safari Chrome Opera
Observed 30 6 4 8 2
Expected 10 10 10 10 10
52
12
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

13
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

High Performance Compact Midsize Full size
FO 20 14 9 9
MR 10 40 30 20
FE 5 (10 of 50) 20 15 10
14
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

Mastery Challenge Fantasy
Male 10 32 8
Female 24 8 18
Observed frequency table
15
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

16
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 (R2, C3)
  • Look up in chi-squared table
  • Chi-squared value 5.99

Mastery Challenge Fantasy
Male 10 32 8
Female 24 8 18
17
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 (FO) 10 32 8 50
Female (FO) 24 8 18 50
Cat total 34 40 26

Male (FE) 17 20 13
Female (FE) 17 20 13
18
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 (FO) 10 32 8
Female (FO) 24 8 18

Male (FE) 17 20 13
Female (FE) 17 20 13
19
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 (FO) 10 32 8
Female (FO) 24 8 18

Male (FE) 17 20 13
Female (FE) 17 20 13
20
Workshop
  • Work on Workshop 7 activities
  • Your journal (Homework)
  • Your Literature Review (Complete/update)

References
  • Dr C. Prices notes 2010
  • Gravetter, F. and Wallnau, L. (2003) Statistics
    for the Behavioral Sciences, New York West
    Publishing Company
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