HYPOTHESIS TESTING

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HYPOTHESIS TESTING Powered By Docstoc
					     A Review of
Widely-Used Statistical
       Methods



                     1
               REVIEW OF FUNDAMENTALS

 When testing hypotheses,
  all statistical methods will
  always be testing the null.

 Null Hypothesis?
   No difference/no relationship


 If we do not reject the null, conclusion?
   Found no difference/no relationship

 If we do decide to reject the null, conclusion?
   A significant relationship/difference is found and reported
      o The observed relationship/difference is too large to be
        attributable to chance/sampling error.                    2
How do we decide to reject/not reject the null?
 Statistical tests of significance always test the null and
  always report a?
    a (sig. level)—probability of erroneously rejecting a
     true null based on sample data.
    a represents the odds of being wrong if we decide to
     reject the null
       the probability that null is in fact true and that
        any apparent relationship/difference is a result
        of chance/sampling error and, thus
       the odds of being wrong if we report a significant
        relationship/difference.
 Rule of thumb for deciding to reject/not reject the null?
                                                          3
       STATITICAL DATA ANALYSIS
COMMON TYPES OF ANALYSIS?

– Examine Strength and Direction of Relationships
   • Bivariate (e.g., Pearson Correlation—r)
    Between one variable and another:   Y = a + b1 x 1
 • Multivariate (e.g., Multiple Regression Analysis)
    Between one dep. var. and an independent variable, while
     holding all other independent variables constant:
           Y = a + b1 x1 + b2 x2 + b3 x3 + … + bk xk
– Compare Groups
   • Between Proportions (e.g., Chi Square Test—2)
          H0:    P1 = P2 = P3 = … = Pk
 • Between Means (e.g., Analysis of Variance)
          H0:   µ1 = µ2 = µ3 = …= µk
 Let’s first review some fundamentals.                      4
Remember: Level of measurement determines choice of
statistical method.

       Statistical Techniques and Levels of Measurement:
                     INDEPENDENT
     NOMINAL/CATEGORICAL             METRIC (ORDERED METRIC or
                                           HIGHER)
 N
 O
 M
 I
 N          * Chi-Square             * Discriminant Analysis
 A          * Fisher’s Exact Prob.   * Logit Regression
 L

 M
 E
 T         * T-Test                  * Correlation (and Covariance) Analysis
 R         * Analysis of Variance    * Regression Analysis
 I
 C                                                                    5
       Correlation and Covariance: Measures of Association
                      Between Two Variables

Often we are interested in the strength and nature of the
relationship between two variables.



Two indices that measure the linear relationship between two
continuous/metric variables are:
  a. Covariance
  b. Correlation Coefficient (Pearson Correlation)
                          Covariance
Covariance is a measure of the linear association between
two metric variables (i.e., ordered metric, interval, or ratio
variables).

   Covariance (for a sample) is computed as follows:

                     ( xi  x )( yi  y )   for
              sxy 
                            n 1             samples



Positive values indicate a positive relationship.

Negative values indicate a negative (inverse) relationship.
          Covariance (sxy ) of Two Variables
 Example: Golfing Study
   A golf enthusiast is interested in investigating
   the relationship, if any, between golfers’
   driving distance (x) and their 18-hole score (y).
   He uses the following sample data (i.e., data
   from n = 6 golfers) to examine the issue:
            x =Average Driving    y = Golfer’s Average
              Distance (yards.)       18-Hole Score

                    277.6                 69
                    259.5                 71
                    269.1                 70
                    267.0                 70
                    255.6                 71
                    272.9                 69
       Covariance (sxy ) of two variables
                                             ( xi  x )( yi  y )
                                    sxy 
 Example: Golfing Study                            n 1

            x      y     ( xi  x ) ( yi  y ) ( xi  x )( yi  y )
          277.6   69       10.65       -1.0            -10.65
          259.5   71       -7.45        1.0             -7.45
          269.1   70        2.15          0                 0
          267.0   70        0.05          0                 0
          255.6   71      -11.35        1.0            -11.35
          272.9   69        5.95       -1.0             -5.95
Average 267.0 70.0                            Total -35.40
Std. Dev. 8.2192 .8944
                                                        n=6
                               Covariance
 Example: Golfing Study


 Covariance:       s xy   
                             ( x  x )( y
                                 i        y )  35.40
                                             i
                                                        7.08
                                     n1        61

• What can we say about the relationship between the two variables?

   The relationship is negative/inverse.
   That is, the longer a golfer’s driving distance is, the lower
    (better) his/her score is likely to be.
• How strong is the relationship between x and y?

  Hard to tell; there is no standard metric to judge it by!
  Values of covariance depend on units of measurement for x and y.
                      WHAT DOES THIS MEAN?
                           Covariance
                            s xy   
                                      (x   i    x )( y i  y )
                                                                   
                                                                        35.40
                                                                                   7.08
  It means:                            n1                              61
  If driving distance (x) were measured in feet, rather than yards,
  even though it is the same relationship (using the same data),
  the covariance sxy would have been much larger. WHY?

  Because x-values would be much larger, and thus ( xi  x )
   values will be much larger which, in turn, will make
                                 (

   ( xi  x )( yi  y ) much larger.
SOLUTION: Correlation Coefficient comes to the rescue!

• Correlation Coefficient (r) is a standard measure/metric for
  judging strength of linear relationship that, unlike covariance,
  is not affected by the units of measurement for x and y.
   This is why correlation coefficient (r) is much more widely
     used that covariance.
                Correlation Coefficient

Correlation Coefficient rxy (Pearson/simple correlation)
is a measure of linear association between two variables.
   It may or may not represent causation.


The correlation coefficient rxy (for sample data) is
computed as follows:



     for                  sxy       sxy = Covariance of x & y
   samples
                 rxy               sx = Std. Dev. of x
                         sx s y     sy = Std Dev. of y
 Correlation Coefficient = r
                                   Francis Galton
                                   (English researcher,
                                   inventor of
                                   fingerprinting, and
                                   cousin of Charles
                                   Darwin)

 In1888, plotted lengths of forearms and head sizes to see to what
  degree one could be predicted by the other.
 Stumbled upon the mathematical properties of correlation plots
  (e.g., y intercept, size of slope, etc.).
 RESULT: An objective measure of how two variables are
  “co-related“--CORRELATION COEFFICIENT (Pearson
  Correlation), r.
    Assesses the strength of a relationship based strictly on
      empirical data, and independent of human judgment or opinion
                                                                 13
Correlation Coefficient (Pearson Correlation) = r

What do you use it for?          Karl Pearson, a Galton
                                 Student & the Founder
                                 of Modern Statistics


To examine:
 a. Whether a relationship exists between two metric
    variables
     • e.g., income and education, or
       workload and job satisfaction and
 b. What the nature and strength of that relationship
    may be.

Range of Values for r?
                                                          14
      Correlation Coefficient (Pearson Correlation) rxy

                               -1 < r < +1.

•
• r-values closer to -1 or +1 indicate stronger linear relationships.
• r-values closer to zero indicate a weaker relationship.



NOTE: Once rxy is calculated, we need to see whether it is
      statistically significant (if using sample data).

• Null Hypothesis when using r?
    H0: r = 0
     There is no relationship between the two variables.
16
  Correlation Coefficient (Pearson Correlation) rxy
 Example: Golfing Study
    A golf enthusiast is interested in investigating
    the relationship, if any, between golfers’
    driving distance (x) and their 18-hole score (y).
    He uses the following sample data (i.e., data
    from n = 6 golfers) to examine the issue:

             x =Average Driving     y =Average
               Distance (yards.)   18-Hole Score

                    277.6               69
                    259.5               71
                    269.1               70
                    267.0               70
                    255.6               71
                    272.9               69
Correlation Coefficient (Pearson Correlation) rxy

 Example: Golfing Study

            x      y     ( xi  x ) ( yi  y ) ( xi  x )( yi  y )
          277.6   69       10.65       -1.0           -10.65
          259.5   71       -7.45        1.0            -7.45
          269.1   70        2.15          0                0
          267.0   70        0.05          0                0
          255.6   71      -11.35        1.0           -11.35
          272.9   69        5.95       -1.0            -5.95
Average 267.0 70.0                            Total -35.40
Std. Dev. 8.2192 .8944
Correlation Coefficient (Pearson Correlation) rxy

 Example: Golfing Study

  We had calculated sample Covariance sxy to be:

           s xy   
                     (x   i    x )( y i  y )
                                                  
                                                     35.40
                                                              7.08
                               n1                   61
  Correlation Coefficient (Pearson Correlation) rxy
                    sxy    7.08
          rxy                        -.9631
                sx sy (8.2192)(.8944)

  Conclusion?
            Not only is the relationship negative, but also
            extremely strong!
Correlation Coefficient (Pearson Correlation):
                ( x  x)( y  y )                    s xy
      r                                      
             ( x  x)        ( y  y)           ( s x ).(s y )
                         2                2



To understand the practical meaning of r, we can square it.

   • What would r2 mean/represent?
      • e.g., r = 0.96 r2 = 92%

 r2 Represents the proportion (%) of the total/combined variation
 in both x and y that is accounted for by the joint variation
 (covariation) of x and y together (x with y and y with x)

  • r2 always represents a %
  • Why do we show more interest in r, rather than r2?
                                                                   20
Correlation Coefficient: Computation
 r2 = (Covariation of X and Y together) / (All of variation of X & Y combined)

                                 ( x  x)( y  y )                  s xy
                     r                                      
                            ( x  x)        ( y  y)           ( s x ).(s y )
                                        2                2

     Blood
 Age Pressure    _          _           _      _                 _                _
 X     Y    X–X         Y–Y        (X – X) (Y – Y)       (X – X)2    (Y – Y)2
 4     12   -3         -4               12                 9            16
 6     19   -1          3               -3                 1             9
 9     14    2         -2               -4                 4             4
 .     .     .          .               .                   .              .
 .     .     .          .               .                   .              .
 _     _                                  _      _               _           _
 X=7   Y=16                        ∑(X – X) (Y – Y)       ∑ (X – X)2 ∑ (Y – Y)2

 NOTE: Once r is calculated, we need to see if it is statistically
       significant (if sample data). That is, we need to test H0: r = 021
Correlation Coefficient?
  Suppose the correlation between X (say, Students’ GMAT Scores)
  and Y (their 1st year GPA in MBA program) is r = +0.48 and
  is statistically significant. How would we interpret this?
 a) GMAT score and 1st year GPA are positively related so that as
    values of one variable increase, values of the other also tend to
    increase, and

 b) R2 = (0.48)2 = 23% of variations/differences in students’ GPAs
    are explained by (or can be attributed to) variations/
    differences in their GMAT scores.
          Lets now practice on SPSS
 Menu Bar: Analyze, Correlate, Bivariate, Pearson

 EXAMPLE: Using data in SPSS File Salary.sav
 we wish to see if beginning salary is related to seniority,
                                                                  22
 age, work experience, and education
       STATITICAL DATA ANALYSIS
COMMON TYPES OF ANALYSIS:

– Examine Strength and Direction of Relationships
   • Bivariate (e.g., Pearson Correlation—r)
    Between one variable and another:   Y = a + b1 x 1

 • Multivariate (e.g., Multiple Regression Analysis)
    Between one dep. var. and an independent variable, while
     holding all other independent variables constant:
           Y = a + b1 x1 + b2 x2 + b3 x3 + … + bk xk
– Compare Groups
   • Between Proportions (e.g., Chi Square Test—2)
          H0:    P1 = P2 = P3 = … = Pk

 • Between Means (e.g., Analysis of Variance)
                                                           23
          H0:   µ1 = µ2 = µ3 = …= µk
            STATITICAL DATA ANALYSIS
 Chi-Square Test of Independence?
 Developed by Karl Pearson in 1900.
 Is used to compare two or more groups regarding a categorical
  characteristic.
 That is, to compare proportions/percentages:
   – Examines whether proportions of different groups of subjects
      (e.g., managers vs professionals vs operatives) are equal/
      different across two or more categories (e.g., males vs females).
 Examines whether or not a relationship exists between
  two categorical/nominal variables
   (e.g., employee status and gender)
   – A categorical DV and a categorical IV.
   – EXAMPLE?
 Is smoking a function of gender? That is, is there a difference
  between the percentages of males and females who smoke? 24
       •Chi-Square Test of Independence
 Research Sample (n=100):
        ID     Gender      Smoking Status
        1      0 = Male    1 = Smoker
        2      1 = Female  0 = Non-Smoker
        3      1           1
        4      1           0
        5      0           0
        .      .           .
        .      .           .
        .      .           .
        100 1              0
 Dependent variable (smoking status) and the independent variable (gender)
  are both categorical.
 Null Hypothesis?
 H0: There is no difference in the percentages of males and females
     who smoke/don’t smoke (i.e., Smoking is not a function of gender).

 QUESTION: Logically, what would be the first thing you would do? 25
     •Chi-Square Test of Independence
H0: There is no difference in the percentages of males and females who
    smoke (Smoking is not a function of gender).
H1: The two groups are different with respect to the proportions who smoke.


TESTING PROCEDURE AND THE INTUITIVE LOGIC:

 Construct a contingency Table: Cross-tabulate the
   observations and compute Observed (actual) Frequencies (Oij ):

                 Male          Female         TOTAL
Smoker           O11 = 15      O12 = 25       40
Nonsmoker        O21 = 5       O22 = 55       60

TOTAL              20             80          n = 100

                                                                    26
                                •
     Chi-Square Test of Independence
 Next, ask yourself: What numbers would you expect to find
  in the table if you were certain that there was absolutely no
  difference between the percentages of males and females who
  smoked (i.e., if you expected the Null to be true)? That is,
  compute the Expected Frequencies (Eij ).

 Hint: What % of all the subjects are smokers/non-smokers?

                     Male             Female      TOTAL
Smoker               O11 = 15         O12 = 25    40

Nonsmoker            O21 = 5          O22 = 55    60

TOTAL                  20               80        n = 100
                                                             27
                                 •
     Chi-Square Test of Independence
 If there were absolutely no differences between the two groups
  with regard to smoking, you would expect 40% of individuals
  in each group to be smokers (and 60% non-smokers).
 Compute and place the Expected Frequencies (Eij ) in the
  appropriate cells:

                      Male             Female     TOTAL
Smoker                O11 = 15         O12 = 25   40
                      E11 = 8          E12 = 32

Nonsmoker             O21 = 5          O22 = 55   60
                      E21 = 12         E22 = 48
TOTAL                       20               80   n = 100

NOW WHAT? What is the next logical step?                     28
                                 •
      Chi-Square Test of Independence
Compare the Observed and Expected frequencies—i.e.,
examine the (Oij – Eij) discrepancies.
                      Male            Female     TOTAL
 Smoker               O11 = 15        O12 = 25   40
                      E11 = 8         E12 = 32

 Nonsmoker            O21 = 5         O22 = 55   60
                      E21 = 12        E22 = 48

 TOTAL                     20              80    n = 100


QUESTION: What can we infer if the observed/actual frequencies
happen to be reasonably close (or identical) to the expected
frequencies?                                                 29
      •Chi-Square Test of Independence
So, the key to answering our original question lies in the size of the
discrepancies between observed and expected frequencies.

If the observed frequencies were reasonably close to the expected
frequencies:
 – Reasonably certain that no difference exists between percentages of males
   and females who smoke,
 – Good chance that H0 is true
      • That is, we would be running a large risk of being wrong if we
        decide to reject it.

On the other hand, the farther apart the observed frequencies
happen to be from their corresponding expected frequencies:
 – The greater the chance that percentages of males and females who smoke
   would be different,
 – Good chance that H0 is false and should be rejected
      •   That is, we would run a relatively small risk of being wrong if
          we decide to reject it.
                                                                               30
                What is, then, the next logical step?
         Chi-Square Test of Independence
Compute an Overall Discrepancy Index: One way to quantify the total
discrepancy between observed (Oij) and expected (Eij) frequencies is to
add up all cell discrepancies--i.e., compute S (Oij – Eij).
• Problem?
   Positive and negative values of (Oij – Eij) RESIDUALS for different cells
   will cancel out.

• Solution?
   Square each (Oij – Eij) and then sum them up--compute S(Oij – Eij)2.

• Any Other Problems?
   Value of S(Oij – Eij)2 is impacted by sample size (n).
    – For example, if you double the number of subjects in each cell, even though cell
       discrepancies remain proportionally the same, the above discrepancy index will

       be much larger and may lead to a different conclusion.    Solution?      31
     Chi-Square Test of Independence

• Divide each (Oij – Eij)2 value by its corresponding Eij value
  before summing them up across all cells
   • That is, compute an index for average discrepancy per subject.
                      (Oij – Eij)2
                  S
                          Eij

                                                                    (Oij – Eij)2
 You have just developed the formula   for 2 Statistic:
                                                           2 = S
                                                                       Eij

2 can be intuitively viewed as:
   An index that shows how much the observed frequencies are in agreement
   with (or apart from) the expected frequencies (for when the null is
   assumed to be true).

So, let’s compute 2 statistic for our example:                              32
                                        •
       Chi-Square Test of Independence

                          Male                   Female          TOTAL
Smoker                    O11 = 15               O12 = 25        40
                          E11 = 8                E12 = 32

Nonsmoker                 O21 = 5                O22 = 55        60
                          E21 = 12               E22 = 48

TOTAL                          20                   80           n = 100


       (15 – 8)2       (25 – 32)2       (5 – 12)2        (55 – 48)2
2 =               +                +               +                 = 12.76
          8               32                12              48

                                                                                33
   •Chi-Square Test of Independence
Let’s Review:
 Obtaining a small 2 value means?
   – Observed frequencies are in close agreement with what we would expect
     them to be if there were no differences between our comparison groups.

   – That is, there is a strong likelihood that no difference exists between the
     percentages of males and females who smoke.

   – Hence, we would be running a significant risk of being wrong if we were to
     reject the null hypothesis. That is, a is expected to be relatively large.
      • Therefore, we should NOT reject the null.

      › NOTE: Smaller 2 values result in larger a levels (if n remains the same).


 A large 2 value means?
                                                                                 34
    Chi-Square Test of Independence
 A large 2 value means:
 – Observed frequencies are far apart from what they ought
   to be if the null hypothesis were true.

 – That is, there is a strong likelihood for existence of a difference
   in the percentages of male and female smokers.

 – Hence, we would be running a small risk of being wrong if we
   were to reject the null hypothesis. That is, a is likely to be small.
    • Thus, we should reject the null.
    › NOTE: larger 2 values result in smaller a levels (if n remains the same).

But, how large is large?
For example, does 2 = 12.76 represent a large enough departure (of
observed frequencies) from expected frequencies to warrant
rejecting the null? Check out the associated a level!
                                                                          35
a reflects whether 2 is large enough to warrant rejecting the null.
  •Chi-Square Test of Independence
 Answer:
  – Consult the table of probability distribution for 2 statistic to see
    what the actual value of a is (i.e., what is the probability that our
    2 value is not large enough to be considered significant).

  – That is, look up the alevel associated with your 2 value (under
    appropriate degrees of freedom).

      • Degrees of Freedom: df = (r-1) (c-1)

                              df = (2 – 1) (2 – 1) = 1
        where r and c are # of rows and columns of the contingency
        table.


                                                                    36
     a




37
          •Chi-Square Test of Independence
    From the table, the a level for 2 = 10.83 (with df = 1) is 0.001 .

      Our 2 = 12.76 > 10.83
 QUESTION: for our 2 = 12.76 will a be smaller or greater than 0.001?
           • Smaller than 0.001
           • Therefore, If we reject the null, the odds of being wrong will be even
             smaller than 1 in 1000.

     Can we afford to reject the null? Is it safe to do so?

     CONCLUSION?
        –% of males and females who smoke are not equal.
        –That is, smoking is a function of gender.
        –Can we be more specific?
            »Percentage of males who smoke is significantly larger than that of the
            females (75% vs. 31%, respectively)

                                                                              38
• CAUTION: Select the appropriate percentages to report (Row% vs. Column%)
                                         •
        Chi-Square Test of Independence

                              Male              Female          TOTAL
 Smoker                       O11 = 15          O12 = 25        40



 Nonsmoker                    O21 = 5           O22 = 55        60



 TOTAL                          20                 80           n = 100



                              15 / 20 = 75%     25 / 80 = 31%

Phi (a non-parametric correlation for categorical data):
        Φ=       χ2 / N   =      12.76 / 100 = 0.357 (Note: sign is NA)   39
     Chi-Square Test of Independence
VIOLATION OF ASSUMTIONS:
  2 test requires expected frequencies (Eij) to be reasonably large. If
    this requirement is violated, the test may not be applicable.

  SOLUTION:
  – For 2 x 2 contingency tables (df = 1), use the Fisher’s Exact
    Probability Test results (automatically reported by SPSS).
      That is, look up a of the Fisher’s exact test to arrive at your conclusion.

  – For larger tables (df > 1), eliminate small cells by combining
    their corresponding categories in a meaningful way.
      That is, recode the variable that is causing small cells into a
       new variable with fewer categories and then use this new
       variable to redo the Chi-Square test.                      40
•Chi-Square Test of Independence

Let’s now use SPSS to do the same analysis!
                     Menu Bar: Analyze, Descriptive
                     Statistics, Crosstabs
                     Statistics: Chi-Square, Contingency
                     Coefficient.
                     Cells: Observed, Row/Column
                     percentages (for the independent
                     variable)

                     SPSS File: smoker

                     SPSS File: GSS93 Subset

                                                      41
    Chi-Square Test of Independence

 Suppose we wish to examine the validity of the “gender
  gap hypothesis” for the 1992 presidential election between
  Bill Clinton, George Bush, and Ross Perot.

             SPSS File: Voter




                                                           42
Correlation Coefficient (Pearson Correlation) = r

What do you use it for?           Karl Pearson, a Galton
                                  Student & the Founder
                                  of Modern Statistics
To examine whether a relationship exists between
two metric variables (e.g., income and education, or
workload and job satisfaction) and what the nature
and strength of that relationship may be.

Range of Values for r?

-1 < r < +1
Null Hypothesis when using r?
r = 0 (There is no relationship between the two variables.)
                                                              43
44
Correlation Coefficient:
To understand the practical meaning of r, we can square it.
   • What would r2 mean/represent?
 r2 Represents the proportion (%) of the total/combined variation in
    both x and y that is accounted for by the joint variation (covariation)
    of x and y together (x with y and y with x)

 • How is it calculated?
 r2 = (Covariation of X and Y together) / (Total variation of X & Y combined)
 How do we measure/quantify variations?
                                               2
                ( x  x)( y  y) / n  1
               
                                         
                                          
  r2 
         [ ( x  x) 2 / n  1][  ( y  y) 2 / n  1]


  r 
              ( x  x)( y  y)                     • r2 always represents a %
                         2                 2        • Why do we show more interest in
            ( x  x)  ( y  y )                     r , rather than r2?               45
Correlation Coefficient: Computation
 r2 = (Covariation of X and Y together) / (All of variation of X & Y combined)
                    (x    x)( y  y)
           r 
                  (x    x) 2   (y    y) 2
                  _               _             _       _          _            _
 X      Y    X–X              Y–Y           (X – X) (Y – Y)    (X – X)2    (Y – Y)2
 4      12   -3              -4                  12              9            16
 6      19   -1               3                  -3              1             9
 9      14    2              -2                  -4              4             4
 .      .     .               .                  .                .              .
 .      .     .               .                  .                .              .
 .      .     .               .                  .                .              .
 _      _                                          _      _            _           _
 X=7    Y=16                                ∑(X – X) (Y – Y)    ∑ (X – X)2 ∑ (Y – Y)2

 NOTE: Once r is calculated, we need to see if it is statistically
       significant (if sample data). That is, we need to test H0: r = 046
Correlation Coefficient?
  Suppose the correlation between X (say, Students’ GMAT Scores)
  and Y (their 1st year GPA in MBA program) is r = +0.48 and is
  statistically significant. How would we interpret this?
 a) GMAT score and 1st year GPA are positively related
    so that as values of one variable increase, values of
    the other also tend to increase, and

 b) 23% of variations/differences in students’ GPAs are
    explained by (or can be attributed to) variations/
    differences in their GMAT scores.
            Lets now practice on SPSS
   Menu Bar: Analyze, Correlate, Bivariate, Pearson

   Using data in SPSS File Salary.sav we wish
   to see if beginning salary is related to seniority,
                                                             47
   age, work experience, and education
    STATITICAL DATA ANALYSIS
COMMON TYPES OF ANALYSIS:

 – Examine Strength and Direction of
   Relationships
   • Bivariate (e.g., Pearson Correlation—r)
     Between one variable and another: Y = a + b1 x1
   • Multivariate (e.g., Multiple Regression Analysis)
     Between one dep. var. and an independent variable, while
      holding all other independent variables constant:
             Y = a + b1 x1 + b2 x2 + b3 x3 + … + bk xk
 – Compare Groups
   • Proportions (e.g., Chi Square Test—2)
   • Means (e.g., Analysis of Variance)                     48
          STATITICAL DATA ANALYSIS

Chi-Square Test of Independence?
 To examine whether proportions of different groups
  of subjects (e.g., managers vs operatives) are
  equal/different across two or more categories (e.g.,
  males vs females).
 To examine whether or not a relationship exists
  between two categorical/nominal variables (e.g.,
  employee status and gender)--categorical dependent
  variable, categorical independent variable.
   – EXAMPLE?
 Is smoking a function of gender? That is, is there a
  difference between the percentages of males and
  females who smoke?

                                                         49
       •Chi-Square Test of Independence
 Research Sample:
                 ID     Gender           Smoking Status
                 1      0 = Male         1 = Smoker
                 2      1 = Female       0 = Non-Smoker
                 3      1                1
                 4      1                0
                 5      0                0
                 .      .                .
                 .      .                .
                 .      .                .
                 100    1                0
 dependent variable (smoking status) and the independent variable (gender)
  are both categorical.
 Null Hypothesis?
   H0: There is no difference in the percentages of males and females
       who smoke (Smoking is not a function of gender).
                                                                  50
   QUESTION: Logically, what would be the first thing you would do?
     •Chi-Square Test of Independence
H0: There is no difference in the percentages of males and females who
smoke (Smoking is not a function of gender).
H1: The two groups are different with respect to the proportions who smoke.


TESTING PROCEDURE AND THE INTUITIVE LOGIC:

Construct a contingency Table: Cross-tabulate the
 observations and compute Observed (actual) Frequencies (Oij ) :

                 Male          Female         TOTAL
Smoker           O11 = 15      O12 = 25       40
Nonsmoker        O21 = 5       O22 = 55       60

TOTAL              20             80          n = 100

                                                                    51
                                •
     Chi-Square Test of Independence
 Next, ask yourself: What numbers would you expect to find in
  the table if you were certain that there was absolutely no
  difference between the percentages of males and females who
  smoked? That is, compute the Expected Frequencies (Eij ).
 Hint: What % of all the subjects are smokers/non-smokers?

                     Male            Female      TOTAL
Smoker               O11 = 15        O12 = 25    40

Nonsmoker            O21 = 5         O22 = 55    60

TOTAL                 20               80        n = 100



                                                           52
                                 •
     Chi-Square Test of Independence
 If there were absolutely no differences between the two groups
  with regard to smoking, you would expect 40% of individuals
  in each group to be smokers (and 60% non-smokers).
 Compute and place the Expected Frequencies (Eij ) in the
  appropriate cells:

                      Male             Female     TOTAL
Smoker                O11 = 15         O12 = 25   40
                      E11 = 8          E12 = 32

Nonsmoker             O21 = 5          O22 = 55   60
                      E21 = 12         E22 = 48
TOTAL                   20                80      n = 100

NOW WHAT? What is the next logical step?                     53
                                 •
      Chi-Square Test of Independence
Compare the Observed and Expected frequencies—i.e.,
examine the (Oij – Eij) discrepancies.
                      Male             Female      TOTAL
 Smoker               O11 = 15         O12 = 25    40
                      E11 = 8          E12 = 32

 Nonsmoker            O21 = 5          O22 = 55    60
                      E21 = 12         E22 = 48

 TOTAL                  20               80        n = 100


QUESTION: What can we infer if the observed frequencies happen
to be reasonably close (or identical) to the expected frequencies?
                                                             54
       •Chi-Square Test of Independence
If the observed frequencies were reasonably close to the expected
frequencies:
 – Reasonably certain that no difference exists between percentages of
   males and females who smoke,
 – Good chance that H0 is true
      • That is, we would be running a large risk of being wrong if
        we decide to reject it.

On the other hand, the farther apart the observed frequencies happen to be
from their corresponding expected frequencies:
 – The greater the chance that percentages of males and females who smoke
     would be different,
 – Good chance that H0 is false and should be rejected
      •    That is, we would run a relatively small risk of being wrong
          if we decide to reject it.

 So, the key to answering our original question lies in the size of the
   discrepancies between observed and expected frequencies.
   What is, then, the next logical step?
                                                                             55
        Chi-Square Test of Independence
 Compute an Overall Discrepancy Index: To quantify the overall
discrepancy between observed (Oij). and expected (Eij). frequencies,
we can add up all our cell discrepancies--i.e., compute S (Oij – Eij).
• Problem?
  Positive and negative values of (Oij – Eij) RESIDUALS for different cells will
  cancel out.

• Solution?
  Square each (Oij – Eij) and then sum them up--compute S(Oij – Eij)2.

• Any Other Problems?
  Value of S(Oij – Eij)2 is impacted by sample size (n).
   – For example, if you double the number of subjects in each cell, even though cell
      discrepancies remain proportionally the same, the above discrepancy index will

      be much larger and may lead to a different conclusion.    Solution?      56
     Chi-Square Test of Independence

• Divide each (Oij – Eij)2 value by its corresponding Eij value
  before summing them up across all cells
   • That is, compute the total discrepancy per subject index.
                        (Oij – Eij)2
                    S
                            Eij

                                                                      (Oij – Eij)2
 You have just developed the formula     for 2 Statistic:
                                                             2 = S
                                                                         Eij

2 can be intuitively viewed as an index that shows how much the
observed frequencies are in agreement with (or apart from) the
expected frequencies (when the null is assumed to be true).

    So, let’s compute 2 statistic for our example:                            57
                                        •
       Chi-Square Test of Independence

                          Male                   Female          TOTAL
Smoker                    O11 = 15               O12 = 25        40
                          E11 = 8                E12 = 32

Nonsmoker                 O21 = 5                O22 = 55        60
                          E21 = 12               E22 = 48

TOTAL                          20                   80           n = 100


       (15 – 8)2       (25 – 32)2       (5 – 12)2        (55 – 48)2
2 =               +                +               +                 = 12.76
          8               32                12              48

                                                                                58
   •Chi-Square Test of Independence
Let’s Review:
 Obtaining a small 2 value means?
   – Observed frequencies are in close agreement with what we would expect
     them to be if there were no differences between our comparison groups.

   – That is, there is a strong likelihood that no difference exists between the
     percentages of males and females who smoke.

   – Hence, we would be running a significant risk of being wrong if we were to
     reject the null hypothesis. That is, a is expected to be relatively large.
      •   Therefore, we should NOT reject the null.


 A large 2 value means?


                                                                          59
    Chi-Square Test of Independence
 A large 2 value means:
 – Observed frequencies are far apart from what they ought
   to be if the null hypothesis were true

 – That is, there is a strong likelihood for existence of a difference
   in the percentages of male and female smokers.

 – Hence, we would be running a small risk of being wrong if we
   were to reject the null hypothesis. That is, a is likely to be
   small.
    • Thus, we should reject the null.

But, how large is large?
For example, does 2 = 12.76 represent a large enough departure (of
observed frequencies) from expected frequencies to warrant
rejecting the null?
                                                                         60
  •Chi-Square Test of Independence
 Answer:
  – Consult the table of probability distribution for 2 statistic to
    see what the actual value of a is (i.e., what is the probability
    that it is not large enough to be considered significant).

  – That is, look up the a level associated with your 2 value
    (under appropriate degrees of freedom).

      • Degrees of Freedom: df = (r-1) (c-1)
                                df = (2 – 1) (2 – 1) = 1
        where r and c are # of rows and columns of the contingency
        table.



                                                                  61
     a




62
          •Chi-Square Test of Independence
    From the table, the a level for 2 = 10.83 (with df = 1) is 0.001 .

      Our 2 = 12.76 > 10.83
 QUESTION: If we decide to reject the Null, will a be smaller or greater than 0.001?
           • Smaller than 0.001
           • Therefore, If we reject the null, the odds of being wrong will be even
             smaller than 1 in 1000.

     Can we afford to reject the null? Is it safe to do so?

     CONCLUSION?
        –% of males and females who smoke are not equal.
        –That is, smoking is a function of gender.
        –Can we be more specific?
            »Percentage of males who smoke is significantly larger than that of the
            females (75% vs. 31%, respectively)

                                                                               63
• CAUTION: Select the appropriate percentages to report (Row% vs. Column%)
                                          •
        Chi-Square Test of Independence

                             Male                 Female          TOTAL
 Smoker                      O11 = 15             O12 = 25        40



 Nonsmoker                   O21 = 5              O22 = 55        60



 TOTAL                         20                   80            n = 100



                             15 / 20 = 75%        25 / 80 = %31

Phi (a non-parametric correlation for categorical data):
         Φ=      χ2 / N                                                     64
     Chi-Square Test of Independence
VIOLATION OF ASSUMTIONS:

   2 test requires expected frequencies (Eij) to be reasonably large.
     If this requirement is violated, the test may not be applicable.
SOLUTION:

   – For 2 x 2 contingency tables (df = 1), use the Fisher’s Exact
     Probability Test results (automatically reported by SPSS).
      • That is, look up a of the Fisher’s exact test



   – For larger tables (df > 1), eliminate small cells by combining
     their corresponding categories in a meaningful way.         65
•Chi-Square Test of Independence

Let’s now use SPSS to do the same analysis!
                     Menu Bar: Analyze, Descriptive
                     Statistics, Crosstabs
                     Statistics: Chi-Square, Contingency
                     Coefficient.
                     Cells: Observed, Row/Column
                     percentages (for the independent
                     variable)

                     SPSS File: smoker

                     SPSS File: GSS93 Subset

                                                      66
    Chi-Square Test of Independence

 Suppose we wish to examine the validity of the “gender
  gap hypothesis” for the 1992-93 presidential elections
  between Bill Clinton, George Bush, and Ross Perot.

            SPSS File: Voter




                                                           67
                        Assignment #3
1. As a population demographer, you have long suspected that women’s fertility rate in
   different countries (fertility--average number of children born to a woman) would be
   related to male and female literacy rates (lit-male and lit_fema), access to health care, as
   characterized by number of hospital beds per 10,000 people (hospbed) and number of
   doctors per 10,000 people(docs), infant mortality rate (babymort--number of deaths per
   1,000 live births), as well as male and female life expectancies(lifeexpm and lifeexpf).
   The data file “World95.sav” (in the U drive) contains 1995 population and socio-
   economic statistics from 109 different countries, including statistics on all of the above-
   mentioned variables. Please use the data to test your suspicion (i.e., see fertility rate is
   correlated with which of the above variables and in what way).




                                                                                           68
                      Assignment #3
2. Suppose you are a medical researcher and you wish to examine whether there is a
relationship between incidents of coronary heart disease (CHD) and family history of
CHD. Specifically, if CHD has in part a genetic component. In other words, you wish to
know if incidents of CHD is proportionally higher among men with a family history of
CHD, than among men without such family history. Researchers in the Western Electric
Study have collected data on such issues using a sample of 240 men, ½ with and ½ without
prior incidents of CHD (variable chd). All these men are by now deceased. Included in
the data set is information on whether or not each subject has had a history of CHD in his
immediate family (variable famhxcvr), as well as the day of the week when the subject’s
death has occurred (variable dayofwk). The data is available in the electric.sav SPSS data
file. (Note: Due to SPSS site license restrictions, this hyperlink will not work if you are
off campus).

a) Please conduct the appropriate statistical test to address the above research objective.

b) Suppose you have been noticing that for most people earlier working days of the week
   (especially Mondays, Tuesdays, and Wednesday) appear to be more stressful in
   comparison with Fridays, Saturdays, and Sundays. Also, suppose you have come
   across prior research that indicates stress and CHD tend to create a deadly
   combination. As such, you have recently begun to suspect (hypothesize) that a larger
   percentage of men with CHD, compared with those without CHD, tend to die during
   Mondays, Tuesdays, and Wednesdays (as opposed to Fridays, Saturdays, and
   Sundays). That is, you suspect that having CHD increases likelihood of dying during
                                                                                     69
   the more stressful days of the week. Please perform the necessary analysis to verify
   the validity of your suspicion (plausibility of your hypothesis).
                             Assignment #3
NOTE:
 If you examine the value labels for the variable daysofwk, you will see that it is coded
 as 1=Sunday, 2=Monday, 3=Tuesday, 4=Wednesday, 5=Thursday, 6=Friday, and
 7=Saturday. Therefore, for part (b), you will need to create a new variable--i.e., Recode
 daysofwk into a new dichotomous variable (say, deathday), that would represent death
 during Mondays, Tuesdays, and Wednesdays vs. Fridays, Saturdays and Sundays.
 Notice that the subjects who died on Thursdays should not be included in the analysis
 (i.e., should not be represented in any of the two categories of days represented by the
 new variable) Also, make sure you properly define the attributes (e.g., label, value
 label, etc.) of this new variable (i.e., deathday).

REMINDERS:
 For each analysis, include the Notes in the printout. Also, edit the first page of your
 first analysis output to include your name. Make sure that on your printout you
 explain your findings and conclusions. Be specific as to what parts of the output you
 have used, and how you have used them, to reach your conclusions.

  Make sure that you tell the whole story and that your explanations of the findings are
  complete. For example, it is not enough to say that there is a significant relationship
  between characteristic A and characteristic B. You have to go on to indicate how the
  two characteristics are related and what that relationship really means.

                                                                                     70
  HYPOTHESIS TESTING




QUESTIONS OR
  COMMENTS

     ?
                       71

				
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