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									      Economics 105: Statistics
• Any questions?
• Please read Ch 5 of Freakonomics
                      Two-Sample Tests

                        Two-Sample Tests


 Population          Population        Population
   Means,                             Proportions,         Population
                      Means,                               Variances
Independent           Related         Independent
  Samples             Samples           Samples
Examples:
Population 1 vs.   Same population    Proportion 1 vs.   Variance 1 vs.
independent        before vs. after   independent        Variance 2
Population 2       treatment          Proportion 2
      2 Test for Differences Among
       More Than Two Proportions

• Extend the 2 test to the case with more
  than two independent populations:
     H0: π1 = π2 = … = πc
     H1: Not all of the πj are equal (j = 1, 2, …, c)
 The Chi-Square Test Statistic
The Chi-square test statistic is:

                              (fo  fe )2
                   
                  2

                    all cells     fe
• where:
   fo = observed frequency in a particular cell of the 2 x c table
   fe = expected frequency in a particular cell if H0 is true
  2 for the 2 x c case has (2-1)(c-1) = c - 1 degrees of freedom
(Assumed: each cell in the contingency table has expected frequency of
   at least 1)
               Computing the
              Overall Proportion
    The overall        X1  X 2    Xc X
                    p                  
     proportion is:    n1  n2    nc   n

• Expected cell frequencies for the c categories
  are calculated as in the 2 x 2 case, and the
  decision rule is the same:
  Decision Rule:            Where 2U is from the
  If 2 > 2U, reject H0,   chi-squared distribution
  otherwise, do not         with c – 1 degrees of
  reject H0                 freedom
     The Marascuilo Procedure
• Used when the null hypothesis of equal
  proportions is rejected
• Enables you to make comparisons between
  all pairs
• Start with the observed differences, pj – pj’,
  for all pairs (for j ≠ j’) . . .
• . . .then compare the absolute difference to a
  calculated critical range
            The Marascuilo Procedure                                    (continued)
    • Critical Range for the Marascuilo Procedure:
                                           p j (1 p j ) p j' (1 p j' )
            Critical range           2
                                       U                
                                                nj             n j'

    • (Note: the critical range is different for each pairwise comparison)
   • A particular pair of proportions is significantly
   different if
         | pj – pj’| > critical range for j and j’
           Marascuilo Procedure Example
A University is thinking of switching to a trimester academic
calendar. A random sample of 100 administrators, 50 students,
and 50 faculty members were surveyed

 Opinion    Administrators   Students   Faculty
 Favor           63             20        37
 Oppose          37             30        13
 Totals         100             50        50


  Using a 1% level of significance, which groups have a
  different attitude?
             Marascuilo Procedure: Solution
                                        Excel Output:
                                                             compare
Marascuilo Procedure

       Sample      Sample              Absolute Std. Error Critical
Group Proportion    Size    Comparison Difference of Difference Range Results
    1        0.63      100 1 to 2             0.23 0.08444525 0.256 Means are not different
    2          0.4       50 1 to 3            0.11 0.07860662 0.239 Means are not different
    3        0.74        50 2 to 3            0.34 0.09299462 0.282 Means are different

                                   Other Data
Level of significance       0.01                Chi-sq Critical Value   9.21
d.f                            2
Q Statistic             3.034856



   At 1% level of significance, there is evidence of a difference
   in attitude between students and faculty
        2 Test of Independence
• Similar to the 2 test for equality of more than two
  proportions, but extends the concept to
  contingency tables with r rows and c columns



H0: The two categorical variables are
 independent (i.e., there is no relationship between them)
H1: The two categorical variables are dependent
      (i.e., there is a relationship between them)
      2 Test of Independence                           (continued)


The Chi-square test statistic is:

                             (fo  fe )2
               2  
                   all cells     fe
• where:
   fo = observed frequency in a particular cell of the r x c
   table
   fe = expected frequency in a particular cell if H0 is true
  2 for the r x c case has (r-1)(c-1) degrees of freedom
(Assumed: each cell in the contingency table has expected
   frequency of at least 1)
     Expected Cell Frequencies
• Expected cell frequencies:
                      row total  column total
                 fe 
                                  n

 where:
 row total = sum of all frequencies in the row
 column total = sum of all frequencies in the column
  n = overall sample size
               Decision Rule
• The decision rule is
           If 2 > 2U, reject H0,
           otherwise, do not reject H0


    Where 2U is from the chi-squared distribution
    with (r – 1)(c – 1) degrees of freedom
                      Example
• The meal plan selected by 200 students is shown below:



    Class   Number of meals per week
  Standing 20/week 10/week     none               Total
  Fresh.      24      32        14                 70
  Soph.       22      26        12                 60
  Junior      10      14         6                 30
  Senior      14      16        10                 40
    Total     70      88        42                 200
                    Example
                                            (continued)

• The hypothesis to be tested is:

H0: Meal plan and class standing are
 independent (i.e., there is no relationship
 between them)
H1: Meal plan and class standing are dependent
     (i.e., there is a relationship between them)
                            Example:
                     Expected Cell Frequencies
                                                                             (continued)
               Observed:
 Class          Number of meals
                   per week
Standin                                                Expected cell
   g         20/wk 10/wk      none   Total
                                                       frequencies if H0 is true:
Fresh.         24     32      14      70
Soph.          22     26      12      60                   Number of meals
                                               Class          per week
Junior         10     14       6      30
Senior         14     16      10      40     Standing 20/wk 10/wk       none     Total
 Total         70     88      42     200     Fresh.       24.5   30.8   14.7      70
  Example for one cell:                      Soph.        21.0   26.4   12.6      60
             row total  column total        Junior       10.5   13.2   6.3       30
    fe 
                         n                   Senior       14.0   17.6   8.4       40
             30  70                           Total      70     88     42        200
                     10.5
              200
      Example: The Test Statistic (continued)
• The test statistic value is:
                 ( fo  fe )2
   2  
       all cells      fe

         (24  24.5)2 (32  30.8)2     (10  8.4)2
                                               0.709
             24.5         30.8             8.4

   2U = 12.592 for  = 0.05 from the chi-squared
   distribution with (4 – 1)(3 – 1) = 6 degrees of
   freedom
                   Example:
           Decision and Interpretation(continued)
The test statistic is  2  0.709 , U with 6 d.f.  12.592
                                     2



                                   Decision Rule:
                                   If 2 > 12.592, reject H0,
                                   otherwise, do not reject H0

                                          Here,
                        
                                          2 = 0.709 < 2U = 12.592,
                                          so do not reject H0
  0
      Do not           Reject H0   2    Conclusion: there is not
      reject H0                           sufficient evidence that meal
                  2U=12.592
                                          plan and class standing are
                                          related at  = 0.05
        Two-Sample Tests in Excel
For independent samples:
• Independent sample Z test with variances known:
    – Data | data analysis | z-test: two sample for means
• Pooled variance t test:
    – Data | data analysis | t-test: two sample assuming equal variances
• Separate-variance t test:
    – Data | data analysis | t-test: two sample assuming unequal variances


For paired samples (t test):
    – Data | data analysis | t-test: paired two sample for means

For variances:
• F test for two variances:
    – Data | data analysis | F-test: two sample for variances
    Wilcoxon Rank-Sum Test for
     Differences in 2 Medians
• Test two independent population medians
• Populations need not be normally distributed
• Distribution free procedure
• Used when only rank data are available
• Must use normal approximation if either of
  the sample sizes is larger than 10
       Wilcoxon Rank-Sum Test:
            Small Samples
• Can use when both n1 , n2 ≤ 10
• Assign ranks to the combined n1 + n2 sample
  observations
   –   If unequal sample sizes, let n1 refer to smaller-sized
       sample
   –   Smallest value rank = 1, largest value rank = n1 + n2
   –   Assign average rank for ties
• Sum the ranks for each sample: T1 and T2
• Obtain test statistic, T1 (from smaller sample)
       Checking the Rankings
• The sum of the rankings must satisfy the
  formula below
• Can use this to verify the sums T1 and T2

                      n(n  1)
            T1  T2 
                         2

              where n = n1 + n2
            Wilcoxon Rank-Sum Test:
           Hypothesis and Decision Rule
      M1 = median of population 1; M2 = median of population 2

        Test statistic = T1 (Sum of ranks from smaller sample)

   Two-Tail Test                Left-Tail Test           Right-Tail Test
    H0: M1 = M2                 H0: M1  M2               H0: M1  M2
    H1: M1  M2                 H1: M1 < M2               H1: M1 > M2

Reject Do Not      Reject   Reject    Do Not Reject   Do Not Reject      Reject
       Reject
     T1L       T1U                T1L                                 T1U
 Reject H0 if T1 < T1L        Reject H0 if T1 < T1L     Reject H0 if T1 > T1U
         or if T1 > T1U
   Wilcoxon Rank-Sum Test:
    Small Sample Example
Sample data are collected on the capacity rates (% of
  capacity) for two factories.
Are the median operating rates for two factories the
  same?
• For factory A, the rates are 71, 82, 77, 94, 88
• For factory B, the rates are 85, 82, 92, 97
Test for equality of the population medians
  at the 0.05 significance level
           Wilcoxon Rank-Sum Test:
            Small Sample Example
                      Capacity                    Rank           (continued)
  Ranked
  Capacity       Factory A   Factory B   Factory A   Factory B
  values:           71                      1
                    77                      2
Tie in 3rd and      82                     3.5
   4th places                   82                       3.5
                                85                        5
                    88                      6
                                92                        7
                    94                      8
                              97                          9
                         Rank Sums:        20.5          24.5
    Wilcoxon Rank-Sum Test:
     Small Sample Example (continued)
Factory B has the smaller sample size, so
  the test statistic is the sum of the
  Factory B ranks:
                T1 = 24.5

          The sample sizes are:
                n1 = 4 (factory B)
                n2 = 5 (factory A)
          The level of significance is  = .05
                 Wilcoxon Rank-Sum Test:
                  Small Sample Example
                                                           (continued)
   Lower and
    Upper
                                                  n1
    Critical       n2   One- Two-
                                            4              5
    Values for          Tailed Tailed
    T1 from         4
    Appendix
    table E.8,           .05        .10   12, 28         19, 36
    page 829            .025        .05   11, 29         17, 38
    in BLK          5
                         .01        .02   10, 30         16, 39
    10e:
                        .005        .01    --, --        15, 40
                    6
                               T1L = 11 and T1U = 29
                 Wilcoxon Rank-Sum Test:
                  Small Sample Solution
                                                     (continued)

  = .05                   Test Statistic (Sum of
• n1 = 4 , n 2 = 5          ranks from smaller sample):
   Two-Tail Test            T1 = 24.5
    H0: M1 = M2
    H1: M1  M2
                            Decision:
Reject Do Not     Reject
                            Do not reject at  = 0.05
       Reject
T1L=11        T1U=29        Conclusion:
                            There is not enough evidence to
 Reject H0 if T1 < T1L=11   conclude that the pop medians are
        or if T1 > T1U=29   different.
             Wilcoxon Rank-Sum Test
                 (Large Sample)
     • For large samples, the test statistic T1 is
       approximately normal with mean T and standard
                                               1

       deviation  T :
                    1




                n1(n 1)
          T1                    T  n1 n 2 (n  1)
                                  1
                 2                          12
       – Must use the normal approximation if either n1
         or n2 > 10
     – Assign n1 to be smaller of the two sample sizes
                         the
       – Could use the normal approximation for small samples
            Wilcoxon Rank-Sum Test
                (Large Sample)
                                         (continued)

     • The Z test statistic is
                            n1(n  1)
              T1  T1 T1       2
          Z          
                 T1     n1n 2 (n  1)
                              12
     • Where Z ~ N(0,1)

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                                                    Interpreting Electoral Polls*                                                      FF( n
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                                       •     WASHINGTON (Reuters - 09/09/00) - A new Newsweek poll on
                                             Saturday showed Vice President Al Gore maintaining a strong lead
                                             over Texas Gov. George Bush in the presidential race, but a
                                             CNN/USA Today survey found the candidates virtually tied.

                                       •     According to the Newsweek poll, Democratic nominee Gore leads
                                             Republican nominee Bush 47 percent to 39 percent among registered
                                             voters, with Green Party candidate Ralph Nader at 3 percent and
                                             Reform Party candidate Pat Buchanan at 1 percent.

                                       •     Among likely voters, Gore led Bush 49 percent to 41 percent, the
                                             same margin as among registered voters.

                                       •     The poll was conducted by Princeton Survey Research Associates
                                             Sept. 7-8 among 756 registered voters, including 595 who said they
                                             were likely to vote in the election.

                                       •     The margin of error was 4 percentage points for the survey of
                                             registered voters and 5 percentage points for likely voters.
                                       *Source: http://www.kellogg.northwestern.edu/faculty/weber/decs-433/Presidential_Polls.htm
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                                             Interpreting Electoral Polls
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                                       • Where do the 4% points and 5% points come from?
                                       • Recall …
                                                                 p(1 p)
                                                        p  1.96
                                                                    n
                                                                                       1
                                       • A rough calculation for margin of error is
                                                                                        n
                                       • 
                                          Sample Size        Margin of Error          1
                                              10,000                .01                    .0364
                                               2,500                .02               756
                                               1,112                .03
                                                                                    1
                                                 625                .04
                                                                                           .0410
                                                 400                .05               595
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                                           Interpreting Electoral Polls
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        • What are the precise margins of error on each candidate’s
          support?
        • Candidate Sample p        Precise Margin of error
          Gore          47                  3.55 % points
          Bush          39                  3.47                  p(1 p)
          Nader          3                   1.21         1.96
                                                                    756
          Buchanan       1                    .71
        • Conclusions
           1 is upper bound for margin of error at 95% confidence level
                                       n                    

        • But it is much too big for proportions far from 50%
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                                              Interpreting Electoral Polls
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• Is Gore statistically ahead?
• Rule of thumb to use when watching Fox/MSNBC/etc:
                                       – Double the margin of error reported in the news article &
                                         compare that to the difference in sample proportions
                            if Gore
                            1,                                                  n
                           
                     X i   if neither
                            0,                                                 X       i

                            if Bush                                     D   i1
                            1,
                                                                                  n

• D will then be the difference between the proportion of voters
  supporting Gore and the proportion supporting Bush
                            
• Find E[D] and Var[D]
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                                               Interpreting Electoral Polls
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• Is Gore statistically ahead? Yes …
• Rule of thumb to use when watching Fox/MSNBC/etc:
                                        – Double the margin of error reported in the news article &
                                          compare that to the difference in sample proportions
                                                   E[D]  .47.39  .08
                                      .47(1 .47)  .39(1 .39)  2(.47)(.39)
                          StdDev[D]                                           .0336
                                                        756
 95% CI for the “lead” .08  1.96 * (.0336)
•


                         (.0141, .1458)

								
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