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```									      Economics 105: Statistics
• Any questions?
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
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

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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TI FF( U Q uipr issed) anom pr essr
e         c          o
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ar e neded t osee t hi pt ur e.

• 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)
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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• 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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