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Module III
Lecture 4
Test of Hypotheses For More Than Two Independent Samples
Suppose that you recently read an article that indicated that the tempo of
music can affect a consumer’s behavior. In particular, it is conjectured that the
faster the tempo (speed) of the music, the more likely a consumer is to make a
purchase.
Since you are a V.P at George Giant Food Stores, you pick a random sample
of 15 stores. You divide the stores into three set of five. At five stores you play no
music. At five other stores you play pleasant but slow music. At the remaining five
stores you play fast light music. For randomly chosen days, you measure the
volume of sales (in purchases not dollars) at the 15 stores.
The resulting data is given below:
Daily Supermarket Purchases
Music
Slow
Fast
None
14,140
13,991
14,772
13,266
14,040
15,029
14,656
16,029
14,783
14,700
12,874
13,165
13,140
11,245
12,400
Is there evidence that the tempo of the music has an effect?
This is an example of a situation where we are interested in whether or not
groups have different means but we have more than two groups.
One approach is to just look at the groups in pairs. That is, use the results of
the previous lecture and compare the groups two at a time. In our case we would
look at the No Music Group versus the Slow Music Group, the No Music Group
versus the Fast Music Group, and the Fast Music Group versus the Slow Music
Group. The results are shown below:
t-Test: Two-Sample Assuming Unequal VariancesSlow vs None
Variable 1 Variable 2
Mean
14041.8 12564.8
Variance
286821.2 638932.7
Observations
5
5
Hypothesized Mean Difference
0
df
7
t Stat
3.432557
P(T<=t) one-tail
0.005474
t Critical one-tail
1.894578
P(T<=t) two-tail
0.010947
t Critical two-tail
2.364623
Using an alpha level of .05, this would indicate that there is a significant
difference between sales in stores that play no music and stores that play fast music.
When comparing the No Music Group and the Fast Music Group, one
obtains:
t-Test: Two-Sample Assuming Unequal Variances Fast vs None
Variable 1 Variable 2
Mean
15039.4 12564.8
Variance
326836.3 638932.7
Observations
5
5
Hypothesized Mean Difference
0
df
7
t Stat
5.630583
P(T<=t) one-tail
0.000395
t Critical one-tail
1.894578
P(T<=t) two-tail
0.00079
t Critical two-tail
2.364623
This indicates at the .05 alpha level, that there is a difference between the No
Music Group and the Fast Music Group.
Finally, we compare the Slow Music Group and the Fast Music Group
obtaining the output:
t-Test: Two-Sample Assuming Unequal Variances Slow vs Fast
Variable 1 Variable 2
Mean
14041.8 15039.4
Variance
286821.2 326836.3
Observations
5
5
Hypothesized Mean Difference
0
df
8
t Stat
-2.8476
P(T<=t) one-tail
0.010779
t Critical one-tail
1.859548
P(T<=t) two-tail
0.021559
t Critical two-tail
2.306006
Again, at the .05 level, this indicates that there is a difference between the Fast
Group and the Slow Group.
Unfortunately, things are not quite so simple. As the number of groups
increases, the number of groups goes up quite rapidly. If there are k groups, the
number of pair wise comparisons which must be made is:
k( k  1 ) / 2
This number increases quite quickly as the table below shows:
k
Pairwise
Comparisons
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
3
6
10
15
21
28
36
45
55
66
78
91
105
Now recall that in the logic of statistical testing, there is an 100 % chance
that we will say that there is a difference when in fact there is none.
Consider performing two independent tests under circumstances where there
are no differences. The correct conclusion is that both tests should accept the null
hypothesis. The probability of this happening is
( 1   )2
Therefore the probability of making at least one error, that is rejecting either one or
both of the true null hypotheses, would be:
1  ( 1   )2
If  = .05, this probability is 1 – (.95)2 = .0975. So if we do two tests at alpha level
.05, we have almost a 1 in ten chance of rejecting a at least one true null hypothesis.
Let E be the probability of rejecting at least one true null hypothesis when
we make k(k-1)/2 independent tests, each at level .
Specifically, E is the probability of rejecting the null hypothesis in the
following situation:
H0: 1 = 2 = . . . . = k
HA: at least one difference i - j different from zero
While  represents the probability of rejecting one of the k(k-1)/2 hypotheses
of the form:
H0 : i   j
H A : i   j
Then from the basic rules of probability, one has that:
 E  1  ( 1   )k ( k 1 ) / 2
Although the pair wise tests are not independent, the following table gives an
indication of how the probability of at least one error is related to the number of
groups, k, if we apply the previous results:
Probability of
Number of Groups at Least One Error
k
alpha = .05
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.0500
0.1426
0.2649
0.4013
0.5367
0.6594
0.7622
0.8422
0.9006
0.9405
0.9661
0.9817
0.9906
0.9954
Thus, even with three groups, as in our example, we have an unacceptably
high probability of rejecting at least one true null hypothesis.
One way around this problem is to use a result due to the Italian probabilist,
Bonferroni. He showed that for m comparisons, whether independent or dependent,
the following relationship always holds:
 E  m
This implies that if we pick

E
m
we could control the overall error of rejecting at least one true null hypothesis.
The procedure is as follows:
1) Pick your overall alpha level of rejecting at least one true null hypothesis,
I will typically choose E = .05 ;
2) If I have k groups and will be comparing the means of the groups two at a
time use:

(since m = k (k-1)/2 in this case).
2(.05 )
k( k  1 )
In our case we have k = 3 and we are comparing each group to each of the
others. I would therefore use:

2(.05 )
 .01667
3( 3  1 )
Recalling the p-values of the pair wise t-tests, the table below shows the
results of testing the various pair wise hypotheses using the Bonferroni approach
which gives us only a .05 chance of making any pair wise error:
Comparison
Two-Sided p-value
None vs Slow
None vs Fast
Slow vs Fast
0.0109
0.0008
0.0216
Any p-value less than .01667 would be declared significant. In this case the
None vs Slow and None vs Fast comparisons are significant. This seems to support
the argument that music makes a difference but there is no difference between slow
and fast music.
For the pair wise differences which show significance using the Bonferroni
approach, a confidence interval on the difference in the means is given by the
equation:
( x i  x j )  t / 2
2
s i2 s j

ni n j
where the t value is chosen based on the computed degrees of freedom given earlier.
First I compute the mean and standard deviation of each group using the
usual EXCEL functions “average” and “stdev”. This results in the table below:
Daily Supermarket Sales
Music
Slow
14,140
13,991
14,772
13,266
14,040
Average
Stan Dev
Size
Fast
15,029
14,656
16,029
14,783
14,700
None
12,874
13,165
13,140
11,245
12,400
14,041.80 15,039.40 12,564.80
535.56
571.70
799.33
5
5
5
Then I use the template from the EXCEL file “twosamp.xls” .
For the Slow vs No Music comparison, the confidence interval is:
Template for Confidence Interval on Positive Difference in Means
mean
sd
n
Approx Degrees of Freedom =
Slow
Music
No
Music
14041.8000
535.5600
5
12564.8000
799.3300
5
Alpha
0.01667
7
Confidence Interval on Positive Difference in Means
131.3042
to
2822.6958
The comparison of Fast Music and No Music is given as:
Template for Confidence Interval on Positive Difference in Means
mean
sd
n
Fast
Music
No
Music
15039.4000
571.7000
5
12564.8000
799.3300
5
Approx Degrees of Freedom =
Alpha
0.01667
7
Confidence Interval on Positive Difference in Means
1100.1275
to
3849.0725
Finally, even though not significant, the comparison between Fast and Slow Music
is:
Template for Confidence Interval on Positive Difference in Means
mean
sd
n
Approx Degrees of Freedom =
Fast
Music
Slow
Music
15039.4000
571.7000
5
14041.8000
535.5600
5
Alpha
0.01667
8
Confidence Interval on Positive Difference in Means
-58.8741
Notice that the confidence interval contains the value of zero.
to
2054.0741
Sometimes when one has performed an analysis like the one above, you
arrive at a set of seemingly conflicting conclusions. For example you might accept
the hypothesis that A = B. You might also accept the hypothesis that B = C. And
then reject the hypothesis A = C !!! This seems to violate logic, but it violates
mathematical logic not statistical logic.
Consider the three probability distributions below:
Comparison of Three Groups
0.5
freq
0.4
A
B
C
0.3
0.2
0.1
6.9
6
5.1
4.2
3.3
2.4
1.5
0.6
-0.3
-1.2
-2.1
-3
0
x
Clearly Groups A and B overlap quite a bit. Groups B and C also overlap
quite a bit. But Groups A and C overlap very little.
With the picture in mind, what the statistical results are saying is that the
data does not provide enough evidence to say that Groups A and B are different.
Similarly, the data does not provide enough evidence to say that Groups A and C
are different, but the data does provide enough evidence to indicate that Groups A
and C are different.
The seeming inconsistency then disappears.
The Analysis of Variance
The Bonferroni procedure described in the previous section is the most
general approach to the multiple sample analysis of group differences in that it
makes few assumptions on the data. If one is willing to make more assumptions
about the data, other methods exist for analyzing the data which if the assumptions
are valid is more “powerful” than the Bonferroni procedure. By more powerful we
mean that they have a higher chance of rejecting the null hypothesis when it is false.
The most used of these alternative procedures is called the “Analysis of
Variance”. It is used in exactly the same situation as the Bonferroni method except
that it only applies when the standard deviations in each of the groups can be
assumed to be the same!
Examining our data below, we see that although the standard deviations in
the Fast Music and Slow Music groups are approximate the same, the standard
deviation of the No Music group is approximately 60% higher.
Average
Stan Dev
Size
Music
Slow
14,140
13,991
14,772
13,266
14,040
Fast
15,029
14,656
16,029
14,783
14,700
None
12,874
13,165
13,140
11,245
12,400
14,042
535.56
5
15,039
571.70
5
12,565
799.33
5
It is somewhat debatable as to whether this method can be used on this data,
but we will use it to illustrate the procedure.
The basic data structure of this problem is given below:
Group
1
Group
2
.
.
.
Group
k
x11
x21
.
.
.
xk1
x12
x22
.
.
.
xk2
.
.
.
.
.
.
.
.
.
.
x1n1
.
.
.
.
.
xknk
x2n2
nk
Size
n1
n2
Mean
x1
x2
.
.
.
xk
Standard
Deviation
s1
s2
.
.
.
sk
The basic hypothesis being tested is:
H0: 1 = 2 = . . . . = k
HA: at least one difference i - j different from zero
The basis of the Analysis of Variance Method (ANOVA) is the fact that one
can estimate the assumed common variance in two ways.
For each group we can form an estimate of the standard deviation by using
the formula:
ni
si 
( x
j 1
ij
 x i )2
ni  1
where,
ni
x i   x ij / ni
j 1
Now since all groups are assumed to have the same variance, we can pool all
k estimates of the sample variance to get one estimate of the common variance as:
k
̂ W2 
(n
i 1
i
 1 )s i2
k
n
i 1
i
k
We will call this the “Within Group” estimate of the common variance since it
is based on the deviations, within each group, of the observations from the group
mean.
The second estimate of the common variance is based on the central
limit theorem. Remember that:
Var( x ) 
2
n
Since we have a mean for each group, it should be possible to estimate the variance
by looking at the variability of the group means “between” the groups. One can
show theoretically that:
k
ˆ B2 
n ( x
i 1
i
i
 x )2
k1
Where,
k
x   ni x i / n
i 1
One can also show that if the null hypothesis that all the groups come from
populations with the same mean is true, then
E ( ˆ W2 )  E ( ˆ B2 )   2
On the other hand, if the null hypothesis is false , then:
E ( ˆ W2 )   2
k
E ( ˆ B2 )   2 
n (
i 1
i
i
  )2
k 1
where,
k
k
i 1
i 1
   ni  i /  ni
Define,
Fobs 
ˆ B2
ˆ W2
If the null hypothesis is true, then this F-ratio should be close to one. On the
other hand if the null hypothesis is false, then this F-ratio should be shifted upward
by an amount that increases as the means of the groups differ more from one
another.
This forms the basis of the so called “F-Test” of the null hypothesis that all
the groups have the same mean as with the alternative hypothesis that at least one
pair of groups have means that differ (actually the alternative is a bit more
complicated, but in practice the above alternative hypothesis will suffice).
The procedure for testing the basic null hypothesis is:
a) Pick E (usually .05);
b) Compute Fobs and find its one-sided p-value;
c) Reject the null hypothesis of the one-sided p-value is less than E, otherwise
accept the null hypothesis that all groups have the same mean.
Fortunately, almost all of the proceeding computations are done
automatically in EXCEL. To perform the analysis, first label each column (group)
in the cell immediately above the first number in the group. Then click on “Tools”,
“Data Analysis”, and then “ANOVA: Single Factor”. You screen should look
something like:
The data (including the group labels) is in cells B6:D11. Be sure to check the
box “Labels in First Row” and finally enter alpha. Then hit “OK”.
The output will look something like:
Anova: Single Factor
SUMMARY
Groups
Slow
Fast
None
Count
5
5
5
Sum
70209
75197
62824
Average
14041.8
15039.4
12564.8
Variance
286821.2
326836.3
638932.7
ANOVA
Source of Variation
SS
Between Groups
15500633
Within Groups
5010361
df
2
12
MS
7750317
417530.1
F
18.5623
Total
14
20510994
P-value
0.000212468
F crit
3.88529
I have highlighted the important variables:
ˆ B2  7 ,750 ,317
ˆ W2  417530.1
Fobs  18.5623
The one-sided p-value is .000212. Since this is much smaller than .05, we
reject the null hypothesis that all the groups have the same mean in favor of the
alternative that at least one pair of groups have different means.
Since we suspect that there are differences between the groups, we need a
procedure to find out which group mean differ from which.
A confidence interval for the difference in means between groups i and j is
given by:
( x i  x j )  t / 2
ˆ W2
ni

ˆ W2
nj
  i   j  ( x i  x j )  t / 2
ˆ W2
ni

ˆ W2
nj
using the same value of alpha that we had used to perform the F-test. The degrees of
freedom for the t distribution is given in the EXCEL ANOVA output in the row
labeled “Within Groups”. In our case this value for the degrees of freedom is shown
below:
ANOVA
Source of Variation
SS
Between Groups
15500633
Within Groups
5010361
df
2
12
Total
14
20510994
MS
7750317
417530.1
F
18.5623
P-value
0.000212468
F crit
3.88529
In our case  = .05, so the appropriate value of t with 12 degrees of freedom
is given by the EXCEL function “tinv” as:
t .025  tinv(.05 ,12 )  2.178813
Further, in our case all the groups are the same size so we only have to compute the
+/- limits once as:
( x i  x j )  2.178813
417530.1 417530.1

5
5
which gives:
( x i  x j )  890.42
This is all summarized in the table below:
Difference in Means
Confidence Interval
Conslusion
Fast vs Slow
997.60
107.18
to
1,888.02
Significant
Slow vs None
1477.00
586.58
to
2367.42
Significant
Fast vs None
2474.60
1584.18
to
3365.02
Significant
The ANOVA analysis thus indicates that all groups are different from one
another leading us to maximize sales by playing fast music with numbers of sales
increasing by between 1,584 to 3,365 per day.
These results differ from the Bonferroni approach only in the comparison of
the Fast vs Slow group. By making an assumption that seems dubious for the data,
the analysis has been changed.
In this case I would stay with the Bonferroni analysis. Further study of the
problem might indicate whether or not the difference between Fast and Slow Music
is real.
The Binomial Distribution with Multiple Groups
The following data was collected by the marketing department at your
company:
Effect of Advertising on Random Sample of Consumers
No Ads
TV
Paper
TV and Paper
Total
Buy
Didn't buy
Total
15
25
12
30
85
90
53
90
100
115
65
120
82
318
400
Here we have four forms of advertising (groups) and a binomial response of "buy"
or "didn’t buy" for each group. We are interested in whether or not the probability
of buying or not buying differs based on the form of advertising seen by the
consumer. Before performing a formal analysis, I computed the probability of
buying or not buying for the four groups with the results shown below:
Prob
Buy
No Ads
TV
Paper
TV and Paper
Total
Prob
Didn't buy
0.150
0.217
0.185
0.250
0.850
0.783
0.815
0.750
0.205
0.795
The results look encouraging since any form of advertising is associated with a
higher probability of buying. Further it looks like advertising both on TV and in
the newspaper is the most effective technique.
Before rushing to judgment however, we need to apply statistical methods to
see if these effects justify major advertising expenditures.
The structure of the data in this situation is:
Successes
Failures
Total
Estimate
Group
1
Group
2
.
.
.
.
.
.
Group
k
x1
x2
.
.
.
xk
n2 – x2
______
.
.
.
.
.
.
n2
.
.
.
p̂2  x 2 / n2 .
.
.
n1 – x1
_______
n1
p̂1  x 1 / n1
nk – xk
_______
nk
p̂ k  x k / nk
The null hypothesis being tested is:
H0 : p1 = p2 = . . . pk = p
HA: at least one pair pi and pj not equal
Under the null hypothesis, all the groups have the same probability of
success, so the natural estimate of the common value of p is:
k
p̂ 
x
i
n
i
i 1
k
i 1
Then for each group we can find the expected number of successes and
expected number of failures by the following formulae:
Expected Successes in Group i if Null Hypothesis is true = ni p̂
Expected Failures in Group i if Null Hypothesis is true = ni ( 1  p̂ )
Now since ni is the total for group i (i.e. the column total), and p̂ is the total
number of successes divided by the grand total of all observations, we arrive at the
fact that just as in the two sample case:
EXPij = (Total for Column i) x (Total for Row j) / (Grand Total)
Now if ni p̂  5 for all i, then we can use the Chi-Square distribution with
(2 –1) x (k – 1) degrees of freedom to test the null hypothesis. Specifically, we would
compute the test statistic:

k
2
obs
 
i 1
j
( OBS ij  EXPij ) 2
EXPij
Then we compute the one-sided p-value using the EXCEL function "chidist" as:
2
pvalue  chidist(  obs
,k  1 )
If the one-sided p-value is less than  we reject the null hypothesis. Otherwise we
accept the null hypothesis.
If the null hypothesis is rejected, we again examine the contributions to the
chi-square statistic for values of magnitude greater than 3.5 as the cells providing
the greatest deviation of observed values from expectation.
For our data let us work with alpha = .05. We begin with the original table:
Observed
No Ads
TV
Paper
TV and Paper
Total
Buy
Didn't buy
Total
15
25
12
30
85
90
53
90
100
115
65
120
82
318
400
Next we compute our Expected values as shown below:
Computation of Expected Entries
No Ads
TV
Paper
TV and Paper
Total
Buy
Didn't buy
Total
100 x 82 / 400
115 x 82 / 400
65 x 82 / 400
120 x 82 / 400
100 x 318 / 400
115 x 318 / 400
65 x 318 / 400
120 x 318 / 400
100
115
65
120
82
318
400
The resultant computation gives the expected table as:
Expected
No Ads
TV
Paper
TV and Paper
Total
Buy
Didn't buy
Total
20.5
23.575
13.325
24.6
79.5
91.425
51.675
95.4
100
115
65
120
82
318
400
Next we compute for each cell the square difference between the observed and
expected values, divided by the expected value. This gives the "Contributions" to
chi-square values given below:
Contributions to Chi-Square
Buy
No Ads
TV
Paper
TV and Paper
Didn't buy
1.47561 0.380503
0.086135 0.022211
0.131754 0.033974
1.185366 0.30566
Observed Chi-Square = 3.621213
The one-sided p-value is computed as:
one-sided p-value = chidist(3.621213, 3) = .305378.
Since this value exceeds .05, we would accept the null hypothesis that the
probability of purchasing does not change from group to group. That is, the data is
insufficient to support what seemed to be a clear pattern in the data.
Multiple Group Structural Hypotheses
Assume that you are responsible for choosing the health care plan for your
company. You wish to choose a plan with provides excellent health care for your
employees but at the same time minimizing the cost to the firm. In studying various
health plans, you note that the percentage of a hospital bill covered by the plan is
one of your major choices. In discussing this issue with your colleagues, one of them
points out an article that showed how the length of the hospital stay varied with the
percentage of the hospital bill paid for a random sample of 311 patients. The article
claims that the greater the percentage of the hospital bill paid by insurance
coverage, the longer patients stayed in the hospital. They presented the following
data:
Company Paid
Hospital
Coverage
<5
Hospital Stay in Days
6 to 10
11 to 15
>15
Total
<25%
25-50%
51-75%
>75%
26
21
25
11
30
30
25
32
6
11
45
17
5
7
9
11
67
69
104
71
Total
83
117
79
32
311
Does this data indicate that the pattern of hospital stay changes depending on
the coverage plan? Does it indicate that the greater the percentage of the hospital
bill covered by the health plan, the longer patients tend to stay in the hospital?
The basic question is whether or not the proportionate distribution of
patients by length of stay is the same for the four hospital coverage categories.
The basic structure of the problem is:
Category
1
Category
2
.
.
.
Category
c
Group 1
p11
p12
.
.
.
p1c
Group 2
p21
p22
.
.
.
p2c
.
.
.
.
.
.
.
.
.
.
.
.
.
.
pk2
.
.
.
pkc
Group k
pk1
The hypothesis testing situation is:
H0 : p1j = p2j = . . . . = pkj = pj, for j = 1, 2, . . . , c
HA: at least one pair pij not equal to pmj for some j
All the null hypothesis says is that the probability of falling in a category is the same
for all groups. The alternative hypothesis indicates that there are at least two
groups where the probability of falling in a category differ.
Let me illustrate this for our data, our original data gives the following
probabilities of length of stay by percentage coverage:
Company Paid
Hospital
Coverage
<5
Hospital Stay in Days
6 to 10
11 to 15
>15
Total
<25%
25-50%
51-75%
>75%
38.81%
30.43%
24.04%
15.49%
44.78%
43.48%
24.04%
45.07%
8.96%
15.94%
43.27%
23.94%
7.46%
10.14%
8.65%
15.49%
100.00%
100.00%
100.00%
100.00%
Total
26.69%
37.62%
25.40%
10.29%
100.00%
Notice that in the less than 5 days stay column, the percentage varies from
38.81% to 15.49%. Overall 26.69% of all people had stays less than 5 days. Is the
variability in this column consistent with chance or does it indicate that it is more
likely to be changing in some way associated with the hospital coverage? The
hypothesis simultaneously asks this question for the all the columns.
The basic structure of the observed data is given below:
Category
1
Group 1
Group 2
.
.
Group k
Column
Total
x11
Category .
2
.
x12
.
.
.
.
.
Category
c
.
.
x1c
Row
Total
|
x1+
|
x21
xp22
.
.
.
x2c
|
x2+
|
.
.
.
.
.
.
|
.
|
.
.
.
.
.
.
|
.
|
xk1
xk2
.
.
.
xkc
|
xk+
_______________________________________________________
|
x+1
x+2
.
.
.
x+c
|
x++
To get the expected values under the null hypothesis consider the entry for category
j and group i. Since under the null hypothesis all the groups have the same chance
of falling into category j, the estimate of the expected number would be just the
number in group i times the percentage of all the data in category j. Symbolically
the formula would be:
EXPij 
( x i  )( x  j )
x 
which is the same as the formula we have used before of:
EXPij = (Total for Row i) x (Total for Column j) / (Grand Total)
Then we would compute our chi-square statistic with (k – 1) x (c – 1) degrees
of freedom as:

2
obs
 
i
j
( OBS ij  EXPij ) 2
EXPij
We would then use EXCEL to compute the one-sided p-value using the
function "chidist", in the format
2
p  value  chidist(  obs
, ( k  1 )( c  1 )
If the p-value is less than  then we reject the null hypothesis and look for
cells with contributions to the chi-square statistic with values higher than 3.5.
If the p-value is greater than  we accept the null hypothesis.
Let us work at the .01 level of alpha. The observed table is:
Observed
Company Paid
Hospital
Coverage
<5
Hospital Stay in Days
6 to 10
11 to 15
>15
Total
<25%
25-50%
51-75%
>75%
26
21
25
11
30
30
25
32
6
11
45
17
5
7
9
11
67
69
104
71
Total
83
117
79
32
311
I then compute the expected values to obtain the expected table of:
Company Paid
Hospital
Coverage
<25%
25-50%
51-75%
>75%
Total
Expected
Hospital Stay in Days
<5
6 to 10
11 to 15
>15
Total
17.88
18.41
27.76
18.95
25.21
25.96
39.13
26.71
17.02
17.53
26.42
18.04
6.89
7.10
10.70
7.31
67
69
104
71
83
117
79
32
311
This was computed using the formula on the previous page. For example the
entry in the Group "<25%" and Hospital Stay "<5" was computed as:
(67) x (83) / 311 = 17.88 .
The contributions to chi-square and the observed chi-square statistic are then given
by:
Contributions to Chi-Square
Company Paid
Hospital
Coverage
<5
<25%
25-50%
51-75%
>75%
3.69
0.36
0.27
3.33
Hospital Stay in Days
6 to 10
11 to 15
>15
0.91
0.63
5.10
1.05
7.13
2.43
13.07
0.06
0.52
0.00
0.27
1.87
Chisquare =
40.70
The degrees of freedom in this problem are (4 –1) x (4 –1) = 9. Therefore, the
one-sided p-value is given by:
p-value = chidist( 40.70, 9) = .00000567.
Since this value is much lower than .01, we reject the null hypothesis, and say
that the distribution of the length of stay is changes by insurance coverage.
Since we have rejected the null hypothesis, I have highlighted below the cells
that seem to show the greatest deviation between the observed and expected values:
Contributions to Chi-Square
Company Paid
Hospital
Coverage
<5
<25%
25-50%
51-75%
>75%
Hospital Stay in Days
6 to 10
11 to 15
>15
3.69
0.36
0.27
3.33
0.91
0.63
5.10
1.05
7.13
2.43
13.07
0.06
0.52
0.00
0.27
1.87
Chisquare =
40.70
By comparing the observed to the expected values, we arrive at the following
structure:
Company Paid
Hospital
Coverage
<25%
25-50%
51-75%
>75%
Observed Compared to Expected
Hospital Stay in Days
<5
6 to 10
11 to 15
>15
Higher
x
x
Lower
x
x
Lower
x
Lower
x
Higher
x
x
x
x
x
The yellow highlighted cells seem to indicate that the lower the coverage, the
quicker you will leave the hospital. In the "<25%" coverage category that is
reinforced by the fact that the observed number in the "11 to 15" day stay period is
lower than would be expected. Interestingly, in the "51-75%" coverage group, the
lower than expected number in the "6 – 10" day category and the much higher than
expected number in the "11 – 15" day category also tend to reinforce the argument
that the hospital coverage is related to length of hospital stay with increasing
coverage associated with increasingly long stays.