Daniel W.W. Biostatistics: A Foundation for Analysis in the Health Sciences

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respectively. The subscript i runs from 1 to a and j runs from 1 to b. The total num-

ber of observations is nab.

To show that Table 8.5.3 represents data from a completely randomized

design, we consider that each combination of factor levels is a treatment and that

we have n observations for each treatment. An alternative arrangement of the

data would be obtained by listing the observations of each treatment in a sepa-

rate column. Table 8.5.3 may also be used to display data from a two-factor

randomized block design if we consider the first observation in each cell as

belonging to block 1, the second observation in each cell as belonging to block

2, and so on to the nth observation in each cell, which may be considered as

belonging to block n.

Note the similarity of the data display for the factorial experiment as shown

in Table 8.5.3 to the randomized complete block data display of Table 8.3.1. The

factorial experiment, in order that the experimenter may test for interaction, requires

at least two observations per cell, whereas the randomized complete block design

requires only one observation per cell. We use two-way analysis of variance to ana-

lyze the data from a factorial experiment of the type presented here.

2. Assumptions. We assume a ﬁxed-effects model and a two-factor completely ran-

domized design. For a discussion of other designs, consult the references at the

end of this chapter.

The Model

The ﬁxed-effects model for the two-factor completely randomized design

may be written as

(8.5.1)

where is a typical observation, is a constant, represents an effect due to factor A,

represents an effect due to factor B, represents an effect due to the interaction of fac-

tors A and B, and represents the experimental error.

Assumptions of the Model

a. The observations in each of the ab cells constitute a random independent sam-

ple of size n drawn from the population deﬁned by the particular combination

of the levels of the two factors.

b. Each of the ab populations is normally distributed.

c. The populations all have the same variance.

3. Hypotheses. The following hypotheses may be tested:

not all

not all 1ab2

= 0H

i = 1, 2, Á , a; j = 1, 2, Á , bH

: 1ab2

= 0

= 0H

j = 1, 2, Á , bH

: b

= 0

= 0H

i = 1, 2, Á , aH

: a

= 0

ijk

1ab2

bamx

ijk

k = 1, 2, Á , nj = 1, 2, Á , b ;i = 1, 2, Á , a;

ijk

= m + a

+ b

+ 1ab2

ijk

8.5 THE FACTORIAL EXPERIMENT

357

Before collecting data, the researchers may decide to test only one of the possible

hypotheses. In this case they select the hypothesis they wish to test, choose a signiﬁ-

cance level and proceed in the familiar, straightforward fashion. This procedure is free

of the complications that arise if the researchers wish to test all three hypotheses.

When all three hypotheses are tested, the situation is complicated by the fact that

the three tests are not independent in the probability sense. If we let be the signiﬁ-

cance level associated with the test as a whole, and and the signiﬁcance lev-

els associated with hypotheses 1, 2, and 3, respectively, we ﬁnd

(8.5.2)

If then or This means that the

probability of rejecting one or more of the three hypotheses is something less than .143

when a significance level of .05 has been chosen for the hypotheses and all are true.

To demonstrate the hypothesis testing procedure for each case, we perform all three

tests. The reader, however, should be aware of the problem involved in interpreting the

results.

4. Test statistic. The test statistic for each hypothesis set is V.R.

5. Distribution of test statistic. When is true and the assumptions are met, each

of the test statistics is distributed as F.

6. Decision rule. Reject if the computed value of the test statistic is equal to or

greater than the critical value of F.

7. Calculation of test statistic. By an adaptation of the procedure used in partition-

ing the total sum of squares for the completely randomized design, it can be shown

that the total sum of squares under the present model can be partitioned into two

parts as follows:

(8.5.3)

(8.5.4)

The sum of squares for treatments can be partitioned into three parts as follows:

(8.5.5)

i =1

j =1

k =1

ij#

- x

i##

- x

#j#

+ x

###

i =1

j =1

k =1

# j #

- x

###

i =1

j =1

k =1

- x

...

i =1

j =1

k =1

- x

...

SST = SSTr + SSE

i =1

j =1

k =1

ijk

- x

...

i =1

j =1

k =1

- x

...

i =1

j =1

k =1

ijk

- x

a 6 .143.a 6 1 - 1.952

,a¿=a–=a‡=.05,

a 6 1 - 11 - a¿211 - a–211 - a‡2

a‡a–,a¿,

358 CHAPTER 8 ANALYSIS OF VARIANCE

The ANOVA Table

The results of the calculations for the ﬁxed-effects model for

a two-factor completely randomized experiment may, in general, be displayed as shown

in Table 8.5.4.

8. Statistical decision. If the assumptions stated earlier hold true, and if each

hypothesis is true, it can be shown that each of the variance ratios shown in Table

8.5.4 follows an F distribution with the indicated degrees of freedom. We reject

if the computed V.R. values are equal to or greater than the corresponding crit-

ical values as determined by the degrees of freedom and the chosen signiﬁcance

levels.

9. Conclusion. If we reject we conclude that is true. If we fail to reject

we conclude that may be true.

10. p value.

EXAMPLE 8.5.2

In a study of length of time spent on individual home visits by public health nurses,

data were reported on length of home visit, in minutes, by a sample of 80 nurses. A

record was made also of each nurse’s age and the type of illness of each patient vis-

ited. The researchers wished to obtain from their investigation answers to the follow-

ing questions:

1. Does the mean length of home visit differ among different age groups of nurses?

2. Does the type of patient affect the mean length of home visit?

3. Is there interaction between nurse’s age and type of patient?

Solution:

1. Data. The data on length of home visit that were obtained during the

study are shown in Table 8.5.5.

SSTr = SSA + SSB + SSAB

8.5 THE FACTORIAL EXPERIMENT 359

TABLE 8.5.4 Analysis of Variance Table for a Two-Factor Completely Randomized

Experiment (Fixed-Effects Model)

Source

SS d.f.

MS V.R.

ASSA

B SSB

AB SSAB

Treatments

SSTr

Residual

SSE

Total

SST

abn - 1

MSE = SSE>ab1n - 12ab1n - 12

ab - 1

MSAB>MSEMSAB = SSAB>1a - 121b - 121a - 121b - 12

MSB>MSEMSB = SSB>1b - 12b - 1

MSA>MSEMSA = SSA>1a - 12a - 1

2. Assumptions. To analyze these data, we assume a ﬁxed-effects model

and a two-factor completely randomized design.

For our illustrative example we may test the following hypotheses subject

to the conditions mentioned above.

a. not all

b. not all

c. all not all

Let

3. Test statistic. The test statistic for each hypothesis set is V.R.

4. Distribution of test statistic. When is true and the assumptions are

met, each of the test statistics is distributed as F.

a = .05

1ab2

= 0H

:1ab2

= 0H

: b

= b

= 0

= 0H

: a

= a

= 0

360 CHAPTER 8 ANALYSIS OF VARIANCE

TABLE 8.5.5 Length of Home Visit in Minutes by Public Health Nurses by

Nurse’s Age Group and Type of Patient

Factor

(Nurse’s Age Group) Levels

Factor

(Type of Patient) 1 2 3 4

Levels (20 to 29) (30 to 39) (40 to 49) (50 and Over)

1 (Cardiac) 20 25 24 28

25 30 28 31

22 29 24 26

27 28 25 29

21 30 30 32

2 (Cancer) 30 30 39 40

45 29 42 45

30 31 36 50

35 30 42 45

36 30 40 60

3 (C.V.A.) 31 32 41 42

30 35 45 50

40 30 40 40

35 40 40 55

30 30 35 45

4 (Tuberculosis) 20 23 24 29

21 25 25 30

20 28 30 28

20 30 26 27

19 31 23 30

5. Decision rule. Reject if the computed value of the test statistic is

equal to or greater than the critical value of F. The critical values of F

for testing the three hypotheses of our illustrative example are 2.76,

2.76, and 2.04, respectively. Since denominator degrees of freedom

equal to 64 are not shown in Appendix Table G, 60 was used as the

denominator degrees of freedom.

6. Calculation of test statistic. We use MINITAB to perform the calcu-

lations. We put the measurements in Column 1, the row (factor A) codes

in Column 2, and the column (factor B) codes in Column 3. The result-

ing column contents are shown in Table 8.5.6. The MINITAB output is

shown in Figure 8.5.3.

8.5 THE FACTORIAL EXPERIMENT

361

TABLE 8.5.6 Column Contents for MINITAB Calculations,

Example 8.5.2

Row C1 C2 C3 Row C1 C2 C3

12011 413131

22511 423031

32211 434031

42711 443531

52111 453031

62512 463232

73012 473532

82912 483032

92812 494032

10 30 1 2 50 30 3 2

11 24 1 3 51 41 3 3

12 28 1 3 52 45 3 3

13 24 1 3 53 40 3 3

14 25 1 3 54 40 3 3

15 30 1 3 55 35 3 3

16 28 1 4 56 42 3 4

17 31 1 4 57 50 3 4

18 26 1 4 58 40 3 4

19 29 1 4 59 55 3 4

20 32 1 4 60 45 3 4

21 30 2 1 61 20 4 1

22 45 2 1 62 21 4 1

23 30 2 1 63 20 4 1

24 35 2 1 64 20 4 1

25 36 2 1 65 19 4 1

26 30 2 2 66 23 4 2

27 29 2 2 67 25 4 2

28 31 2 2 68 28 4 2

29 30 2 2 69 30 4 2

(

Continued

)

362 CHAPTER 8 ANALYSIS OF VARIANCE

Row C1 C2 C3 Row C1 C2 C3

30 30 2 2 70 31 4 2

31 39 2 3 71 24 4 3

32 42 2 3 72 25 4 3

33 36 2 3 73 30 4 3

34 42 2 3 74 26 4 3

35 40 2 3 75 23 4 3

36 40 2 4 76 29 4 4

37 45 2 4 77 30 4 4

38 50 2 4 78 28 4 4

39 45 2 4 79 27 4 4

40 60 2 4 80 30 4 4

7. Statistical decision. The variance ratios are

14.7  67.86, and

14.7  4.60. Since the three computed values of V.R. are all greater

than the corresponding critical values, we reject all three null

hypotheses.

8. Conclusion. When is rejected, we conclude

that there are differences among the levels of A, that is, differences in

the average amount of time spent in home visits with different types of

patients. Similarly, when is rejected, we con-

clude that there are differences among the levels of B, or differences in

the average amount of time spent on home visits among the different

nurses when grouped by age. When is rejected, we con-

clude that factors A and B interact; that is, different combinations of lev-

els of the two factors produce different effects.

9. p value. Since 67.86, 27.24, and 4.60 are all greater than the criti-

cal values of for the appropriate degrees of freedom, the p value

for each of the tests is less than .005. When the hypothesis of no

interaction is rejected, interest in the levels of factors A and B usu-

ally become subordinate to interest in the interaction effects. In other

words, we are more interested in learning what combinations of lev-

els are significantly different.

Figure 8.5.4 shows the SAS

output for the analysis of Example 8.5.2. ■

We have treated only the case where the number of observations in each cell is the same.

When the number of observations per cell is not the same for every cell, the analysis

becomes more complex.

In such cases the design is said to be unbalanced. To analyze these designs with

MINITAB we use the general linear (GLM) procedure. Other software packages such as

SAS

also will accommodate unequal cell sizes.

.995

: 1ab2

= 0

: b

= b

: a

= a

V.R.1AB2= 67.6>V.R.1B2= 400.4>14.7 = 27.24,

V.R. 1A2= 997.5>

8.5 THE FACTORIAL EXPERIMENT 363

Dialog box: Session command:

Stat ➤ ANOVA ➤ Twoway MTB > TWOWAY C1 C2 C3;

SUBC > MEANS C2 C3.

Type C1 in Response. Type C2 in Row factor and

check Display means. Type C3 in Column factor and

check Display means. Click OK.

Output:

Two-Way ANOVA: C1 versus C2, C3

Analysis of Variance for C1

Source DF SS MS F P

C2 3 2992.4 997.483 67.94 0.000

C3 3 1201.1 400.350 27.27 0.000

Interaction 9 608.5 67.606 4.60 0.000

Error 64 939.6 14.681

Total 79 5741.5

Individual 95% CI

C2 Mean -+---------+---------+---------+---------+

1 26.70 (----

---)

2 38.25 (----

---)

3 38.30 (----

---)

4 25.45 (----

---)

-+---------+---------+---------+---------+

24.00 28.00 32.00 36.00 40.00

Individual 95% CI

C3 Mean ------+---------+---------+---------+-----

1 27.85 (----

---)

2 29.80 (----

---)

3 32.95 (----

---)

4 38.10 (----

---)

------+---------+---------+---------+-----

28.00 31.50 35.00 38.50

FIGURE 8.5.3 MINITAB procedure and ANOVA table for Example 8.5.2.

EXERCISES

For Exercises 8.5.1 to 8.5.4, perform the analysis of variance, test appropriate hypotheses at the

.05 level of signiﬁcance, and determine the p value associated with each test.

8.5.1 Uryu et al. (A-21) studied the effect of three different doses of troglitazone on neuro cell

death. Cell death caused by stroke partially results from the accumulation of high concentrations

of glutamate. The researchers wanted to determine if different doses of troglitazone (1.3, 4.5, and

and different ion forms of LY294002, a PI3-kinase inhibitor, would give dif-

ferent levels of neuroprotection. Four rats were studied at each dose and ion level, and the meas-

ured variable is the percent of cell death as compared to glutamate. Therefore, a higher value

implies less neuroprotection. The results are in the table below.

1- and + 213.5

mM2

1mM2

364 CHAPTER 8 ANALYSIS OF VARIANCE

The SAS System

Analysis of Variance Procedure

Dependent Variable: TIME

Source DF Sum of Squares Mean Square F Value Pr  F

Model 15 4801.95000000 320.13000000 21.81 0.0001

Error 64 939.60000000 14.68125000

Corrected Total 79 5741.55000000

R-Square C.V. Root MSE TIME Mean

0.836351 11.90866 3.83161193 32.17500000

Source DF Anova SS Mean Square F Value Pr  F

FACTORB 3 1201.05000000 400.35000000 27.27 0.0001

FACTORA 3 2992.45000000 997.48333333 67.94 0.0001

FACTORB*FACTORA 9 608.450000000 67.60555556 4.60 0.0001

FIGURE 8.5.4 SAS

output for analysis of Example 8.5.2.

Percent Compared to Troglitazone

Glutamate Dose ( )

73.61 Negative 1.3

130.69 Negative 1.3

118.01 Negative 1.3

140.20 Negative 1.3

MM-LY294002 vs + LY294002

(Continued)

8.5.2 Researchers at a trauma center wished to develop a program to help brain-damaged trauma vic-

tims regain an acceptable level of independence. An experiment involving 72 subjects with the

same degree of brain damage was conducted. The objective was to compare different combi-

nations of psychiatric treatment and physical therapy. Each subject was assigned to one of 24

different combinations of four types of psychiatric treatment and six physical therapy programs.

There were three subjects in each combination. The response variable is the number of months

elapsing between initiation of therapy and time at which the patient was able to function inde-

pendently. The results were as follows:

EXERCISES 365

Percent Compared to Troglitazone

Glutamate Dose ( )

97.11 Positive 1.3

114.26 Positive 1.3

120.26 Positive 1.3

92.39 Positive 1.3

26.95 Negative 4.5

53.23 Negative 4.5

59.57 Negative 4.5

53.23 Negative 4.5

28.51 Positive 4.5

30.65 Positive 4.5

44.37 Positive 4.5

36.23 Positive 4.5

Negative 13.5

25.14 Negative 13.5

20.16 Negative 13.5

34.65 Negative 13.5

Positive 13.5

5.36 Positive 13.5

Source: Shigeko Uryu. Used with permission.

-19.08

-7.93

-35.80

-8.83

MM-LY294002 vs + LY294002

Physical Psychiatric Treatment

Therapy

Program ABCD

11.0 9.4 12.5 13.2

I 9.6 9.6 11.5 13.2

10.8 9.6 10.5 13.5

10.5 10.8 10.5 15.0

II 11.5 10.5 11.8 14.6

12.0 10.5 11.5 14.0

12.0 11.5 11.8 12.8

III 11.5 11.5 11.8 13.7

11.8 12.3 12.3 13.1

(Continued)

Can one conclude on the basis of these data that the different psychiatric treatment programs have

different effects? Can one conclude that the physical therapy programs differ in effectiveness? Can

one conclude that there is interaction between psychiatric treatment programs and physical ther-

apy programs? Let for each test.

Exercises 8.5.3 and 8.5.4 are optional since they have unequal cell sizes. It is recommended that

the data for these be analyzed using SAS

or some other software package that will accept unequal

cell sizes.

8.5.3 Main et al. (A-22) state, “Primary headache is a very common condition and one that nurses

encounter in many different care settings. Yet there is a lack of evidence as to whether advice

given to sufferers is effective and what improvements may be expected in the conditions.” The

researchers assessed frequency of headaches at the beginning and end of the study for 19 sub-

jects in an intervention group (treatment 1) and 25 subjects in a control group (treatment 2).

Subjects in the intervention group received health education from a nurse, while the control

group did not receive education. In the 6 months between pre- and post-evaluation, the sub-

jects kept a headache diary. The following table gives as the response variable the difference

(prepost) in frequency of headaches over the 6 months for two factors: (1) treatment with two

levels (intervention and control), and (2) migraine status with two levels (migraine sufferer and

nonmigraine sufferer).

a = .05

366 CHAPTER 8 ANALYSIS OF VARIANCE

Physical Psychiatric Treatment

Therapy

Program ABCD

11.5 9.4 13.7 14.0

IV 11.8 9.1 13.5 15.0

10.5 10.8 12.5 14.0

11.0 11.2 14.4 13.0

V 11.2 11.8 14.2 14.2

10.0 10.2 13.5 13.7

11.2 10.8 11.5 11.8

VI 10.8 11.5 10.2 12.8

11.8 10.2 11.5 12.0

Change in Change in

Frequency of Migraine Sufferer Frequency of Migraine Sufferer

Headaches ( ) Treatment Headaches ( ) Treatment

11 22

221 22

33 1 1 11 1 2

2164 12

6216512

98 1 1 14 1 2

221812

621622

-6

-3-2

1  No, 2  Yes1  No, 2  Yes

(Continued)