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One-Way Analysis of Variance: A Guide to Testing Differences Between Multiple Groups REQUIREMENTS: ONEWAY ANOVA tests the equality of group means for a single specified variable. For example, The F ratio tests the statistical significance between means. Mathematical Formulations:
i) Sum of Squares for Treatment (SST):
k
SST = ∑ (Ti2/ni) - CM
i=1
where:Ti = Total of all observations receiving the treatment i (or of the ith population) ni = Number of observations receiving the treatment i (or of the ith population) CM= Correction for the mean = T2/n T = Total of all observations = ( T1 + T2 + T3 + ....... + Tk ) n = Total number of Observations = ( n1 + n2 + n3 + ....... + nk ) ii) Sum of Squares for Error (SSE)
n1 n1 n1
SSE is usually computed in a simplified way from the equation;SSERROR = SSTOTAL - SSTREATMENT B) The Degrees of Freedom: The degrees of freedom for the Total Sum of Squares is always (n - 1); where n = Total number of observations in all samples = ( n1 + n2 + n3 + ....... + nk ) The degrees of freedom of the Model (Treatment) is always (k - 1); where k = Total number of populations being analysed. The degrees of freedom of the Error is always (n - k). The following relationship always holds: D.F.(Treatment) + D.F.(ERROR) = (k-1) + (n-k) = (n-1) = D.F.(TOTAL SS) C) The Mean Square: The mean square gives an estimate of the s˛ based on the variation among the sample means (corresponding to the model) and the variation within the samples (corresponding to the error). These estimates are calculated by dividing the sum of squares by the corresponding degrees of freedom. Thus, i) The Mean Square for Treatment (Model) = MST = (SST)/(k-1) ii) The Mean Square of the Error = MSE = (SSE)/(n-k) ( The MSE is a pooled estimate of s2 based on the sum of squares of deviations of the x-values about their respective sample means and is also denoted by s2 ) D) The F Statistic: The F statistic is used for comparing the estimate of s2 (MS(Treatment)) and the s2 (MS(Error)) and is given by F = MS(Treatment)/MS(Error) E) The ANALYSIS : The ANOVA is done with the Ho: µ1 = µ2 = µ3 = .....= µk Next, using the tables, the F-value with degrees of freedom v1 (v1 = D.F. of the numerator i.e. of MS(Treatment) = k-1) and v2 (v2 = D.F. of the denominator i.e. of MS(Error) = n-k), and for the significance level used in the analysis, is obtained. This F-value is compared with the F statistic computed. If the F-value obtained is greater than or equal to the F-Statistic Computed; then we say that THERE IS INSUFFICIENT EVIDENCE TO REJECT THE NULL HYPOTHESIS AT THE GIVEN LEVEL OF SIGNIFICANCE. But, if the F-value obtained is less than the F-Statistic Computed; then we say that THERE IS SUFFICIENT EVIDENCE TO REJECT THE NULL HYPOTHESIS AT THE GIVEN LEVEL OF SIGNIFICANCE and that leads to the conclusion that at least one of the population means (µi) is different from the others. The observed significance level is the significance level for which the F-value obtained from the table, corresponding to degrees of freedom v1 and v2, is equal to the F statistic computed. Another way of testing the null hypothesis is by using this observed significance level. If this significance level is less than or equal to the significance level set for the test, then the null hypothesis is rejected. +------------------------------------+ ¦Example One: University Selection ¦ +------------------------------------+ The table below gives the number of students graduating in six areas of study at four Universities.
+--------------------------------------------------------------+
¦ ¦ ¦ ¦
¦ ¦ IIT BYU CSU VPI ¦ TOTAL ¦
¦--------------+-----------------------------------+-----------¦
¦ SCIENCE ¦ 597 280 245 339 ¦ 1461 ¦
¦ ACCOUNTING ¦ 768 260 240 275 ¦ 1543 ¦
¦ BUSINESS ¦ 776 284 257 304 ¦ 1621 ¦
¦ LAW ¦ 739 334 262 317 ¦ 1652 ¦
¦ HUMANITIES ¦ 562 338 250 335 ¦ 1485 ¦
¦ ENGINEERING¦ 696 315 330 350 ¦ 1691 ¦
¦--------------+-----------------------------------+-----------¦
¦ Sum ¦ 4138 1811 1584 1920 ¦ 9453 ¦
¦ Average ¦ 690 302 264 320 ¦ 1576 ¦
¦ Sum of Sq ¦2894790 551681 423718 618176 ¦14936821 ¦
+--------------------------------------------------------------+
n1 = n2 = n3 = n4 = 6 ;
k = 4 ;
n = (n1 + n2 + n3 + n4) = 24
CM (Correction for the Mean) = T˛/N = 94532 / 24 = 3,723,300
Total Sum of Sq. Deviation (TSS)= SSx
= Total Sum of Square CM
= 1,49,36,821 - 3,723,300
= 11,213,520
Sum of Squares for the TREATMENT = SST
4
=(S Ti˛)/n CM
i=1
= (4138)˛ + (1811)˛
= (4,433,036 3,723,300)
= 709,736
Sum of Squares for the ERROR = SSERROR
= SSTOTAL SSTREATMENT
= 112,13,520 709,736
= 105,03,784
Mean Square for the TREATMENT = MS(Treatment)
= SST/(k 1)
= 236,578.8
Mean Square for the ERROR = MS(Error) = SSE/(n k) = 525,189.2
F statistic = (MS(Treatment)/MS(Error)) = 0.450463
The ANOVA Table is given below:
+----------------------------------------------------+
¦ Source d.f. SS MS F ¦
¦ Treatment 3 709736.4 236578.8 0.450463 ¦
¦ Error 20 10503784 525189.2 ¦
¦ Total 23 11213520 ¦
+----------------------------------------------------+
+---------------------------------------+
¦Example Two: HOSPITAL STAFF EVALUATION ¦
+---------------------------------------+
Using the standard tables, for a=0.05, v1(d.f.1)=3, v2(d.f.2)=20; the value of the
F-statistic is 2.38. The Computed F-statistic is much less than the F-statistic from
the tables. We can say that there insufficient evidence to conclude that there is a
difference in the mean number of students graduating from the four universities.
PATIENT SURVEY, FOUR LOCAL HOSPITALS
VARIABLE V18 COMPETENCE OF NURSING STAFF
GROUP COUNT MEAN STD. DEV. STD. ERR. 95 PCT CONF INT FOR MEAN
1 107 1.009 .096 .009 .991 TO 1.028
2 46 1.022 .146 .022 .979 TO 1.065
3 11 1.000 .000 .000 1.000 TO 1.000
4 2 1.000 .000 .000 1.000 TO 1.000
TOTAL 166 1.012 .109 .008 .995 TO 1.029
GROUP MINIMUM MAXIMUM
1 1.00 2.00
2 1.00 2.00
3 1.00 1.00
4 1.00 1.00
VARIABLE 1 V18 COMPETENCE OF NURSING STAFF
BY VARIABLE 5 V5 PERCENT HOSPITAL COVERAGE
ANALYSIS OF VARIANCE
SUM OF MEAN F F
SOURCE DF1 SQUARES SQUARE2 RATIO 3 PROB. 4
BETWEEN GROUPS 3 .007 .002 .192 .902
WITHIN GROUPS 162 1.969 .012
TOTAL 165 1.976
TESTS FOR HOMOGENIETY OF VARIANCES
COCHRANS C = MAX. VARIANCE/SUM VARIANCES = .697
MAXIMUM VARIANCE / MINIMUM VARIANCE = 999999.000
******************************************************************************
1.The Degree of Freedom for the Regression Model also called the explained model
is given by k, where k = number of independent variables in the regression
equation. For the Residual, the error unexplained by the regression model,
the Degree of Freedom is given by (n-k-1), where n = number of counts of the
independent variable in the data set.
2.Mean Square = (Sum of Squares)/(DF)
3.F Ratio = (Mean Square of the Regression)/(Mean Square of the Residual)
4.F-Prob = Level of significance corresponding to the F Value
PATIENT SURVEY, LOCAL HOSPITAL
VARIABLE V19 PROMPTNESS OF RESPONSE
GROUP COUNT MEAN STD. DEV. STD. ERR. 95 PCT CONF INT FOR MEAN
1 107 1.000 .193 .019 .963 TO 1.037
2 48 1.125 1.166 .168 .788 TO 1.462
3 11 1.818 2.289 .690 .438 TO 3.199
4 2 1.000 .000 .000 1.000 TO 1.000
TOTAL 168 1.089 .892 .069 .952 TO 1.227
GROUP MINIMUM MAXIMUM
1 .00 2.00
2 .00 9.00
3 1.00 9.00
4 1.00 1.00
VARIABLE 2 V19 PROMPTNESS OF RESPONSE
BY VARIABLE 5 V5 PERCENT HOSPITAL COVERAGE
ANALYSIS OF VARIANCE
SUM OF MEAN F F
SOURCE D.F. SQUARES SQUARES RATIO PROB.
BETWEEN GROUPS 3 6.774 2.258 2.919 .036
WITHIN GROUPS 164 126.886 .774
TOTAL 167 133.661
TESTS FOR HOMOGENIETY OF VARIANCES
COCHRANS C = MAX. VARIANCE/SUM VARIANCES = .790
MAXIMUM VARIANCE / MINIMUM VARIANCE = 999999.000
******************************************************************************
PATIENT SURVEY, LOCAL HOSPITAL
VARIABLE V20 FRIENDLINESS OF STAFF
GROUP COUNT MEAN STD. DEV. STD. ERR. 95 PCT CONF INT FOR MEAN
1 107 1.019 .135 .013 .993 TO 1.045
2 48 1.167 1.161 .168 .832 TO 1.502
3 11 1.000 .000 .000 1.000 TO 1.000
4 2 1.000 .000 .000 1.000 TO 1.000
TOTAL 168 1.060 .633 .049 .962 TO 1.157
GROUP MINIMUM MAXIMUM
1 1.00 2.00
2 .00 9.00
3 1.00 1.00
4 1.00 1.00
VARIABLE 3 V20 FRIENDLINESS OF STAFF
BY VARIABLE 5 V5 PERCENT HOSPITAL COVERAGE
ANALYSIS OF VARIANCE
SUM OF MEAN F F
SOURCE D.F. SQUARES SQUARES RATIO PROB.
BETWEEN GROUPS 3 .775 .258 .636 .593
WITHIN GROUPS 164 66.629 .406
TOTAL 167 67.405
TESTS FOR HOMOGENIETY OF VARIANCES
COCHRANS C = MAX. VARIANCE/SUM VARIANCES = .987
MAXIMUM VARIANCE / MINIMUM VARIANCE = 999999.000
******************************************************************************
PATIENT SURVEY, LOCAL HOSPITAL
VARIABLE V21 NURSE EXPLAINING
GROUP COUNT MEAN STD. DEV. STD. ERR. 95 PCT CONF INT FOR MEAN
1 107 1.047 .286 .028 .991 TO 1.102
2 48 1.125 1.166 .168 .788 TO 1.462
3 11 1.091 .287 .087 .918 TO 1.264
4 2 1.000 .000 .000 1.000 TO 1.000
TOTAL 168 1.071 .669 .052 .968 TO 1.175
GROUP MINIMUM MAXIMUM
1 .00 2.00
2 .00 9.00
3 1.00 2.00
4 1.00 1.00
VARIABLE 4 V21 NURSE EXPLAINING
BY VARIABLE 5 V5 PERCENT HOSPITAL COVERAGE
ANALYSIS OF VARIANCE
SUM OF MEAN F F
SOURCE D.F. SQUARES SQUARES RATIO PROB.
BETWEEN GROUPS 3 .217 .072 .159 .924
WITHIN GROUPS 164 74.925 .457
TOTAL 167 75.143
TESTS FOR HOMOGENIETY OF VARIANCES
COCHRANS C = MAX. VARIANCE/SUM VARIANCES = .892
MAXIMUM VARIANCE / MINIMUM VARIANCE = 999999.000
******************************************************************************
END OF ONEWAY ANALYSIS OF VARIANCE
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