Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Analysis of Variance and Co-variance 267 Solution: As the given problem is a two-way design of experiment without repeated values, we shall adopt all the above stated steps while setting up the ANOVA table as is illustrated on the following page. ANOVA table can be set up for the given problem as shown in Table 11.5. From the said ANOVA table, we find that differences concerning varieties of seeds are insignificant at 5% level as the calculated F-ratio of 4 is less than the table value of 5.14, but the variety differences concerning fertilizers are significant as the calculated F-ratio of 6 is more than its table value of 4.76. (b) ANOVA technique in context of two-way design when repeated values are there: In case of a two-way design with repeated measurements for all of the categories, we can obtain a separate independent measure of inherent or smallest variations. For this measure we can calculate the sum of squares and degrees of freedom in the same way as we had worked out the sum of squares for variance within samples in the case of one-way ANOVA. Total SS, SS between columns and SS between rows can also be worked out as stated above. We then find left-over sums of squares and left-over degrees of freedom which are used for what is known as ‘interaction variation’ (Interaction is the measure of inter relationship among the two different classifications). After making all these computations, ANOVA table can be set up for drawing inferences. We illustrate the same with an example. Table 11.4: Computations for Two-way Anova (in a design without repeated values) bg 2 × Step (i) T = 60, n = 12, ∴ Correction factor = = = T 60 60 300 n 12 Step (ii) Total SS = (36 + 25 + 25 + 49 + 25 + 16 + 9 + 9 + 9 + 64 + 49 + 16) – = 332 – 300 = 32 × × × Step (iii) SS between columns treatment = + + QP − 24 24 20 20 16 16 4 4 4 L NM L NM = 144 + 100 + 64 – 300 = 8 F 60 × 60I HG 12 KJ Step (iv) SS between rows treatment = × × × × + + + QP − 16 16 16 16 9 9 19 19 3 3 3 3 Step (v) = 85.33 + 85.33 + 27.00 + 120.33 – 300 = 18 SS residual or error = Total SS – (SS between columns + SS between rows) = 32 – (8 + 18) = 6 O L NM 60 × 60 12 O L NM O QP 60 × 60 12 O QP
268 Research Methodology Table 11.5: The Anova Table Source of variation SS d.f. MS F-ratio 5% F-limit (or the tables values) Between columns (i.e., between varieties of seeds) 8 (3 – 1) = 2 8/2 = 4 4/1 = 4 F(2, 6) = 5.14 Between rows (i.e., between varieties of fertilizers) 18 (4 – 1) = 3 18/3 = 6 6/1 = 6 F(3, 6) = 4.76 Residual or error 6 (3 – 1) × (4 – 1) = 6 6/6=1 Total 32 (3 × 4) – 1 = 11 Illustration 3 Set up ANOVA table for the following information relating to three drugs testing to judge the effectiveness in reducing blood pressure for three different groups of people: Amount of Blood Pressure Reduction in Millimeters of Mercury Drug X Y Z Group of People A 14 10 11 15 9 11 B 12 7 10 11 8 11 C 10 11 8 11 11 7 Do the drugs act differently? Are the different groups of people affected differently? Is the interaction term significant? Answer the above questions taking a significant level of 5%. Solution: We first make all the required computations as shown below: We can set up ANOVA table shown in Table 11.7 (Page 269).
- Page 234 and 235: Testing of Hypotheses I 217 Solutio
- Page 236 and 237: Testing of Hypotheses I 219 Hence t
- Page 238 and 239: Testing of Hypotheses I 221 of succ
- Page 240 and 241: Testing of Hypotheses I 223 Thus, q
- Page 242 and 243: Testing of Hypotheses I 225 the val
- Page 244 and 245: Testing of Hypotheses I 227 and 2 X
- Page 246 and 247: Testing of Hypotheses I 229 LIMITAT
- Page 248 and 249: Testing of Hypotheses I 231 20. Ten
- Page 250 and 251: Chi-square Test 233 10 Chi-Square T
- Page 252 and 253: Chi-square Test 235 S. No. 1 2 3 4
- Page 254 and 255: Chi-square Test 237 As a test of go
- Page 256 and 257: Chi-square Test 239 (i) First of al
- Page 258 and 259: Chi-square Test 241 The expected fr
- Page 260 and 261: Chi-square Test 243 Show that the s
- Page 262 and 263: Chi-square Test 245 Events or Expec
- Page 264 and 265: Chi-square Test 247 c h χ 2 N ⋅
- Page 266 and 267: Chi-square Test 249 from a number o
- Page 268 and 269: Chi-square Test 251 Questions 1. Wh
- Page 270 and 271: Chi-square Test 253 No. of boys 5 4
- Page 272 and 273: Chi-square Test 255 23. For the dat
- Page 274 and 275: Analysis of Variance and Co-varianc
- Page 276 and 277: Analysis of Variance and Co-varianc
- Page 278 and 279: Analysis of Variance and Co-varianc
- Page 280 and 281: Analysis of Variance and Co-varianc
- Page 282 and 283: Analysis of Variance and Co-varianc
- Page 286 and 287: Analysis of Variance and Co-varianc
- Page 288 and 289: Analysis of Variance and Co-varianc
- Page 290 and 291: Analysis of Variance and Co-varianc
- Page 292 and 293: Analysis of Variance and Co-varianc
- Page 294 and 295: Analysis of Variance and Co-varianc
- Page 296 and 297: Analysis of Variance and Co-varianc
- Page 298 and 299: Analysis of Variance and Co-varianc
- Page 300 and 301: Testing of Hypotheses-II 283 12 Tes
- Page 302 and 303: Testing of Hypotheses-II 285 Illust
- Page 304 and 305: Testing of Hypotheses-II 287 Table
- Page 306 and 307: Testing of Hypotheses-II 289 fact,
- Page 308 and 309: Testing of Hypotheses-II 291 The te
- Page 310 and 311: Testing of Hypotheses-II 293 Pair B
- Page 312 and 313: Testing of Hypotheses-II 295 Table
- Page 314 and 315: Testing of Hypotheses-II 297 Illust
- Page 316 and 317: Testing of Hypotheses-II 299 Soluti
- Page 318 and 319: Testing of Hypotheses-II 301 r = nu
- Page 320 and 321: Testing of Hypotheses-II 303 where
- Page 322 and 323: Testing of Hypotheses-II 305 Table
- Page 324 and 325: Testing of Hypotheses-II 307 We now
- Page 326 and 327: Testing of Hypotheses-II 309 (ii) I
- Page 328 and 329: Testing of Hypotheses-II 311 all po
- Page 330 and 331: Testing of Hypotheses-II 313 CONCLU
- Page 332 and 333: Multivariate Analysis Techniques 31
268 <strong>Research</strong> <strong>Methodology</strong><br />
Table 11.5: The Anova Table<br />
Source of variation SS d.f. MS F-ratio 5% F-limit (or the<br />
tables values)<br />
Between columns<br />
(i.e., between varieties<br />
of seeds)<br />
8 (3 – 1) = 2 8/2 = 4 4/1 = 4 F(2, 6) = 5.14<br />
Between rows<br />
(i.e., between varieties<br />
of fertilizers)<br />
18 (4 – 1) = 3 18/3 = 6 6/1 = 6 F(3, 6) = 4.76<br />
Residual or error 6 (3 – 1) ×<br />
(4 – 1) = 6<br />
6/6=1<br />
Total 32 (3 × 4) – 1 = 11<br />
Illustration 3<br />
Set up ANOVA table for the following information relating to three drugs testing to judge the<br />
effectiveness in reducing blood pressure for three different groups of people:<br />
Amount of Blood Pressure Reduction in Millimeters of Mercury<br />
<strong>Dr</strong>ug<br />
X Y Z<br />
Group of People A 14 10 11<br />
15 9 11<br />
B 12 7 10<br />
11 8 11<br />
C 10 11 8<br />
11 11 7<br />
Do the drugs act differently?<br />
Are the different groups of people affected differently?<br />
Is the interaction term significant?<br />
Answer the above questions taking a significant level of 5%.<br />
Solution: We first make all the required computations as shown below:<br />
We can set up ANOVA table shown in Table 11.7 (Page 269).