Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Chi-square Test 243 Show that the sampling technique of at least one research worker is defective. Solution: Let us take the hypothesis that the sampling techniques adopted by research workers are similar (i.e., there is no difference between the techniques adopted by research workers). This being so, the expectation of A investigator classifying the people in (i) Poor income group = (ii) Middle income group = 200 300 × 500 200 150 × 500 = 120 = 60 200 50 (iii) Rich income group = × = 20 500 Similarly the expectation of B investigator classifying the people in (i) Poor income group = (ii) Middle income group = (iii) Rich income group = 300 300 × 500 300 150 × 500 300 50 × 500 = 180 = 30 = 90 We can now calculate value of χ 2 as follows: Table 10.6 Groups Observed Expected O – E ij ij (O – E ) ij ij 2 Eij frequency frequency Investigator A Oij Eij classifies people as poor classifies people as 160 120 40 1600/120 = 13.33 middle class people 30 60 –30 900/60 = 15.00 classifies people as rich Investigator B 10 20 –10 100/20 = 5.00 classifies people as poor classifies people as 140 180 –40 1600/180 = 8.88 middle class people 120 90 30 900/90 = 10.00 classifies people as rich 40 30 10 100/30 = 3.33
244 Research Methodology Hence, χ 2 dOij −Eiji =∑ E Q Degrees of freedom = (c – 1) (r – 1) = (3 – 1) (2 – 1) = 2. ij 2 = 55.54 The table value of χ 2 for two degrees of freedom at 5 per cent level of significance is 5.991. The calculated value of χ 2 is much higher than this table value which means that the calculated value cannot be said to have arisen just because of chance. It is significant. Hence, the hypothesis does not hold good. This means that the sampling techniques adopted by two investigators differ and are not similar. Naturally, then the technique of one must be superior than that of the other. Illustration 8 Eight coins were tossed 256 times and the following results were obtained: Numbers of heads 0 1 2 3 4 5 6 7 8 Frequency 2 6 30 52 67 56 32 10 1 Are the coins biased? Use χ 2 test. Solution: Let us take the hypothesis that the coins are not biased. If that is so, the probability of any one coin falling with head upward is 1/2 and with tail upward is 1/2 and it remains the same whatever be the number of throws. In such a case the expected values of getting 0, 1, 2, … heads in a single throw in 256 throws of eight coins will be worked out as follows * . Table 10.7 Events or Expected frequencies No. of heads 0 8 1 8 2 8 0 1 2 0 8 1 2 256 1 C F H G I K J F H G I K J × = 1 1 2 1 7 1 2 256 8 C F H G I K J F H G I K J × = 2 1 2 2 6 1 2 256 28 C F H G I K J F H G I K J × = ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ contd. * The probabilities of random variable i.e., various possible events have been worked out on the binomial principle viz., through the expansion of (p + q) n where p = 1/2 and q = 1/2 and n = 8 in the given case. The expansion of the term n Cr p r q n–r has given the required probabilities which have been multiplied by 256 to obtain the expected frequencies.
- Page 210 and 211: Testing of Hypotheses I 193 MEASURI
- Page 212 and 213: Testing of Hypotheses I 195 We can
- Page 214 and 215: Testing of Hypotheses I 197 HYPOTHE
- Page 216 and 217: 1 2 3 4 5 z OR X − X 1 2 2 2 p p
- Page 218 and 219: 1 2 3 4 5 z = p q 0 0 p$ − p$ F H
- Page 220 and 221: Testing of Hypotheses I 203 to have
- Page 222 and 223: Testing of Hypotheses I 205 S. No.
- Page 224 and 225: Testing of Hypotheses I 207 S. No.
- Page 226 and 227: Testing of Hypotheses I 209 nX + nX
- Page 228 and 229: Testing of Hypotheses I 211 (Since
- Page 230 and 231: Testing of Hypotheses I 213 Table 9
- Page 232 and 233: Testing of Hypotheses I 215 σ diff
- 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 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 284 and 285: 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
244 <strong>Research</strong> <strong>Methodology</strong><br />
Hence,<br />
χ 2<br />
dOij −Eiji<br />
=∑<br />
E<br />
Q Degrees of freedom = (c – 1) (r – 1)<br />
= (3 – 1) (2 – 1) = 2.<br />
ij<br />
2<br />
= 55.54<br />
The table value of χ 2 for two degrees of freedom at 5 per cent level of significance is 5.991.<br />
The calculated value of χ 2 is much higher than this table value which means that the calculated<br />
value cannot be said to have arisen just because of chance. It is significant. Hence, the hypothesis<br />
does not hold good. This means that the sampling techniques adopted by two investigators differ and<br />
are not similar. Naturally, then the technique of one must be superior than that of the other.<br />
Illustration 8<br />
Eight coins were tossed 256 times and the following results were obtained:<br />
Numbers of heads 0 1 2 3 4 5 6 7 8<br />
Frequency 2 6 30 52 67 56 32 10 1<br />
Are the coins biased? Use χ 2 test.<br />
Solution: Let us take the hypothesis that the coins are not biased. If that is so, the probability of any<br />
one coin falling with head upward is 1/2 and with tail upward is 1/2 and it remains the same whatever<br />
be the number of throws. In such a case the expected values of getting 0, 1, 2, … heads in a single<br />
throw in 256 throws of eight coins will be worked out as follows * .<br />
Table 10.7<br />
Events or Expected frequencies<br />
No. of heads<br />
0 8<br />
1 8<br />
2<br />
8<br />
0<br />
1<br />
2<br />
0 8<br />
1<br />
2<br />
256 1<br />
C F H G I K J F H G I K J × =<br />
1<br />
1<br />
2<br />
1 7<br />
1<br />
2<br />
256 8<br />
C F H G I K J F H G I K J × =<br />
2<br />
1<br />
2<br />
2 6<br />
1<br />
2<br />
256 28<br />
C F H G I K J F H G I K J × =<br />
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○<br />
contd.<br />
* The probabilities of random variable i.e., various possible events have been worked out on the binomial principle viz.,<br />
through the expansion of (p + q) n where p = 1/2 and q = 1/2 and n = 8 in the given case. The expansion of the term<br />
n Cr p r q n–r has given the required probabilities which have been multiplied by 256 to obtain the expected frequencies.