Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Chi-square Test 235 S. No. 1 2 3 4 5 6 7 8 9 10 Weight (kg.) 38 40 45 53 47 43 55 48 52 49 Can we say that the variance of the distribution of weight of all students from which the above sample of 10 students was drawn is equal to 20 kgs? Test this at 5 per cent and 1 per cent level of significance. 2 Solution: First of all we should work out the variance of the sample data or σ s and the same has been worked out as under: Table 10.1 S. No. X i (Weight in kgs.) (X i – X ) (X i – X ) 2 1 38 –9 81 2 40 –7 49 3 45 –2 04 4 53 +6 36 5 47 +0 00 6 43 –4 16 7 55 +8 64 8 48 +1 01 9 52 +5 25 10 49 +2 04 d i i 2 n = 10 ∑ Xi = 470 ∑ X − X = 280 ∴ σ s or σ s 2 X X = n ∑ = ∑ − i 470 = = 10 dXiXi 2 n − 1 = 3111 . . 47 kgs. = 280 = 10 − 1 3111 . 2 2 Let the null hypothesis be H0: σp = σs . In order to test this hypothesis we work out the χ 2 value as under: χ 2 2 σ s = n − 1 2 σ p b g
236 Research Methodology 3111 = 10 − 1 20 . b g = 13.999. Degrees of freedom in the given case is (n – 1) = (10 – 1) = 9. At 5 per cent level of significance the table value of χ 2 = 16.92 and at 1 per cent level of significance, it is 21.67 for 9 d.f. and both these values are greater than the calculated value of χ 2 which is 13.999. Hence we accept the null hypothesis and conclude that the variance of the given distribution can be taken as 20 kgs at 5 per cent as also at 1 per cent level of significance. In other words, the sample can be said to have been taken from a population with variance 20 kgs. Illustration 2 A sample of 10 is drawn randomly from a certain population. The sum of the squared deviations from the mean of the given sample is 50. Test the hypothesis that the variance of the population is 5 at 5 per cent level of significance. Solution: Given information is ∴ σ s 2 n = 10 d i i 2 dXi 2 Xi ∑ X − X = 50 = ∑ − n − 1 2 2 Take the null hypothesis as H0: σp = σs . In order to test this hypothesis, we work out the χ 2 value as under: Degrees of freedom = (10 – 1) = 9. χ 2 2 p b g 50 b g = 50 9 σ s 9 50 1 9 = n − 1 = 10 − 1 = × × = 10 2 σ 5 9 5 1 The table value of χ 2 at 5 per cent level for 9 d.f. is 16.92. The calculated value of χ 2 is less than this table value, so we accept the null hypothesis and conclude that the variance of the population is 5 as given in the question. CHI-SQUARE AS A NON-PARAMETRIC TEST Chi-square is an important non-parametric test and as such no rigid assumptions are necessary in respect of the type of population. We require only the degrees of freedom (implicitly of course the size of the sample) for using this test. As a non-parametric test, chi-square can be used (i) as a test of goodness of fit and (ii) as a test of independence.
- Page 202 and 203: Testing of Hypotheses I 185 Charact
- Page 204 and 205: Testing of Hypotheses I 187 when th
- Page 206 and 207: Testing of Hypotheses I 189 Mathema
- Page 208 and 209: Testing of Hypotheses I 191 PROCEDU
- 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 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 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
236 <strong>Research</strong> <strong>Methodology</strong><br />
3111<br />
= 10 − 1<br />
20<br />
. b g = 13.999.<br />
Degrees of freedom in the given case is (n – 1) = (10 – 1) = 9. At 5 per cent level of significance<br />
the table value of χ 2 = 16.92 and at 1 per cent level of significance, it is 21.67 for 9 d.f. and both<br />
these values are greater than the calculated value of χ 2 which is 13.999. Hence we accept the null<br />
hypothesis and conclude that the variance of the given distribution can be taken as 20 kgs at 5 per<br />
cent as also at 1 per cent level of significance. In other words, the sample can be said to have been<br />
taken from a population with variance 20 kgs.<br />
Illustration 2<br />
A sample of 10 is drawn randomly from a certain population. The sum of the squared deviations from<br />
the mean of the given sample is 50. Test the hypothesis that the variance of the population is 5 at<br />
5 per cent level of significance.<br />
Solution: Given information is<br />
∴ σ s<br />
2<br />
n = 10<br />
d i i 2<br />
dXi 2<br />
Xi<br />
∑ X − X = 50<br />
= ∑ −<br />
n − 1<br />
2 2<br />
Take the null hypothesis as H0: σp = σs<br />
. In order to test this hypothesis, we work out the χ 2<br />
value as under:<br />
Degrees of freedom = (10 – 1) = 9.<br />
χ<br />
2<br />
2<br />
p<br />
b g<br />
50<br />
b g<br />
=<br />
50<br />
9<br />
σ s<br />
9 50 1 9<br />
= n − 1 = 10 − 1 = × × = 10<br />
2<br />
σ<br />
5 9 5 1<br />
The table value of χ 2 at 5 per cent level for 9 d.f. is 16.92. The calculated value of χ 2 is less<br />
than this table value, so we accept the null hypothesis and conclude that the variance of the population<br />
is 5 as given in the question.<br />
CHI-SQUARE AS A NON-PARAMETRIC TEST<br />
Chi-square is an important non-parametric test and as such no rigid assumptions are necessary in<br />
respect of the type of population. We require only the degrees of freedom (implicitly of course the<br />
size of the sample) for using this test. As a non-parametric test, chi-square can be used (i) as a test<br />
of goodness of fit and (ii) as a test of independence.