Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Chi-square Test 251 Questions 1. What is Chi-square text? Explain its significance in statistical analysis. 2. Write short notes on the following: (i) Additive property of Chi-square; (ii) Chi-square as a test of ‘goodness of fit’; (iii) Precautions in applying Chi-square test; (iv) Conditions for applying Chi-square test. 3. An experiment was conducted to test the efficacy of chloromycetin in checking typhoid. In a certain hospital chloromycetin was given to 285 out of the 392 patients suffering from typhoid. The number of typhoid cases were as follows: Typhoid No Typhoid Total Chloromycetin 35 250 285 No chloromycetin 50 57 107 Total 85 307 392 With the help of χ 2 , test the effectiveness of chloromycetin in checking typhoid. (The χ 2 value at 5 per cent level of significance for one degree of freedom is 3.841). (M. Com., Rajasthan University, 1966) 4. On the basis of information given below about the treatment of 200 patients suffering from a disease, state whether the new treatment is comparatively superior to the conventional treatment. No. of patients Treatment Favourable No Response Response New 60 20 Conventional 70 50 For drawing your inference, use the value of χ 2 for one degree of freedom at the 5 per cent level of significance, viz., 3.84. 5. 200 digits were chosen at random from a set of tables. The frequencies of the digits were: Digit 0 1 2 3 4 5 6 7 8 9 Frequency 18 19 23 21 16 25 22 20 21 15 Calculate χ 2 . 6. Five dice were thrown 96 times and the number of times 4, 5, or 6 was thrown were Number of dice throwing 4, 5 or 6 5 4 3 2 1 0 Frequency Find the value of Chi-square. 8 18 35 24 10 1
252 Research Methodology 7. Find Chi-square from the following information: Condition of home Condition of child Total Clean Dirty Clean 70 50 120 Fairly clean 80 20 100 Dirty 35 45 80 Total 185 115 300 State whether the two attributes viz., condition of home and condition of child are independent (Use Chi-square test for the purpose). 8. In a certain cross the types represented by XY, Xy, xY and xy are expected to occur in a 9 : 5 : 4 : 2 ratio. The actual frequencies were: XY Xy xY xy 180 110 60 50 Test the goodness of fit of observation to theory. 9. The normal rate of infection for a certain disease in cattle is known to be 50 per cent. In an experiment with seven animals injected with a new vaccine it was found that none of the animals caught infection. Can the evidence be regarded as conclusive (at 1 per cent level of significance) to prove the value of the new vaccine? 10. Result of throwing die were recorded as follows: Number falling upwards 1 2 3 4 5 6 Frequency 27 33 31 29 30 24 Is the die unbiased? Answer on the basis of Chi-square test. 11. The Theory predicts the proportion of beans, in the four groups A, B, C and D should be 9 : 3 : 3 : 1. In an experiment among 1600 beans, the number in the four groups were 882, 313, 287 and 118. Does the experimental result support the theory? Apply χ 2 test. (M.B.A., Delhi University, 1975) 12. You are given a sample of 150 observations classified by two attributes A and B as follows: A 1 A 2 A 3 Total B 1 40 25 15 80 B 2 11 26 8 45 B 3 9 9 7 25 Total 60 60 30 150 Use the χ 2 test to examine whether A and B are associated. 13. A survey of 320 families with five children each revealed the following distribution: (M.A. Eco., Patiala University, 1975)
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Chi-square Test 251<br />
Questions<br />
1. What is Chi-square text? Explain its significance in statistical analysis.<br />
2. Write short notes on the following:<br />
(i) Additive property of Chi-square;<br />
(ii) Chi-square as a test of ‘goodness of fit’;<br />
(iii) Precautions in applying Chi-square test;<br />
(iv) Conditions for applying Chi-square test.<br />
3. An experiment was conducted to test the efficacy of chloromycetin in checking typhoid. In a certain<br />
hospital chloromycetin was given to 285 out of the 392 patients suffering from typhoid. The number of<br />
typhoid cases were as follows:<br />
Typhoid No Typhoid Total<br />
Chloromycetin 35 250 285<br />
No chloromycetin 50 57 107<br />
Total 85 307 392<br />
With the help of χ 2 , test the effectiveness of chloromycetin in checking typhoid.<br />
(The χ 2 value at 5 per cent level of significance for one degree of freedom is 3.841).<br />
(M. Com., Rajasthan University, 1966)<br />
4. On the basis of information given below about the treatment of 200 patients suffering from a disease,<br />
state whether the new treatment is comparatively superior to the conventional treatment.<br />
No. of patients<br />
Treatment Favourable No Response<br />
Response<br />
New 60 20<br />
Conventional 70 50<br />
For drawing your inference, use the value of χ 2 for one degree of freedom at the 5 per cent level of<br />
significance, viz., 3.84.<br />
5. 200 digits were chosen at random from a set of tables. The frequencies of the digits were:<br />
Digit 0 1 2 3 4 5 6 7 8 9<br />
Frequency 18 19 23 21 16 25 22 20 21 15<br />
Calculate χ 2 .<br />
6. Five dice were thrown 96 times and the number of times 4, 5, or 6 was thrown were<br />
Number of dice throwing<br />
4, 5 or 6 5 4 3 2 1 0<br />
Frequency<br />
Find the value of Chi-square.<br />
8 18 35 24 10 1