48 therefore the null hypothesis is rejected thus the

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freedom is 9.48. Therefore, the null hypothesis is rejected. Thus, the survey indicates flavour preference by consumers. Example - 5 The number of automobile accidents per week in a certain community are as follows : 12,8,20,2,14,10,15,6,9,4. Are these frequencies in agreement with the belief that accident conditions were the same during this 10 week period.
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141 Solution Expected frequency of accidents each week = 100 10 100 Null hypothesis 0 H : The accident conditions were the same during the 10 week period. O E O E 2 O E E 12 8 20 2 14 10 15 6 9 4 10 10 10 10 10 10 10 10 10 10 2 -2 10 -8 4 0 5 -4 -1 -6 0.4 0.4 10.0 6.4 1.6 0.0 2.5 1.6 0.1 3.6 100 26.6 Now 2 2 26.6 O E E 2 calculated 26.6 i.e., Here n observations are given degrees of freedom d.f. =10-1 9 2 2 2 2 9, 16.9 tabulated 16.9 tabulated calculated or The null hypothesis is rejected. i.e., the accident conditions were not the same during the 10 week period. Example - 6 A sample analysis of examination results of 500 students was made. It was found that 220 students had failed, 170 had secured a third class, 90 were placed in second class and 20 got a first class. Do these figures commensurate with the general examination result which is in the ratio of 4 : 3 : 2 : 1 for the various categories respectively.
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142 Solution Null hypothesis 0 H : The observed results commensurate with the general examination results. Expected frequencies are in the ratio of 4 : 3 : 2 : 1 Total frequency = 500 If we divided the total frequency 500 in the ratio 4 : 3 : 2 : 1we get the expected frequencies as 200, 150, 100, 50. Class Observed frequency (O) Expected frequency (E) O E 2 O E E Failed Third Second First 220 170 90 20 200 150 100 50 20 20 -10 -30 2.000 2.667 1.000 18.000 500 500 23.667 Now 2 2 23.667 O E E degrees of freedom d.f. =4-1 3 2 2 2 2 3, 23.667 tabulated 7.81 calculated tabulated or The null hypothesis is rejected. i.e., the observed results are not commensurate with the general examination results. 16.3.4. CONTINGENCY TABLE OF ORDER 2 2 There are many situations in which the contingency table has 2 rows and 2 columns. In case of contingency table of order 2 2 , short cut method i.e., a direct formula for chi-square can be used. This formula is derived from direct approach. Suppose the contingency table of order 2 2 is, 1 B 2 B Total 1 A 2 A a c b d a b c d Total a c b d a b c d n Where n is the sample size. One can of course calculate the value of chi-square by calculating the expected frequencies. But by algebraic multiplication, the direct formula for chi-square statistic is, 2 ( ) 2 ( )( )( )( ) n ad bc a b c d a c b d 2 has 1 d.f
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143 Decision about the independent of the attributes A and B can be taken in the usual way, i.e., 2 2 cal   1, reject 0 H . It means that the attributes A and B are dependent. Again if , 2 2 cal   1, accept 0 H . It leads to the conclusion that the attributes A and B are independent.
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