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By 
Akhil sankar
 The chi-square test is useful measure of 
comparing experimentally obtained results 
with those expected theoretically and based 
one the hypothesis . 
 In general chi-square best is applied to those 
problems with which a given event has 
occurred is significantly different from the 
one as expected theoretically.
 1 or more categories 
 Independent observations 
 A sample size of at least 10 
 Random sampling 
 All observations must be used 
 For the test to be accurate, the expected 
frequency should be at least 5
1) Make a hypothesis based on your basic 
biological question 
2) Determine the expected frequencies 
3) Create a table with observed frequencies, 
expected frequencies, and chi-square values 
using the formula: 
Χ2=(O-E)2 
E 
4) Find the degrees of freedom: (c-1)(r-1) 
5) Find the chi-square statistic in the Chi-Square 
Distribution table 
6) If chi-square statistic > your calculated chi-square 
value, you do not reject your null 
hypothesis and vice versa.
 There is no significant difference between 
the observed and expected frequencies. 
 Chi-Square formula: 
χ2 =  (O-E)2 
E
Package Color 
preference by 
Consumers 
Red 70 
Blue 106 
Green 80 
Pink 70 
Orange 74 
Total 400 
Random sample :400
Package 
Color 
Observed 
Frequencie 
s (O) 
Expected 
Frequencie 
s (E) 
(O-E)² 
χ2 =  (O-E)2 
E 
Red 70 80 100 1.250 
Blue 106 80 676 8.450 
Green 80 80 0 0.000 
Pink 70 80 100 1.250 
Orange 74 80 36 0.450 
Total 400 400 11.400 
value of chi-square ( X²) = 11.400 
Degree of freedom = (c-1)(r-1) 
(2-1)(5-1) 
=4
 null hypothesis: equal preference for all 
colors being true, the expected frequencies 
for all the colors will be equal to 80. 
 The critical value of X2 at 5% level of 
significance for 4 degrees of freedom is 
9.488. 
 So, the null hypothesis is rejected.
 Tossing of coin 100 times. 
 Observation 68 H and 32 T. 
No. O E (O-E) (O-E)2 
1 68 50 18 324 
2 32 50 -18 324 
-------- ---------- 
100 64 8 
Degree of freedom = (c-1)(r-1) 
(2-1)(2-1)=1 
Level of significance = 5%
χ2 = 648/100 
= 6.48 
Critical value is 3.841 
Reject the null hypothesis
THANK U

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Chisquare

  • 2.  The chi-square test is useful measure of comparing experimentally obtained results with those expected theoretically and based one the hypothesis .  In general chi-square best is applied to those problems with which a given event has occurred is significantly different from the one as expected theoretically.
  • 3.  1 or more categories  Independent observations  A sample size of at least 10  Random sampling  All observations must be used  For the test to be accurate, the expected frequency should be at least 5
  • 4. 1) Make a hypothesis based on your basic biological question 2) Determine the expected frequencies 3) Create a table with observed frequencies, expected frequencies, and chi-square values using the formula: Χ2=(O-E)2 E 4) Find the degrees of freedom: (c-1)(r-1) 5) Find the chi-square statistic in the Chi-Square Distribution table 6) If chi-square statistic > your calculated chi-square value, you do not reject your null hypothesis and vice versa.
  • 5.  There is no significant difference between the observed and expected frequencies.  Chi-Square formula: χ2 =  (O-E)2 E
  • 6. Package Color preference by Consumers Red 70 Blue 106 Green 80 Pink 70 Orange 74 Total 400 Random sample :400
  • 7. Package Color Observed Frequencie s (O) Expected Frequencie s (E) (O-E)² χ2 =  (O-E)2 E Red 70 80 100 1.250 Blue 106 80 676 8.450 Green 80 80 0 0.000 Pink 70 80 100 1.250 Orange 74 80 36 0.450 Total 400 400 11.400 value of chi-square ( X²) = 11.400 Degree of freedom = (c-1)(r-1) (2-1)(5-1) =4
  • 8.  null hypothesis: equal preference for all colors being true, the expected frequencies for all the colors will be equal to 80.  The critical value of X2 at 5% level of significance for 4 degrees of freedom is 9.488.  So, the null hypothesis is rejected.
  • 9.  Tossing of coin 100 times.  Observation 68 H and 32 T. No. O E (O-E) (O-E)2 1 68 50 18 324 2 32 50 -18 324 -------- ---------- 100 64 8 Degree of freedom = (c-1)(r-1) (2-1)(2-1)=1 Level of significance = 5%
  • 10. χ2 = 648/100 = 6.48 Critical value is 3.841 Reject the null hypothesis