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*
*Chi- square test is the test of significance.
*It was first of all used by Karl Pearson in the year 1900.
*Chi-square test is a useful measure of comparing
experimentally obtained result with those expected
theoretically and based on the hypothesis.
*It is denoted by the Gr. sign-
*Following is the formula.
*It is a mathematical expression, representing the ratio between
experimentally obtained result (O) and the theoretically expected
result (E) based on certain hypothesis. It uses data in the form of
frequencies (i.e., the number of occurrence of an event).
*Chi-square test is calculated by dividing the square of the overall
deviation in the observed and expected frequencies by the
expected frequency.
*If there is no difference between actual and
observed frequencies, the value of chi-square is
zero.
*If there is a difference between observed and
expected frequencies, then the value of chi-
square would be more than zero.
*But the difference in the observed frequencies
may also be due to sampling fluctuations and it
should be ignored in drawing inference.
*
*In test, while comparing the calculated value
of with the table value, we have to calculated
the degree of freedom. The degree of freedom
is calculated from the no. of classes. Therefore,
the no. of degrees of freedom in a test is equal
to the no. of classes minus one.
*If there are two classes, three classes, and four classes, the
degree of freedom would be 2-1, 3-1, and 4-1, respectively. In
a contingency table, the degree of freedom is calculated in a
different manner:
d.f. = (r-1) (c-1)
where- r = number of row in a table,
c = number of column in a table.
*Thus in a 2×2 contingency table, the degree of freedom is
(2-1 ) (2-1) = 1. Similarly, in a 3×3 contingency table, the
number of degree of freedom is (3-1) (3-1) = 4. Likewise in 3×4
contingency table the degree of freedom is (3-1) (4-1) = 6, and
so on.
*
*The term contingency table was first used by Karl
Pearson.
*A contingency table is a type of table in a matrix
format that displays the (multivariate) frequency
distribution of the variables. They are heavily used in
survey research, business intelligence, engineering and
scientific research. They provide a basic picture of the
interrelation between two variables and can help find
interactions between them.
*The value of depends on the no. of classes or in
other words on the number of degrees of freedom
(d. f.) and the critical level of probability.
*2×2 table when there are only two sample, each
divided into classes and a 2×2 contingency table is
prepared. It is also known ad four fold or four
cell table.
Column 1 Column 2 ROW Total
ROW 1
+ + RT 1
ROW 2
+ + RT 2
Column Total CT 1 CT 2
Degree of freedom = (r-1) (c-1)
= (2-1) (2-1)
= 1×1
= 1
2 rows × 2 columns
The chi-square distribution has some important characteristics
1. This test is based on frequencies, whereas, in theoretical distribution the
test is based on mean and standard deviation.
2. The other distribution can be used for testing the significance of the
difference between a single expected value and observed proportion.
However this test can be used for testing difference between the entire set
of the expected and the observed frequencies.
3. A new chi-square distribution is formed for every increase in is the
number of degree of freedom.
4. This rest is applied for testing the hypothesis but is not useful for
estimation
*
*
There are a few assumptions for the validity of chi-square test.
1. All the observation must be independent. No individual item should be
included twice or a number of items in the sample.
2. The total number of observation should be large. The chi-square test
should not be used if n>50.
3. All the events must be mutually exclusive.
4. For comparison purposes, the data must be in original units.
5. If the theoretical frequencies is less than five, than we pool it with the
preceding or the succeeding frequency, so that the resulting sum is
greater than five.
*
*The chi-square test is applicable to varied problems in agriculture, biology
and medical science-
1. To test the goodness of fit.
2. To test the independence of attributes.
3. To test the homogeneity of independent estimates of the population
variance.
4. To test the detection of linkage.
*
*Following steps are required to calculate the value of chi-square.
1. Identify the problem
2. Make a contingency table and note the observed frequency (O) is each classes of one
event, row wise i.e. horizontally and then the numbers in each group of the other
event, column wise i.e. vertically.
3. Set up the Null hypothesis (Ho); According to Null hypothesis, no association exists
between attributes. This need s setting up of alternative hypothesis (HA). It assumes
that an association exists between the attributes.
4. Calculate the expected frequencies (E).
5. Find the difference between observed and Expected frequency in each cell (O-E).
6. Calculate the chi-square value applying the formula. The value is ranges from zero to
Infinite.
*Solved Example-
Two varieties of snapdragons, one with red flower and other with white flowers were
crossed. The result obtained in F2 generation are: 22 red, 52 pink, and 23 white flower
plants. Now it is desired to ascertain these figures show that segregation occurs in the
simple Mendelian ratio of 1:2:1.
*Solution-
Null hypothesis- H0 : The genes carrying red color and white color
characters are segregating in simple Mendelian ratio of 1:2:1
*Expected frequencies:
Red =
1
4
× 97 = 24.25
Pink =
2
4
× 97 = 48.50
White =
1
4
× 97 = 24.25
Red Pink White Total
Observed frequency (O) 22 52 23 97
Expected frequency (E) 24.25 48.50 24.25 97
Deviation (O-E) -2.25 3.50 -1.25
= (-2.25)2/ 24.25 + (3.50)2/ 48.50 + (-1.25)2/ 24.25
=
5.06
24.25
+
12.25
48.50
+
1.56
24.25
= 0.21 + 0.25 + 0.06
= 0.53
Conclusion –
The calculated Chi-square value (.53) is less than the
tabulated chi-square value (5.99) at 5% level of probability for 2 d.
f. The hypothesis is, therefore, in agreement with the recorded
facts.
*
*Singh, Pandey, Jain, Plant Resource Utilization,
Palynology, Biostatistics.
*Khan and Khanum, Fundamentals of Biostatistics.
*

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Chi -square test

  • 1.
  • 2. * *Chi- square test is the test of significance. *It was first of all used by Karl Pearson in the year 1900. *Chi-square test is a useful measure of comparing experimentally obtained result with those expected theoretically and based on the hypothesis. *It is denoted by the Gr. sign- *Following is the formula.
  • 3. *It is a mathematical expression, representing the ratio between experimentally obtained result (O) and the theoretically expected result (E) based on certain hypothesis. It uses data in the form of frequencies (i.e., the number of occurrence of an event). *Chi-square test is calculated by dividing the square of the overall deviation in the observed and expected frequencies by the expected frequency.
  • 4. *If there is no difference between actual and observed frequencies, the value of chi-square is zero. *If there is a difference between observed and expected frequencies, then the value of chi- square would be more than zero. *But the difference in the observed frequencies may also be due to sampling fluctuations and it should be ignored in drawing inference.
  • 5. * *In test, while comparing the calculated value of with the table value, we have to calculated the degree of freedom. The degree of freedom is calculated from the no. of classes. Therefore, the no. of degrees of freedom in a test is equal to the no. of classes minus one.
  • 6. *If there are two classes, three classes, and four classes, the degree of freedom would be 2-1, 3-1, and 4-1, respectively. In a contingency table, the degree of freedom is calculated in a different manner: d.f. = (r-1) (c-1) where- r = number of row in a table, c = number of column in a table. *Thus in a 2×2 contingency table, the degree of freedom is (2-1 ) (2-1) = 1. Similarly, in a 3×3 contingency table, the number of degree of freedom is (3-1) (3-1) = 4. Likewise in 3×4 contingency table the degree of freedom is (3-1) (4-1) = 6, and so on.
  • 7. * *The term contingency table was first used by Karl Pearson. *A contingency table is a type of table in a matrix format that displays the (multivariate) frequency distribution of the variables. They are heavily used in survey research, business intelligence, engineering and scientific research. They provide a basic picture of the interrelation between two variables and can help find interactions between them.
  • 8. *The value of depends on the no. of classes or in other words on the number of degrees of freedom (d. f.) and the critical level of probability. *2×2 table when there are only two sample, each divided into classes and a 2×2 contingency table is prepared. It is also known ad four fold or four cell table.
  • 9. Column 1 Column 2 ROW Total ROW 1 + + RT 1 ROW 2 + + RT 2 Column Total CT 1 CT 2 Degree of freedom = (r-1) (c-1) = (2-1) (2-1) = 1×1 = 1 2 rows × 2 columns
  • 10. The chi-square distribution has some important characteristics 1. This test is based on frequencies, whereas, in theoretical distribution the test is based on mean and standard deviation. 2. The other distribution can be used for testing the significance of the difference between a single expected value and observed proportion. However this test can be used for testing difference between the entire set of the expected and the observed frequencies. 3. A new chi-square distribution is formed for every increase in is the number of degree of freedom. 4. This rest is applied for testing the hypothesis but is not useful for estimation *
  • 11. * There are a few assumptions for the validity of chi-square test. 1. All the observation must be independent. No individual item should be included twice or a number of items in the sample. 2. The total number of observation should be large. The chi-square test should not be used if n>50. 3. All the events must be mutually exclusive. 4. For comparison purposes, the data must be in original units. 5. If the theoretical frequencies is less than five, than we pool it with the preceding or the succeeding frequency, so that the resulting sum is greater than five.
  • 12. * *The chi-square test is applicable to varied problems in agriculture, biology and medical science- 1. To test the goodness of fit. 2. To test the independence of attributes. 3. To test the homogeneity of independent estimates of the population variance. 4. To test the detection of linkage.
  • 13. * *Following steps are required to calculate the value of chi-square. 1. Identify the problem 2. Make a contingency table and note the observed frequency (O) is each classes of one event, row wise i.e. horizontally and then the numbers in each group of the other event, column wise i.e. vertically. 3. Set up the Null hypothesis (Ho); According to Null hypothesis, no association exists between attributes. This need s setting up of alternative hypothesis (HA). It assumes that an association exists between the attributes. 4. Calculate the expected frequencies (E). 5. Find the difference between observed and Expected frequency in each cell (O-E). 6. Calculate the chi-square value applying the formula. The value is ranges from zero to Infinite.
  • 14. *Solved Example- Two varieties of snapdragons, one with red flower and other with white flowers were crossed. The result obtained in F2 generation are: 22 red, 52 pink, and 23 white flower plants. Now it is desired to ascertain these figures show that segregation occurs in the simple Mendelian ratio of 1:2:1. *Solution- Null hypothesis- H0 : The genes carrying red color and white color characters are segregating in simple Mendelian ratio of 1:2:1 *Expected frequencies: Red = 1 4 × 97 = 24.25 Pink = 2 4 × 97 = 48.50 White = 1 4 × 97 = 24.25
  • 15. Red Pink White Total Observed frequency (O) 22 52 23 97 Expected frequency (E) 24.25 48.50 24.25 97 Deviation (O-E) -2.25 3.50 -1.25 = (-2.25)2/ 24.25 + (3.50)2/ 48.50 + (-1.25)2/ 24.25
  • 16. = 5.06 24.25 + 12.25 48.50 + 1.56 24.25 = 0.21 + 0.25 + 0.06 = 0.53 Conclusion – The calculated Chi-square value (.53) is less than the tabulated chi-square value (5.99) at 5% level of probability for 2 d. f. The hypothesis is, therefore, in agreement with the recorded facts.
  • 17. * *Singh, Pandey, Jain, Plant Resource Utilization, Palynology, Biostatistics. *Khan and Khanum, Fundamentals of Biostatistics.
  • 18. *