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Test of goodness of fit Test of homogeneity Test of independence Purpose It is a type of Hypothesis testing, where population data with unknown distribution is being tested to see whether it fits any known distribution or not It is a type of Hypothesis testing, is used to check if 2 population data set with unknown distribution has the same distribution as each other It is a type of Hypothesis testing, is used to decide whether two variables are dependent or independent Data used for testing Single qualitative data or single survey question or single outcome is used for test Single qualitative survey question or experiment given to two different population Two qualitative survey questions or Experiments Test Population is uniform, Normal, Same as another with the known distribution Check whether the 2 data sets has same distribution as each other To see if population is related or unrelated H0 & H1 H0 - Population fits the given distribution H0 - Two population follow the same distribution H0 - Two variables are independent H1 - Population doesn’t fit the given distribution H1 - Two population doesn't follow same distribution H1 - Two variables are dependent

Chi-Square Test The test which helps us to measure the differences between the observed and expected value according to an assumed hypothesis is called Chi-Square Test. It was developed by Karl Pearson in 1900 and is the most important test amongst the several test of significance. It is a non-parametric test which is not based on any assumption or distribution of any variable There are 3 applications of Chi-Square Tests 1. Goodness of fit 2. Test for independence 3. Test of homogeneity Goodness of Fit The Chi-square goodness of fit test is used to compare a randomly collected sample containing a single categorical variable to a larger population. For example – If we want to test if 70% of ladies take medical, then we must use Goodness of Fit. This is measured using the below formula. Where: O = Observed Value E = Expected value If X2 (Calculated) > X2 (Tabulated) with (n-1) degree of freedom, then null hypothesis is rejected otherwise accepted. Let’s try to understand this with an example. In a medical college of 1000 students, there are 650 female students. Does it follow the theory that 70% of female students take medical? Observed Expected Male 350 300 Female 650 700 Total 1000 1000 The hypotheses for a Chi-square test of independence are as follows: Null Hypothesis (HO): The collected data is consistent with the population distribution. Alternative Hypothesis (HA): The collected data is not consistent with the population distribution. Observed Expected (O-E) (O-E)^2 (O-E)^2/E Male 350 300 50 2500 8.333333 Female 650 700 -50 2500 3.571429 Total 1000 1000 11.90476 X2 (Calculated) = 11.9 X2 (Tabulated) with (n-1) degree of freedom = 3.8 Decision: - If X2 (Calculated) > X2 (Tabulated) with (n-1) degree of freedom, then null hypothesis is rejected otherwise accepted. 11.9 > 3.8 and therefore the null hypothesis is rejected Test for independence This test helps us to identify if there is an association between two categorical variables within the same population. Let us understand this with an example: Table with Observed Value Qualification / City or Village Middle school High School Bachelors Masters PHD Total City 18 36 21 9 21 105 Village 12 36 45 36 6 135 Total 30 72 66 45 27 240 The hypotheses for a Chi-square test of independence are as follows: Null Hypothesis (HO): There is no association between the qualification and the places the students come from. Alternative Hypothesis (HA): There is association between the qualification and the places the students come from Table with Expected Value Qualification / City or Village Middle school High School Bachelors Masters PHD City (105x30)/240 = 13 (105x72)/240 = 32 29 20 12 Village 17 41 37 25 15 Observed Value (O) Expected Value (E) (O-E) (O-E)^2 (O-E)^2/E 18 13 5 23.8 1.8 36 32 5 20.3 0.6 21 29 -8 62.0 2.1 9 20 -11 114.2 5.8 21 12 9 84.4 7.1 12 17 -5 23.8 1.4 36 41 -5 20.3 0.5 45 37 8 62.0 1.7 36 25 11 114.2 4.5 6 15 -9 84.4 5.6 31.2 Degree of freedom = (Column -1) x (Row-1) = (5-1) x (2-1) =4 x 1 = 4 X2 (Calculated) =31.2 X2 (Tabulated) with 4 degree of freedom = 9.48 Decision: - If X2 (Calculated) > X2 (Tabulated) then null hypothesis is rejected otherwise accepted. 31.2 > 9.48 and therefore the null hypothesis is rejected Test of homogeneity We perform this test to confirm if the event is following uniformity or not. The basic difference from Test for independence is for two categorical variables within the same population and Test of Homogeneity is for single categorical variable within different population. Let us see if the TV watching pattern of males and females are same. The hypotheses for a Chi-square test of homogeneity are as follows: Null Hypothesis (HO): The distribution of watching pattern of TV for Males and females is same. Alternative Hypothesis (HA): The distribution of watching pattern of TV for Males and females is not same. Table with Observed Value Qualification / City or Village Movies Sports Serials Other Total Male 72 84 49 45 250 Female 91 86 88 35 300 Total 163 170 137 80 550 Table with Expected Value Qualification / City or Village Movies Sports Serials Other Male 74 77 62 36 Female 89 93 75 44 Observed Value (O) Expected Value (E) (O-E) (O-E)^2 (O-E)^2/E 72 74 -2 4.4 0.1 84 77 7 45.3 0.6 49 62 -13 176.2 2.8 45 36 9 74.6 2.1 91 89 2 4.4 0.0 86 93 -7 45.3 0.5 88 75 13 176.2 2.4 35 44 -9 74.6 1.7 10.1 Degree of freedom = (Column -1) = 4-1=3 X2 (Calculated) =10.1 X2 (Tabulated) with 3 degree of freedom = 7.8 Decision: - If X2 (Calculated) > X2 (Tabulated) then null hypothesis is rejected otherwise accepted. 10.1 > 7.8 and therefore the null hypothesis is rejected. The watching pattern of TV is not same between males and females.