Question

1) What’s the difference among the chi-square test for goodness of fit, the chi-square test for independence, and the chi-square test for homogeneity

2) State the requirements to perform a chi-square test

Answer 1

Chi-Square test of independence helps us to find whether 2 or more attributes are associated or not.e.g. whether playing chess helps boost the child's math or not. It is not a measure of the degree of relationship between the attributes. it only tells us whether two principles of classification are significantly related or not, without reference to any assumptions concerning the form of relationship.

Chi-Square test of homogeneity is an extension of chi square test of independence...tests of homogeneity are useful to determine whether 2 or more independent random samples are drawn from the same population or from different populations. instead of one sample- as we use with independence problem, here we have two or more samples.

Both the types of tests are concerned with cross classified data. both use the same testing statistics. However they are different from each other.
Test for independence is concerned with whether one attribute is independent of the other and involves a single sample from the population.
On the other hand, test of homogeneity tests whether different samples come from same population. It involves 2 or more independent samples-one from each of the populations in question.

Chi-Square goodness of fit test is a non-parametric test that is used to find out how the observed value of a given phenomena is significantly different from the expected value. In Chi-Square goodness of fit test, the term goodness of fit is used to compare the observed sample distribution with the expected probability distribution. Chi-Square goodness of fit test determines how well theoretical distribution (such as normal, binomial, or Poisson) fits the empirical distribution. In Chi-Square goodness of fit test, sample data is divided into intervals. Then the numbers of points that fall into the interval are compared, with the expected numbers of points in each interval.

Procedure for Chi-Square Goodness of Fit Test:

• Set up the hypothesis for Chi-Square goodness of fit test:

A. Null hypothesis: In Chi-Square goodness of fit test, the null hypothesis assumes that there is no significant difference between the observed and the expected value.

B. Alternative hypothesis: In Chi-Square goodness of fit test, the alternative hypothesis assumes that there is a significant difference between the observed and the expected value.

• Compute the value of Chi-Square goodness of fit test using the following formula:

Where, = Chi-Square goodness of fit test O= observed value E= expected value

Important things to note when considering using the Chi-Square test

First, Chi-Square only tests whether two individual variables are independent in a binary, “yes” or “no” format.

Chi-Square testing does not provide any insight into the degree of difference between the respondent categories, meaning that researchers are not able to tell which statistic (result of the Chi-Square test) is greater or less than the other.

Second, Chi-Square requires researchers to use numerical values, also known as frequency counts, instead of using percentages or ratios. This can limit the flexibility that researchers have in terms of the processes that they use.

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