1. When the null hypothesis for a chi square test for independence is true. a. a large differe

Myah Fuller

Myah Fuller

Answered question

2022-03-01

1. When the null hypothesis for a chi square test for independence is true.
a. a large difference between the observed frequencies and the expected frequencies.
b.little or no difference between the observed frequencies and the expected frequencies.
c. no difference between the observed frequencies and the marginals
d. no differences between the row and the column marginal
2. In the Chi Square test, degrees of freedom are calculated as :
a. N-1
b. N1+N22
c. (r+1)(c+1)
d. (r-1)(c-1)

Answer & Explanation

litoshypinaylh4

litoshypinaylh4

Beginner2022-03-02Added 6 answers

Note:
Hi there! Thank you for posting the question. As your question has more than 1 questions, we have solved only the first question for you. If you need the other question to be answered, please re-submit the question by specifying the question number or name.
Chi-square test for independence:
Chi-square test for independence is used to test whether two categories of interest are independent of each other or not.
The test hypotheses are given below:
Null hypothesis:
H0: The two categories of interest are independent.
Alternative hypothesis:
Hα: The two categories of interest are not independent.
Evidently, the test is to determine whether the values of one category can help to determine the values of another category.
Thus, the study seeks to find whether there is any association between the two categories.
Given information:
Here, it is given that the null hypothesis of chi-square test for independence is true.
That is, the two categories of interest are independent.
Therefore, it can be concluded that the values of one category cannot help to determine the values of another category.
This implies that there is no significant difference between observed frequencies and the expected frequencies.
Therefore, when the null hypothesis for a chi-square test for independence is true, there will be no difference between the observed and expected frequencies.
Hence, option (b) is correct.
Answer: Option(b).

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