How to write a Null Hypothesis and Alternative Hypothesis for two independent nominal Chi Square test.
Example variable:
1. Currently have COPD
2. Visits to the Doctors
How to write a Null Hypothesis and Alternative Hypothesis for two independent nominal Chi Square test....
The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.AnswerTrueFalse
A chi-square test is used to test the null hypothesis that sex and preferred color are independent. Which of the following statements is a correct decision if P-value is 0.23?
Istanbul Aydin Oniversitesi Elektronik Sine S The Alternative Hypothesis (H)) for the Chi-Square test of independence should specify OA) that the two categorical variables are independent B) No of them C) that the two numerical variables are not independent OD) that the two categorical variables are not independent that the two numerical variables are independent
Question 1 In a chi -square test of independence , the null hypothesis states that-----------. a. the two variables of interest are related in the population. b. the column frequencies equal the row frequencies c. the sum of the and the column frequencies equal the total frequency d. the two variables of interest are unrelated in the population Question 2 A survey asked people whether they identified as a morning person, night person, or had no preference. a. 270 replied...
Which of the following could be a null hypothesis for a chi-square goodness-of-fit test? a. The means increase evenly across categories. b. The dependent variable is normally distributed. c. The variances are equal across categories. d. The frequencies in each category are equal
When we carry out a chi- square goodness-of-fit test for a normal distribution, the null hypothesis states that the population Does not have a normal distribution Has a normal distribution Has a chi-square distribution Does not have a chi-square distribution Has k-3 degrees of freedom
If I reject the null hypothesis of a Chi-Square Goodness-of-Fit Test, which of the following statement best summarizes my conclusion? There is sufficient evidence to conclude that the two variables are independent. There is sufficient evidence to conclude that the two variables are associated. There is not sufficient evidence to conclude that my proposed distribution does not fit the population. There is sufficient evidence to conclude that my proposed distribution does not fit the population.
When we carry out a chi-square goodness-of-fit test for a normal distribution, the null hypothesis states that the population: a) does not have a normal distribution. b) has a normal distribution. c) has a chi-square distribution. d) does not have a chi-square distribution. e) has k − 3 degrees of freedom.
MC Question: With Chi-square test the expected values reflect the empirical manifestation of: (a) The alternative hypothesis (b) the null hypothesis (c) the sampling distribution (d) a random distribution
Which step should be taken next if the null hypothesis is rejected in a chi-square test? OA. O B. Create confidence intervals to estimate the differences. Examine the standardized residuals to understand the pattern C. Check that the conditions are satisfied. D. Plot the data on a chart to understand the pattern. Which of the following is NOT one of the tests discussed in Chapter 16? A Chi-square test for independence B. Chi-square goodness-of-fit test OC. CH-square test for contingency...