in chi square test for independence
null hypothesis is H0 : the attributes are statistically independent
then
alternative hypothesis H1 : the attributes are statistically dependent
so option are statistically dependent is true
10.00 points When we carry out a chi-square test of independence, the alternate hypothesis states that...
When we carry out a chi-square test of independence, the chi-square statistic is based on (rxc)-1 degrees of freedom, where r and c denote, respectively, the number of rows and columns in the contingency table. True or false
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.
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
When Chi-square distribution is used as a test of independence, the number of degrees of freedom is related to both the number of rows and the number of columns in the contingency table. Select one: True False Question 2 Answer saved Points out of 1.000 Flag question Question text A goodness of fit test can be used to determine if membership in categories of one variable is different as a function of membership in the categories of a second variable...
The chi-square goodness of fit test can be used when: Select one: a. We conduct a multinomial experiment. b. We perform a hypothesis test to determine if a population has a normal distribution. c. We perform a hypothesis test to determine if two population variances significantly differ from each other. d. We conduct a binomial experiment. The x statistic from a contingency table with 6 rows and five columns will have Select one: a. 24 degrees of freedom b. 50...
The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.AnswerTrueFalse
For a chi-square test of independence, we calculate the degrees of freedom using which formula? A. ??=???? × ??????? B. ??=???? + ??????? C. ??=(????−1)×(???????−1) D. ??=(????−1)+(???????−1)
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...
You observe 100 randomly selected college students to find out whether they arrive on time or late for their classes. The table below gives a two-way classification for these students.GenderOn TimeLateFemale359Male4313For a chi-square test of independence for this contingency table, what is the number of degrees of freedom?
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