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 we carry out a chi- square goodness-of-fit test for a normal distribution, the null hypothesis...
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.
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...
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.
Assume that a Chi-square test was conducted to test the goodness of fit to a 3:1 ratio and that a Chi-square value of 2.62 was obtained (Table value is equal to 3.84). Should the null hypothesis be accepted? How many degrees of freedom would be associated with this test of significance?
In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expected cell frequencies are at least five. The sample is divided into 7 intervals. The second through the sixth intervals all have expected cell frequencies of at least five. The first and the last intervals have expected cell frequencies of 1.5 each. After adjusting the number of intervals, the degrees of freedom for the chi-square statistic is O 2 3...
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 test statistic for goodness of fit has a chi-square distribution with k - 1 degrees of freedom provided that the expected frequencies for all categories are a. 10 or more. b. k or more. c. 2k. d. 5 or more.
When using a chi-square goodness-of-fit test with multinomial probabilities, the rejection of the null hypothesis indicates that at least one of the multinomial probabilities is not equal to the value stated in the null hypothesis. O True False
10.00 points When we carry out a chi-square test of independence, the alternate hypothesis states that the two relevant classifications O are mutually exclusive. O form a contingency table with rrows and c columns O have (1 - 1)(C-1) degrees of freedom are statistically dependent O O are normally distributed
In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expected cell frequencies are at least five. The sample is divided into 7 intervals. The second through the sixth intervals all have expected cell frequencies of at least five. The first and the last intervals have expected cell frequencies of 1.5 each. After adjusting the number of intervals, the degrees of freedom for the chi-square statistic is ____. 2, 3,5,...