For a gooness of fit test, the null and the alternative hypothesis here are given as:
H0: The two variables are independent
Ha: The two variables are not independent.
As we are rejecting the null hypothesis here of a chi square goodness of fit test, therefore the test is significant here and we can conclude here that we have sufficient evidence that the two variables are not independent and associated here.
Therefore B is the correct answer here.
If I reject the null hypothesis of a Chi-Square Goodness-of-Fit Test, which of the following statement...
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 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.
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 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
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?
A chi-square for goodness of fit hypothesis test was conducted using an alpha of .05. The resulting chi-square was as follows: X2(2, n = 30) = 4.20, p > .05. Find the critical value. What decision was made regarding the null hypothesis?
A distribution and the observed frequencies of the values of a variable from a simple random sample of the population are provided below. Use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution Distribution: 0.1875, 0.1875, Observed frequencies: 16, 20, 24, 36 Significance level 0.05 0.3125, 0.3125 Determine the null and alternative hypotheses. Choose the correct answer below. OA. H: The distribution of the variable differs from...
please provide a detailed answer, thank you! You are conducting a Goodness of Fit hypothesis test for the claim that the population data values are not distributed as follows: PA = 0.1; PB = 0.4; Pc = 0.3; pp = 0.2 Give all answers as decimals rounded to 3 places after the decimal point, if necessary. (a) What are the hypotheses for this test? OH:Pa = 0.1, PB = 0.4, pc = 0.3, pp = 0.2 H: all of these...
Which of the statements are correct? A condition for using the chi-square goodness-of-fit test is that all expected counts must be at least two. A condition for using the chi-square goodness-of-fit test is the observations are based on a random sample. The chi-square goodness-of-fit test uses n − 5 degrees of freedom. A) I only B) II only C) I and II only D) II and III only E) I, II, and III
29. manufacturing company uses a chi-square goodness of fit test to determine if a population of all lengths of A fastener -inch bolts it manufactures is distributed according to a normal distribution. If we reasonable to assume that the population distribution is at least eject the null hy approximately normally distributed. O True O False 30. A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants...