42. Degree of Freedom: 10 , Level of significance 0.025, Chi square table value : 20.4831, Computed test Static value : 16.857
On Chi square distribution , the acceptance region for null hypothesis goes to the Chi square value of 20.4831. Since the computed test Static value of 16.857 falls with the acceptance . Therefore we accept the null hypothesis.
So the answer is FALSE
43. The chi-squared test is always a one-sided test. & it's quite possible to be dealing with the upper or lower tails of the chi-square
So the answer is D
44. By checking the Chi-square table the value for 7 degree of freedom is 2.83311
45. Degree of Freedom: 5 , Level of significance 0.05, Chi square table value : 11.0705, Computed test Static value : ?
On Chi square distribution , the acceptance region for null hypothesis goes to the Chi square value of 11.0705. If the computed test Static value falls with the acceptance, we will accept the null hypothesis otherwise reject the null hypothesis.
So the answer is 17.61 as it does not lies in acceptance region.
a) true b) false 42. For a chi-square distributed random variable with 10 degrees of freedom and a level of sigpificanoe computed value of the test statistics is 16.857. This will lead us to reje...
The Chi-Square Table (Chapter 17) The chi-square table: The degrees of freedom for a given test are listed in the column to the far left; the level of significance is listed in the top row to the right. These are the only two values you need to find the critical values for a chi-square test. Increasing k and a in the chi-square table Record the critical values for a chi-square test, given the following values for k at each level...
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.
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...
What is the critical value for a chi-square test with 28 degrees of freedom at the 5 percent level of significance (3 pts)? If the chi-square test statistic were 41.10, what would you conclude regarding the null hypothesis (4 pts)? What would you conclude if the chi-square value were 48.19
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...
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 number of degrees of freedom in a chi-square goodness of fit test depends upon: (1) the number of classes into which the sample observations are classified; (2) the number of observations in the sample; (3) the number of population parameters estimated from the sample data. a. 1 only b. 2 only c. 1 and 2 only d. 1 and 3 only e. none of the above
The following shows the number of individuals in a random sample of 300 adults who indicated they support the new tax proposal. Political Party Support Democrats 100 Republicans 120 Independents 80 We are interested in determining whether the opinions of the individuals of the three groups are uniformly distributed. If the opinions of the individuals of the three groups are uniformly distributed, the expected frequency for each group is _____ 0.333 0.50 50 None Refer to question 1. The calculated...
Consider a Chi-square random variable with 15 degrees of freedom and 0.1 level of significance. Which of the following test statistic values will result in rejection of the null hypothesis? (1) 21.1. (2) 18.5. (3) 19.8. (4) 23.5. (5) 2.7.
What is the chi-square value for 19 degrees of freedom and area of .025 area in the right tail? Select one: a. 30.144 b. 10.117 c. 32.852 d. 8.907