MAT 150 Graded Exam Questions and Answers 2023/2024 - Straighterline

MAT 150 Graded Exam Questions and Answers 2023/2024 - Straighterline. Given the following ANOVA table for three treatments each with six observations: What is the critical value of F at the 5% level of significance? Select one: a. 3.29 b. 3.68 c. 3.59 d. 3.20 Feedback There are three treatments and a total of 18 observations, so k = 3 and n = 18. The treatment degrees of freedom are k - 1 = 3 - 1 = 2, and the error degrees of freedom are n - k = 18 - 3 = 15. Therefore, the F-distribution would have 2 degrees of freedom in the numerator and 15 degrees of freedom in the denominator. Using the "Critical values of the F distribution at a 5 percent level of significance" table, the F critical value is 3.68. MC Qu. 38 Given the following ANOVA table for three tr... AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 12-05 Organize data into appropriate ANOVA tables for analysis. Topic: The ANOVA Test The correct answer is: 3.68 Question 2 Correct 4.60 points out of 4.60 Flag question Question text Three different fertilizers were applied to a field of celery. In computing F, how many degrees of freedom are there in the numerator? Select one: a. 0 b. 1 c. 2 d. 3 Feedback The degrees of freedom in the numerator or the treatment degrees of freedom are k - 1 = 3 - 1 = 2. MC Qu. 18 Three different fertilizers were applied to ... AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 12-05 Organize data into appropriate ANOVA tables for analysis. Topic: The ANOVA Test The correct answer is: 2 Question 3 Correct 4.60 points out of 4.60 Flag question Question text A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B, and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level? Select one: a. Mean hourly wages of unskilled employees of all areas are equal. b. Mean hourly wages in at least two metropolitan areas are different. c. More degrees of freedom are needed. d. None of these is correct. Feedback The critical value is an F-statistic with k - 1 = 3 - 1 = 2 degrees of freedom in the numerator and n - k = 28 - 3 = 25 degrees in the denominator. Using the 0.05 significance level, the critical value is 3.39. The null hypothesis is rejected because the computed value of 4.91 exceeds the critical value of 3.39. In this question, we reject the null hypothesis and conclude that at least one pair of treatment means differs. MC Qu. 25 A preliminary study of hourly wages paid to ... AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 12-05 Organize data into appropriate ANOVA tables for analysis. Topic: The ANOVA Test The correct answer is: Mean hourly wages in at least two metropolitan areas are different. Question 4 Correct 4.60 points out of 4.60 Flag question Question text A manufacturer of automobile transmissions uses two different processes. Management ordered a study of the production costs to see if there is a difference between the two processes. A summary of the findings is shown next. What is the critical value of F at the 1% level of significance? Select one: a. 9.46 b. 8.29 c. 8.18 d. 4.61 Feedback There are two treatments and a total of 20 observations. The F distribution has k - 1 = 2 - 1 = 1 degrees of freedom in the numerator and n - k = 20 - 2 = 18 degrees of freedom in the denominator. From "Critical values of the F distribution at a 1 percent level of significance" table, the critical value is 8.29. MC Qu. 33 A manufacturer of automobile transmissions u... AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 12-05 Organize data into appropriate ANOVA tables for analysis. Topic: The ANOVA Test The correct answer is: 8.29 Question 5 Correct 4.60 points out of 4.60 Flag question Question text The annual dividend rates for a random sample of 16 companies in three different industries, utilities, banking, and insurance were recorded. The ANOVA comparing the mean annual dividend rate among three industries rejected the null hypothesis that the dividend rates were equal. The Mean Square Error (MSE) was 3.36. The following table summarized the results: Based on the comparison between the mean annual dividend rate for companies in utilities and banking, the 95% confidence interval shows an interval of 1.28 to 6.28 for the difference. This result indicates that _____________________. Select one: a. There is no significant difference between the two rates b. The interval contains a difference of 5.00 c. The annual dividend rate in the utilities industry is significantly less than the annual dividend rate in the banking industry