1. Text Exercise 12.2 (a) (b) (c) Construct the completed ANOVA table below. Answer this part by indicating what the f test statistic value is, what the appropriate tabled f value is, what the P-value is, and whether or not multiple comparison is necessary. SOURCEdf SS MSf Treatments 424.76.175 4.914 Error3037.71.2567 Total3462.4 5 f 4,30 = 4.914f 4,30;0.10 = 2.14 P < 0.01 Since H 0 is rejected, multiple comparison is necessary. Homework #18Score____________/ 10Name ______________ Now, check some answers before submitting Homework #18:

2. Text Exercise 12.10 (a) (b) (c) (d) Construct the completed ANOVA table below. Answer this part by summarizing the results (Step 4) of the f test to see if there is sufficient evidence with  = 0.05 of at least one difference in mean hours allocated to accounts receivable. Then, indicate whether or not multiple comparison is necessary. SOURCEdf SS MSfP-value Groups 2 71.51 35.755 4.6710.01 < P < 0.025 Error42321.47 7.654 Total44392.98 Completely Randomized Design The treatments are the different experiences with designing and evaluating audit programs. Since f 2,42 = 4.671 and f 2,42;0.05 = 3.23, we have sufficient evidence to reject H 0, We conclude that there is at least one difference in mean hours allocated to accounts receivable among the three groups (0.01 < P < 0.025). We need multiple comparison,

3. Additional HW Exercise 12.1 (a) (b) The mean breaking strength of rope is being studied for three types named Deluxe, Econ, and Nogood. A 0.05 significance level is chosen for a hypothesis test to see if there is any evidence that mean breaking strength is not the same for the rope types Deluxe, Econ, and Nogood. Pieces of rope are randomly selected from each type, and the breaking strengths in pounds of force are recorded as follows: Deluxe165 162 159 162 Econ156 163 158 Nogood151 154 160 After defining the appropriate dummy variable(s), write a regression model for the prediction of Y = “breaking strength” from “type of rope”. 1 for Deluxe X 1 = 0 otherwise 1 for Econ X 2 = 0 otherwise Y =  0 +  1 X 1 +  2 X 2 +  or E(Y) =  0 +  1 X 1 +  2 X 2 With the model defined in part (a), write the formula for the population mean breaking strength for each type of rope. mean for Deluxe =  0 +  1 mean for Econ =  0 +  2 mean for Nogood = 00

4. (c) (d) Create an SPSS data file named ropedummy containing three variables named X1, X2, and strength. Store the data in this SPSS file by entering each breaking strength together with the corresponding values for each of the dummy variables defined in part (a). Use the Analyze > Regression > Linear options in SPSS to select the variable strength for the Dependent slot, and select the variables X1 and X2 for the Independent(s) section. Title the output to identify the homework exercise (Additional HW Exercise 12.1 - part (c)), your name, today’s date, and the course number (Math 214). Use the File > Print Preview options to see if any editing is needed before printing the output. Attach the printed copy to this assignment before submission. Write the least squares prediction equation for the model in part (a), and use this equation to obtain the sample means for Deluxe, Econ, and Nogood. The least squares prediction equation isstrength = 155 + 7(X 1 ) + 4(X 2 ) With X 1 = 1 and X 2 = 0, we find the mean for Deluxe = 155 + 7(1) + 4(0) =162 With X 1 = 0 and X 2 = 1, we find the mean for Econ = 155 + 7(0) + 4(1) =159 With X 1 = 0 and X 2 = 0, we find the mean for Nogood = 155 + 7(0) + 4(0) =155

5. Title the output to identify the homework exercise (Additional HW Exercise 12.1 - part (e)), your name, today’s date, and the course number (Math 214). Use the File > Print Preview options to see if any editing is needed before printing the output. Attach the printed copy to this assignment before submission. Summarize the results (Step 4) of the f test to see if there is sufficient evidence that mean breaking strength is not the same for the rope types Deluxe, Econ, and Nogood at the 0.05 level. Since f 2,7 = 3.419 and f 2,7;0.05 = 4.74, we do not have sufficient evidence to reject H 0. We conclude that mean breaking strength is the same for the rope types Deluxe, Econ, and Nogood (0.05 < P < 0.10). (f) OR (P = 0.092)

6. Additional HW Exercise 12.1 - continued (g) Based on the results in part (e), decide whether or not a multiple comparison method is necessary. If no, then explain why not; if yes, then use Tukey’s HSD multiple comparison method. Since we have not rejected H 0, we do not need a multiple comparison method to identify significant differences between means.

7. (h) (i) Explain why three contiguous box plots are an appropriate graphical display for the data, and use SPSS to create these. Title the output to identify the homework exercise (Additional HW Exercise 12.1 - part (h)), your name, today’s date, and the course number (Math 214). Attach a printed copy to this assignment before submission. Since the quantitative variable breaking strength is being compared for three rope types, three contiguous box plots is an appropriate graphical display. Summarize the results (Step 4) of Levene’s f test to see if there is sufficient evidence at the 0.05 level that the standard deviation of breaking strength is not the same for the four types of rope. Then, decide whether or not we need to be concerned about whether or not the assumption of equal standard deviations is satisfied. Since we have no evidence of a difference in standard deviation, we believe the assumption of equal standard deviations is satisfied. Since f 2,7 = 0.951 and f 2,7;0.05 = 4.74, we do not have sufficient evidence to reject H 0. We conclude that the standard deviation of breaking strength is not significantly different for the for the rope types Deluxe, Econ, and Nogood (0.10 < P). OR (P = 0.431)

8. Additional HW Exercise 12.2 (a) The mean penetration weight of water is being studied for four brands of paper towels named Tuff, Econ, Cheep, and Super. A 0.05 significance level is chosen for a hypothesis test to see if there is any evidence that mean penetration weight is not the same for the brands Tuff, Econ, Cheep, and Super. Randomly selected observations from each brand are observed, and the penetration weights in grams are recorded as follows: Tuff112 104 110 108 112 114 Econ120 116 122 114 112 118 Cheep106 114 110 108 116 112 Super120 110 112 112 114 116 After defining the appropriate dummy variable(s), write a regression model for the prediction of Y = “penetration weight” from “paper towel brand”. 1 for Tuff X 1 = 0 otherwise 1 for Econ X 2 = 0 otherwise Y =  0 +  1 X 1 +  2 X 2 +  3 X 3 +  or E(Y) =  0 +  1 X 1 +  2 X 2 +  3 X 3 1 for Cheep X 2 = 0 otherwise

9. With the model defined in part (a), write the formula for the population mean penetration weight for each paper towel brand. mean for Tuff =  0 +  1 mean for Econ =  0 +  2 mean for Super = 00 (b) mean for Cheep =  0 +  3

10. Additional HW Exercise 12.3 (a) Consider data for a one-way ANOVA where each of three treatments, labeled A, B, and C, are applied to 4 subjects (that is, the data will consist of 3 random samples each of size 4). Make up a data set for which the SSR will be 0 (zero), but the SSE will not be 0 (zero). (Hint: This means you need a data set for which there is no variation between samples but there is some variation within samples.) A B C 10 12 14 16 (Any data set where the numbers within any sample are not all equal to each other and the sample means are all equal to each other is correct.)