You are given the following ANOVA table, with n = 55. Compute the F statistic for...
You are given the following ANOVA table, with n = 55. Compute the F statistic for the interaction term.
|
Source
of |
Sum
of |
Degrees of |
Mean |
F |
|
Factor A |
127 |
c-1 = 3 |
||
|
Factor B |
784 |
r-1 = 7 |
||
|
Interaction |
253 |
|||
|
Error |
5761 |
|||
|
Total |
|
12.0476 |
||
|
3.608 |
||
|
1.611 |
||
|
0.048 |
||
|
none of the above |
Homework Answers
![Minitab Untitled - [Session] 日 | 盾范 勺 | t ↓鈍鷓 O@E|||幻园倫0E] I File Edit Data Calc Stat Graph Editor Tools Window Help Assistant 04-11-2018 01:04:44 Welcome to Minitab, press F1 for help. MTB > let k1 = 55-1 MTB > print kil Data Display K1 MTB > let k2 =54 _ (3 + 7 + 21) 54.0000 . .. df (Total) MTB > print k2 Data Display K2 MTB > let k3 = 127 + 784 + 253 + 5761 23.0000 .. . df (Error) MTB > print k3 Data Display K3 6925.00 .. . SS (Total) MTB > Let Mean Square = Sum of Squares/ Degrees of Freedom MTB > let k412.048 250.478 MTB > print k4 Data Display K4 0.0481000 .. . F statistic MTB >](http://img.homeworklib.com/questions/b1f420e0-1508-11ec-85ab-4b2682826aca.png?x-oss-process=image/resize,w_560)
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You are given the
following ANOVA table, with n = 55. Compute the F statistic for...
Exhibit 13-7 The following is part of an ANOVA table, which was the result of three...
Exhibit 13-7 The following is part of an ANOVA table, which was the result of three treatments and a total of 18 observations (6 observations per sample). Source of Variation Sum of Mean F Degrees of Freedom Squares Square Between treatments 64 Within treatments (Error) 96 Total 1) Refer to Exhibit 13-7. The number of degrees of freedom corresponding to between treatments is 2) The number of degrees of freedom corresponding to within treatments is 3) The mean square between...
Problem 4: Complete the ANOVA table based on the following data: Replications Standard Deviation ...
Problem 4: Complete the ANOVA table based on the following data: Replications Standard Deviation Factor A Level 1 Level 2 Mean 15 ANOVA Table Degree Freedom of Sum of square Mean sum of F-valuc Source error square error Treatment Error Total NA NA NA Recall that for single-factor ANOVA, we have the following -SSTreatments+SSp where ss,-Σ Σ(y)-T)-total sum of squares n Σ(vi-r-treatment sum of squares i- SSE _ Σ Σ(w-5)2-error sum of squares Given a dataset like the following: Treatment...
Refer to the following partial one-factor ANOVA results from excel (some information is missing. And you...
Refer to the following partial one-factor ANOVA results from excel (some information is missing. And you need to work it out in the questions below.) Now, the F statistic is equal to: Source of Variation Sum of Squares Degrees of Freedom Mean Square F statistic Between Groups 210.2788 Within Groups 1483 74.15 Total 2113.833 4.79 3.56 1.15 2.84 Referring to the table in question 1. The sum of squares for between groups variation is: 129.99 630.83 1233.4 We cannot tell...
Part of an ANOVA table is shown below. Source of Variation Sum of Squares Degrees of...
Part of an ANOVA table is shown below. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Between Treatments 180 3 Within Treatments (Error) TOTAL 480 18 The mean square due to error (MSE) is a. 60. b. 15. c. 20. d. 18.
Part of an ANOVA table is shown below. Source of Variation Sum of Squares Degrees of...
Part of an ANOVA table is shown below. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Between Treatments 64 8 Within Treatments (Error) 2 Total 100 The number of degrees of freedom corresponding to between-treatments is a. 3. b. 4. c. 2. d. 18.
A factorial experiment involving two levels of factor A and three levels of factor B resulted...
A factorial experiment involving two levels of factor A and three levels of factor B resulted in the following data. Factor B Level 1 Level 2 Level 3 135 90 75 Level 1 165 93 Factor A 135 127 120 Level 2 85 105 136 Test for any significant main effects and any interaction. Use α-.05. Round Sum of Squares, F value, Mean Square to two decimals, if necessary. Source of Variation Factor A Factor B Interaction Error Total Sum...
3. Consider the partially completed two-way ANOVA summary table. Source Sum of Squares Degrees of Freedom...
3. Consider the partially completed two-way ANOVA summary table. Source Sum of Squares Degrees of Freedom Mean Sum of Squares Factor B Factor A 600 200 Interaction 144 Error 384 Total 1,288 23 The number of Factor A populations being compared for this ANOVA procedure is _ A) 5 B) 7 C) 4 D) 6
e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest...
e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers. Round all Mean Squares to one decimal places. Round F to two decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total f. At the α-.05 level of significance, test whether the means for the three treatments are equal The p-value is less than.01 What is your conclusion? Select The following data are from a...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three repl...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted the following data: SST 291, SSA 26, SSB 25, SSAB 180. =.05, Show entries to 2 decimals, If necessary Set up the ANOVA table and test for significance using the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value X X Factor A Factor B Interaction 24 Error 35 Total...
Use the following for questions 5,6,7,8, and 9. Part of an ANOVA table is shown below....
Use the following for questions 5,6,7,8, and 9. Part of an ANOVA table is shown below. Source of Sum of Degrees of Mean F p-value Variation Squares Freedom Square Treatments 180 3 Error 16 Total 480 19 The number of treatments (i.e. groups) in the experiment is a.4 b.3 c. 19 d. 16 The mean square between treatments (MSTR) is a. 18.75 b. 60 c. 300 d. 16 The mean square due to error (MSE) is a. 60 b. 16...
Exhibit 13-7 The following is part of an ANOVA table, which was the result of three treatments and a total of 18 observations (6 observations per sample). Source of Variation Sum of Mean F Degrees of Freedom Squares Square Between treatments 64 Within treatments (Error) 96 Total 1) Refer to Exhibit 13-7. The number of degrees of freedom corresponding to between treatments is 2) The number of degrees of freedom corresponding to within treatments is 3) The mean square between...
Problem 4: Complete the ANOVA table based on the following data: Replications Standard Deviation Factor A Level 1 Level 2 Mean 15 ANOVA Table Degree Freedom of Sum of square Mean sum of F-valuc Source error square error Treatment Error Total NA NA NA Recall that for single-factor ANOVA, we have the following -SSTreatments+SSp where ss,-Σ Σ(y)-T)-total sum of squares n Σ(vi-r-treatment sum of squares i- SSE _ Σ Σ(w-5)2-error sum of squares Given a dataset like the following: Treatment...
A factorial experiment involving two levels of factor A and three levels of factor B resulted in the following data. Factor B Level 1 Level 2 Level 3 135 90 75 Level 1 165 93 Factor A 135 127 120 Level 2 85 105 136 Test for any significant main effects and any interaction. Use α-.05. Round Sum of Squares, F value, Mean Square to two decimals, if necessary. Source of Variation Factor A Factor B Interaction Error Total Sum...
3. Consider the partially completed two-way ANOVA summary table. Source Sum of Squares Degrees of Freedom Mean Sum of Squares Factor B Factor A 600 200 Interaction 144 Error 384 Total 1,288 23 The number of Factor A populations being compared for this ANOVA procedure is _ A) 5 B) 7 C) 4 D) 6
e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers. Round all Mean Squares to one decimal places. Round F to two decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total f. At the α-.05 level of significance, test whether the means for the three treatments are equal The p-value is less than.01 What is your conclusion? Select The following data are from a...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted the following data: SST 291, SSA 26, SSB 25, SSAB 180. =.05, Show entries to 2 decimals, If necessary Set up the ANOVA table and test for significance using the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value X X Factor A Factor B Interaction 24 Error 35 Total...
Use the following for questions 5,6,7,8, and 9. Part of an ANOVA table is shown below. Source of Sum of Degrees of Mean F p-value Variation Squares Freedom Square Treatments 180 3 Error 16 Total 480 19 The number of treatments (i.e. groups) in the experiment is a.4 b.3 c. 19 d. 16 The mean square between treatments (MSTR) is a. 18.75 b. 60 c. 300 d. 16 The mean square due to error (MSE) is a. 60 b. 16...
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