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1. The analysis of variance of a completely randomized, two-factor experiment led to the following results....
The following table summarizes a portion of the results for a two-way analysis of variance experiment...
The following table summarizes a portion of the results for a two-way analysis of variance experiment with no interaction. Find the missing values in the ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS" to 4 decimal places and "F" to 3 decimal places.) source of Variation SS df MS F p-value rows 34.63 2 MSB= factor B= 0.042 columns 146.60 3 MSA= Factor A= 0.003 error 18.56 6 MSE= total 199.79 11
A partial result of the computation for a two-factor factorial experiment is given as follows (assume...
A partial result of the computation for a two-factor factorial experiment is given as follows (assume that the number of experimental units in each group is fixed): Source of Variation SS df MS F Factor A 48 4 12 1 Factor B 6 5 Interaction (AB) 24 15 Error 105 - Total 139 - - A) Fill in all the missing results in the ANOVA table. B) At the 0.05 level of significance, is there an interaction effect? (Show your...
The following output summarizes a portion of the results for a two-way analysis of variance experiment...
The following output summarizes a portion of the results for a two-way analysis of variance experiment with no interaction. Factor Aconsists of five different kinds of organic fertilizers, factor B consists of three different kinds of soil acidity levels, and the variable measured is the height (in inches) of a plant at the end of four weeks. a. Find the missing values in the ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS" to 4 decimal...
CH13 Q8 A two-way analysis of variance experiment with interaction was conducted. Factor A had three...
CH13 Q8
A two-way analysis of variance experiment with interaction was conducted. Factor A had three levels (columns), factor B had five levels (rows), and six observations were obtained for each combination. The results include the following sum of squares terms: SST 1548 SSA 1022 SSB 390 SSAB 26 a. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS" to 4 decimal places and "F' to 3 decimal places.) Answer is not complete. ANOVA...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response f...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response for each treatment a. The MINITAB ANOVA printout is shown here. Test for interaction at the a = 0.05 b. level of significance. Analysis of variance for response SS MS df Source 0.11851 0.53777 0.53777 0.55391 5.02708 2.51334 2 1.48678 13.49334 6.74667 2 AB 27.22667 4.53778 6 Error 46.28486 11 Total Does the result warrant...
**** PLEASE ANSWER BY HAND CALCULATIONS***** Question 1 Consider the experimental results for the following randomized...
**** PLEASE ANSWER BY HAND CALCULATIONS*****
Question 1 Consider the experimental results for the following randomized block design. Use these data and hand calculations to complete the analysis of variance table. Show all the calculation steps. Treatments A B C 1 10 8 2 12 6 5 Blocks 3 18 15 14 4 20 18 18 5 8 7 8 Source of SS DF MS Variation Treatments Blocks Error Total 354.93 stLn
Question 1 Consider the experimental results for the...
CH13 Q6 The following output summarizes a portion of the results for a two-way analysis of...
CH13 Q6
The following output summarizes a portion of the results for a two-way analysis of variance experiment with no interaction. Factor A consists of five different kinds of organic fertilizers, factor B consists of three different kinds of soil acidity levels, and the variable measured is the height (in inches) of a plant at the end of four weeks. a. Find the missing values in the ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS"...
3. In a completely randomized design, 5 experimental units were used for each of the four...
3. In a completely randomized design, 5 experimental units were used for each of the four levels of the factor. F Sum of Squares 385.12 Degrees of Freedom Source of Variation Treatment Error Total Mean Square 1563.71 a. Complete the ANOVA table. b. Find the critical value at the 0.05 level of significance from the F table for testing whether the population means for the three levels of the factors are different. c. Use the critical value approach and a...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability of cleaning solutions to remove oil from a piece of cloth. The factors were the concentration of soap (in percent by weight) and the percentage of lauric acid in the solution. The experiment was repeated twice for each combination of factors. The outcome measured was the percentage of oil removed. The results are as follows. 10% acid 25% acid 15% soap 53.1 51.6 56.3...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability of cleaning solutions to remove oil from a piece of cloth. The factors were the concentration of soap (in percent by weight) and the percentage of lauric acid in the solution. The experiment was repeated twice for each combination of factors. The outcome measured was the percentage of oil removed. The results are as follows. 10% acid 25% acid 15% soap 53.1 51.6 56.3...
CH13 Q8
A two-way analysis of variance experiment with interaction was conducted. Factor A had three levels (columns), factor B had five levels (rows), and six observations were obtained for each combination. The results include the following sum of squares terms: SST 1548 SSA 1022 SSB 390 SSAB 26 a. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS" to 4 decimal places and "F' to 3 decimal places.) Answer is not complete. ANOVA...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response for each treatment a. The MINITAB ANOVA printout is shown here. Test for interaction at the a = 0.05 b. level of significance. Analysis of variance for response SS MS df Source 0.11851 0.53777 0.53777 0.55391 5.02708 2.51334 2 1.48678 13.49334 6.74667 2 AB 27.22667 4.53778 6 Error 46.28486 11 Total Does the result warrant...
**** PLEASE ANSWER BY HAND CALCULATIONS*****
Question 1 Consider the experimental results for the following randomized block design. Use these data and hand calculations to complete the analysis of variance table. Show all the calculation steps. Treatments A B C 1 10 8 2 12 6 5 Blocks 3 18 15 14 4 20 18 18 5 8 7 8 Source of SS DF MS Variation Treatments Blocks Error Total 354.93 stLn
Question 1 Consider the experimental results for the...
CH13 Q6
The following output summarizes a portion of the results for a two-way analysis of variance experiment with no interaction. Factor A consists of five different kinds of organic fertilizers, factor B consists of three different kinds of soil acidity levels, and the variable measured is the height (in inches) of a plant at the end of four weeks. a. Find the missing values in the ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "MS"...
3. In a completely randomized design, 5 experimental units were used for each of the four levels of the factor. F Sum of Squares 385.12 Degrees of Freedom Source of Variation Treatment Error Total Mean Square 1563.71 a. Complete the ANOVA table. b. Find the critical value at the 0.05 level of significance from the F table for testing whether the population means for the three levels of the factors are different. c. Use the critical value approach and a...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability of cleaning solutions to remove oil from a piece of cloth. The factors were the concentration of soap (in percent by weight) and the percentage of lauric acid in the solution. The experiment was repeated twice for each combination of factors. The outcome measured was the percentage of oil removed. The results are as follows. 10% acid 25% acid 15% soap 53.1 51.6 56.3...
Cleaning Solution: An experiment was performed to measure the effects of two factors on the ability of cleaning solutions to remove oil from a piece of cloth. The factors were the concentration of soap (in percent by weight) and the percentage of lauric acid in the solution. The experiment was repeated twice for each combination of factors. The outcome measured was the percentage of oil removed. The results are as follows. 10% acid 25% acid 15% soap 53.1 51.6 56.3...
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