F) XXXXXXXXXXgroupAnalysis of VarianceDf Sum Sq Mean Sq F value XXXXXXXXXX257Pr (>F) XXXXXXXXXX ***fertilizerUse the computed values on the side to answer the questions.Residuals414272.7a....

F) 2 0.3282 0.00722 group Analysis of Variance Df Sum Sq Mean Sq F value 2558 1279.0 2 57 Pr (>F) 17.6 0.00000111 *** fertilizer Use the computed values on the side to answer the questions. Residuals 4142 72.7 a. Based on the problem, what is the factor (with levels) and the response variable? b. What is the appropriate test to use? Explain your answer. c. What is the set of hypotheses to answer the objectives of the problem? Signif. codes: 0 ***** 0.001 *** 0.01 0.05 '.' 0.1 ' ' 1 mean sd data:n commercial nanofertilizer 46.50 8.382312 43.05 9.098727 20 20 20 none 31.25 8.058177 d. Is there a difference in the mean chlorophyll content of the coffee plants? Provide the appropriate p-value, decision and conclusion. Kruskal-Wallis Rank Sum Test chlorophyll Kruskal-Wallis chi-squared = 17.871, df = 3, p-value = 0.009184 data: e. Based on the results of the analysis, what will you recommend to CoffeeNatin? "/>
Extracted text: THE COFFEE BEAN PROBLEM commercial Shapiro-Wilk normality data: Shapiro-Wilk normality data: data: nanofertilizer none Problem: The role of chlorophyll in plants is vital. Chlorophyll is responsible for converting carbon dioxide and water, using sunlight, into oxygen and glucose which the plants consume as food. CoffeeNatin is experimenting on a new cheaper fertilizer formulation to increase the quality of their coffee beans. They want to increase the chlorophyll content in the leaves which directly affects the production of glucose in plants. In one experiment, CoffeeNatin wants to compare the new nanofertilizer they created with a commercial fertilizer. They used 60 homogenous coffee plants where 20 of them received the nanofertilizer, 20 received the commercial fertilizer and 20 did not receive fertilizer. After a week, the chlorophyll content of the leaves was obtained. Shapiro-Wilk normality test test test w = 0.94662, p-value - 0.3187 w = 0.94005 W = 0.92052 p-value = 0.1014 p-value = 0.2403 Levene's Test for Homogeneity of Variance (center = "mean") Df F value Pr (>F) 2 0.3282 0.00722 group Analysis of Variance Df Sum Sq Mean Sq F value 2558 1279.0 2 57 Pr (>F) 17.6 0.00000111 *** fertilizer Use the computed values on the side to answer the questions. Residuals 4142 72.7 a. Based on the problem, what is the factor (with levels) and the response variable? b. What is the appropriate test to use? Explain your answer. c. What is the set of hypotheses to answer the objectives of the problem? Signif. codes: 0 ***** 0.001 *** 0.01 0.05 '.' 0.1 ' ' 1 mean sd data:n commercial nanofertilizer 46.50 8.382312 43.05 9.098727 20 20 20 none 31.25 8.058177 d. Is there a difference in the mean chlorophyll content of the coffee plants? Provide the appropriate p-value, decision and conclusion. Kruskal-Wallis Rank Sum Test chlorophyll Kruskal-Wallis chi-squared = 17.871, df = 3, p-value = 0.009184 data: e. Based on the results of the analysis, what will you recommend to CoffeeNatin?