Question

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3. A botanist was testing fertilizers on 149 string bean plants. The researcher used random assignment to create three groups, two with fertilizers (Grow Fast and Super Plant) and one control group (None). After 60 days, the height of each plant was measured. a. What is the response and what is the factor? How many levels? b. Create a box plot with groups of the heights of the plants by fertilizer type. Paste the graph here and observe if there is a difference in heights due to fertilizer. c. The botanist then wants to determine if there is a difference in mean heights due to type of fertilizer by conducting a one factor ANOVA test at the 5% significance level. State the hypotheses in words and parameters. Response data are in one course for all factor levels d. Run the appropriate one factor ANOVA test. Use the command ANOVA>ONEWAY. Make sure you choose the Comparisons option and select the Tukey test. Paste the results here. Besponse: Height Bacter Fertiser One-Way Analysis of Variance: Comparisons trorrate for comparison Comparison procedures assuming equal variances sher e. State a detailed conclusion using the both ANOVA results and the Tukey Test results.

Answer 1

Ans 1 ) From the given output

Response variable is height

Factors is fertilizer

and levels are 3

Ans 2 ) the boxplots of the fertilizer with three levels shows that the three factors are normally distributed which fullfill the assumption of applying one way ANOVA

Ans 3 ) the null and alternatuve hypothesis in word is

Ho: there is no significant difference between population mean of each level

Ha: there is a significant difference at least between two population mean levels

IN parameter form

Ho: u1=u2=u3

Ha: at least two mean are different

the p-value of F stat is less than 0.05 so we reject HO and conclude that t least two mean are different .

For part 4 and 5 output not given