Yield of potato resulting from use of different fertilizer (A, B, C, D, E) from CRDexperiment with 5 (r) replications and 5 (t) treatments.Yield of potato, kg/plotTreatmentIIIVVT118191715T21018191810T314161415T414151315T518171018Required:a. Fill the ANOVA Tableb. Make a decisionnO 0o n 00

Answered: Image /qna-images/answer/30498586-937d-4f08-a439-af8ad698119a.jpg

Answered: 4. Yield of potato resulting from use…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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9. Two independent samples with X, = 5.1, s, = 1.2,n, 20 and x, =4.6, s, =1.5, n, 20. %3D %3D %3D Test if populations 4, > H, at a = 0.01 level of significance.
Solve the ANOVA table Levels of A = 4Levels of B = 4Sample size per condition (n) = 7 Source SS Df MS F Between Columns A 2.99 Between Rows B 6.00 Interaction AB Within Cells 75.00 Total 210.00
The following data were obtained from an independent-measures study comparing three treatment conditions. Treatment 1 2 3 n=6 n=6 n=6 M=1 M=2 M=6 SS=60 SS=65 SS=40 N= 18 G= 54 Sum of squares X^2= 411 Conduct an ANOVA with to determine whether there are any significant differences among the three treatment means. (alpha=.05)
Consider Ho: µ1 = µ2= H3= H4 and sample size are ni= n2= 5, na = 6, n4= 8. a. Complete the ANOVA table. Critical Value / Tabular Value Mean Sum of Source of Variation Sum of Squares df F- value Decision Squares Between - column Within - column 64 240 Total Solutions: b. Using 5% level of significance, would you reject Ho?
7
Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of sweet corn (in tons/acre). Method A 6.51 7.02 6.81 7.27 6.73 6.11 6.17 5.88 6.69 7.12 5.74 6.90 Method B 7.32 7.01 6.66 6.85 5.78 6.48 5.95 6.31 6.50 5.93 6.68 Use a 5% level of significance to test the claim that there is no difference between the yield distributions. (a) What is the level of significance? (b) Compute the sample test statistic. (Round your answer to two decimal places.) (c) Find the P-value of the sample test statistic. (Round your answer to four decimal places.)
8-
A process engineer is trying to improve the life of a cutting tool. He has run a 2³ experiment using cutting speed (A), metal hardness (B), and cutting angle (C) as the factors. The data from two replicates are shown in Table 13E.2. (1) Calculate the main effects of A, B, C, and all interaction effects. (2) Do any of the three factors affect tool life? Make an ANOVA table and draw conclusion from the F-values with α = 0.05. (3) What combination of factor levels produces the longest tool life? ■ TABLE 13E.2 Data for the Experiment in Exercise 13.2. Replicate Run I II (1) 221 311 a 325 435 b 354 348 ab 552 472 с 440 453 སྣ་ ནྟི 406 377 bc 605 500 abc 392 419
no handwriting please.A manufacturer of a samsung phone charges claims that only 4% of his production isdefective. In a random sample of 120 items, the defect rate is 4.99%. At the 1% level ofsignificance, test the manager's claim.i. State the null and alternative hypotheses for the test. ii. Calculate the value of the test statistic for this test. iii. Determine the critical region(s) for this test.
Three different treatments and four different methods are tested for their ability to control acne. What is the value of the LSD (0.05) for the comparison among method means? Treatment 1 Treatment 2 Treatment 3 SUMMARY Treatment 1 Count Sum Average Variance Treatment 2 Count Sum Average Variance Treatment 3 Count Sum Average Variance 3 Anova: Two-Factor With Replication Total Count Sum Average Variance Method 1 5 6 7 ANOVA Source of Variation Sample 6 2 Method 1 2 11 5.5 0.5 2 13 6.5 0.5 2 5 2.5 0.5 6 29 4.833 3.767 Method 2 6 SS 36,75 7 8 7 3 4 Method 2 2 13 6.5 0.5 2 15 7.5 0.5 2 7 3.5 0.5 6 35 5.833 3.767 df 2 Method 3 7 8 9 8 4 6 Method 3 2 15 7.5 0.5 2 17 8.5 0.5 2 10 5 2 6 42 7 3.2 MS 18.4 Method 4 8 9 6 5 6 7 Method 4 2 17 8.5 0.5 2 11 5.5 0.5 2 13 6.5 0.5 6 41 6.833 2.167 F Total 8 56 7 1.714 8 56 7 1.714285714 8 35 4.375 3.125 P-value 2:370761-05 F crit 3.89
13)
2. Consider Ho: H1 H2 H3 = Hμ44= μs a. Complete the ANOVA table. Source of Variation Between - column Within-column Total Solutions: Sum of Squares 118 140 df b. Using 1% level of significance, would you reject Ho? Mean Sum of Squares 15.56 F-value Critical Value / Tabular Value Decision

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