A balanced one-way classification experiment was run to compare the effects of v

Question

A balanced one-way classification experiment was run to compare the effects of various nitrogen sources on sugar beet yields(Y, measured in kg/hectacre). The six nitrogen sources used in the experiment were as follows: 1. Control, no nitrogen 2. ORGANIC nitrogen: CO(NH_2)_2 3. INORGANIC nitrogen, AMMONIUM based: (NH_4)_2SO_4 4. INORGANIC nitrogen, AMMONIUM based: NH_4NO_3 5. INORGANIC nitrogen, NITRATE based: Ca(NO_3)_2 6. INORGANIC nitrogen, NITRATE based: NaNO_3 The data from the experiment is included in R code at the bottom. Treat these as independent samples from the six experimental conditions. This is reasonable because the nitrogen sources were randomly applied to the different plots of land used and also because there were no systematic sources of variation among the plots (at least the experimenter did not think so). There were 10 plots used per nitrogen source. The experimenter's overall goal is to learn about the population mean sugar beet yields across the six experimental conditions (i.e. are they different, how do they compare?, etc.). Prepare a thorough analysis of the data using ANOVA and appropriate follow-up inference procedures (if needed?). Of course, an important part of any statistical analysis is understanding what statistical assumptions are needed and whether they are reasonably satisfied. The experimenter has also asked you to advise him on what experimental conditions) would maximize the population mean yield. What would you him? Defend your recommendations with solid statistical evidence. Nsource1

Explanation / Answer

Solution

Back-up Theory

Suppose we have data of a 1-way classification ANOVA, with r (6) rows (6 Nitrogen sources) and n (10) observations (plots) per cell.

Let xij represent the jth observation in the ith row, k = 1,2,…,n; i = 1,2,……,r

Then the ANOVA model is: xij = µ + i + ij, where µ = common effect, i = effect of ith row, and ij is the error component which is assumed to be Normally Distributed with mean 0 and variance 2.

Now, to work out the solution,

Terminology:

Row total = xi.= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G2/N, where N = total number of observations = r x n =

Total Sum of Squares: SST = (sum over i,j of xij2) – C

Row Sum of Squares: SSR = {(sum over i of xi.2)/(n)} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Error: DF for Total – DF for Row;

Rows: (r - 1);

Fobs: MSSR/MSSE;

Fcrit: upper % point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for Row and n2 is the DF for Error

Significance: Fobs is significant if Fobs > Fcrit

Computations

Summary of Exel Computations are given below:

Grand Tptal G

66163

Correction Factor C

7E+07

SST

2E+06

Raw SS (Row)

7E+08

SSR

738692

SSE

868306

ANOVA TABLE

0.05

Source

DF

SS

MS

F

Fcrit

p-value

Row

5

738692

1E+05

9.1879

2.386

2.3E-06

Error

54

868306

16080

Total

59

2E+06

27237

Since F > Fcrit (5%), the null hypothesis is rejected at 5% level of significance,

Further, p-value being very low, the null hypothesis is rejected even at a very low level of significance.

Conclusion

There is sufficient evidence to suggest that nitrogen sources have definite impact on sugar beet yield.

Further, since the nitrogen source effects are significant, it makes sense to grade the effects.

Comparing means, Nitrogen Source 2 is recommended.

DONE

Grand Tptal G

66163

Correction Factor C

7E+07

SST

2E+06

Raw SS (Row)

7E+08

SSR

738692

SSE

868306

ANOVA TABLE

0.05

Source

DF

SS

MS

F

Fcrit

p-value

Row

5

738692

1E+05

9.1879

2.386

2.3E-06

Error

54

868306

16080

Total

59

2E+06

27237

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.