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

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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 (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 condition(s) would maximize the population mean yield. What would you tell him? Defend your recommendations with solid statistical evidence. Nsource1 leftarrow 0(814.8, 813.2, 974.9, 862.0, 750.8, 769.0, 1026.0, 849.4, 946.3, 997.9) Nsource2 leftarrow 0(1235.3, 1185.9, 1117.0, 1171.8, 1284.7, 1211.5, 1288.9, 1001.4, 1428.4, 1373.6) Nsource3 leftarrow 0(1157.5, 1236.1, 1074.3, 1171.5, 1031.3, 1015.9, 950.1, 1108.5, 1275.8, 999.4) Nsource4 leftarrow 0(955.0, 1039.4, 1318.6, 926.9, 1230.1, 835.3, 1013.8, 1128.3, 1023.7, 1353.5) Nsource5 leftarrow 0(1070.0, 1153.1, 1, 940.1, 998.5, 1264.3, 1351.1, 1117.5, 1389.3, 1037.1, 1047.3) Nsource4 leftarrow 0(1077.2, 1137.7, 1187.4, 1335.8, 1262.6, 1126.7, 1081.6, 1134.6, 1272.0, 1231.3) yeild leftarrow c(Nsource1, Nsource2, Nsource3, Nsource4, Nsource5, Nsource6) N.type leftarrow c(rep("Nsource1", 10), rep("Nsource2", 10), rep("Nsource3", 10), rep("Nsource4", 10), rep("Nsource5", 10), rep("Nsource6", 10), } N type leftarrow factor(N type)

Explanation / Answer

Null Hypothesis H0: The mean sugar beet yields across six experimental conditions are same.

Alternative Hypothesis H1: Atleast one of the mean sugar beet yields across six experimental conditions are different.

Assumptions of ANOVA:
(i) All populations have the same variance (or standard deviation). Given in the problem statement that there is no source of variation among plots.
(ii) The samples are randomly selected and independent of one another. As per the problem statement, the samples are randomly selected and independent.

So, assumptions of ANOVA are satisfied.

I have ran the given R code in the given document to load the data.

All R code is given in bold.

Run the below code to get the mean for all sources (Nsource).

tapply(yield,N.type,mean)
Nsource1 Nsource2 Nsource3 Nsource4 Nsource5 Nsource6
880.43 1229.85 1102.04 1082.46 1136.83 1184.69

Fit a linear model between yield and N.type

model <- lm(yield~N.type)

Now, run the anova on the linear model.

anova(model)

Analysis of Variance Table

Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
N.type 5 738684 147737 9.1894 2.262e-06 ***
Residuals 54 868151 16077
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The ouput shows the p-value as 2.262e-06 which is less than 0.05, so we reject the null hypothesis and conclude that mean sugar beet yields across six experimental conditions are different.

Now we conduct post hoc test (Tukey's test) to check which of the experimental conditions are different from others.

Run anova test again with aov() function.

a1 <- aov(model)

Run Tukey's HSD test to get the difference of yields for all combinations of experimental conditions.

TukeyHSD(a1)

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = model)

$N.type
diff lwr upr p adj
Nsource2-Nsource1 349.42 181.88841 516.95159 0.0000014
Nsource3-Nsource1 221.61 54.07841 389.14159 0.0033938
Nsource4-Nsource1 202.03 34.49841 369.56159 0.0095907
Nsource5-Nsource1 256.40 88.86841 423.93159 0.0004669
Nsource6-Nsource1 304.26 136.72841 471.79159 0.0000248
Nsource3-Nsource2 -127.81 -295.34159 39.72159 0.2309852
Nsource4-Nsource2 -147.39 -314.92159 20.14159 0.1150259
Nsource5-Nsource2 -93.02 -260.55159 74.51159 0.5761005
Nsource6-Nsource2 -45.16 -212.69159 122.37159 0.9669432
Nsource4-Nsource3 -19.58 -187.11159 147.95159 0.9993155
Nsource5-Nsource3 34.79 -132.74159 202.32159 0.9895906
Nsource6-Nsource3 82.65 -84.88159 250.18159 0.6920554
Nsource5-Nsource4 54.37 -113.16159 221.90159 0.9287675
Nsource6-Nsource4 102.23 -65.30159 269.76159 0.4725848
Nsource6-Nsource5 47.86 -119.67159 215.39159 0.9577246

So, the difference between Nsource1 and Nsource2 (Nsource2-Nsource1) is maximum, so Nsource2 experimental condition (Organic Nitrogen) would maximize the mean yield. Also, Nsource1 experimental condition (no Nitrogen) would minimize the mean yield.

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