The following data refers to the yield of tomatoes (in kg/plot) for soils with f
ID: 3315556 • Letter: T
Question
The following data refers to the yield of tomatoes (in kg/plot) for soils with four different types of salinity. Salinity is indirectly measured by measuring the electrical conductivity (EC, in units of nmhos/cm) of each plot of land. From the data below, we can see that there were a total of 18 measurements of tomato yield: 6 at EC level of 1.6 nmhos/cm, 4 at EC level of 3.8 nmhos/cm, 4 at EC level of 6.0 nmhos/cm, and 5 at EC level of 10.2 nmhos/cm. Tomato Yield (kg/plot) 59.5, 53.3, 56.8, 63.1, 58.7 55.2, 59.1, 52.8. 54.5 51.7, 48.8, 53.9, 49.0 44.6, 48.5, 41.0, 47.3, 46.1 EC level (nmhos/cm) 3.8 6.0 10.2 0.05) that the population mean tomato yield is the same Test the null hypothesis ( across the different salinity levels(1.6, 3.8, 6.0, and 10.2 nmhos/cm) versus the alternative that the mean tomato yield is different among at least 2 of the salinity levels. a) Hint: Ignore the actual values of salinity and treat the 4 different salinity levels like 4 different groups. You can use the ANOVA test from chapter 9. Equivalently, you can fit a multiple linear regression model in R with 3 predictors (e.g. an indicator variable for EC level= 3.8, an indicator variable for EC level 6.0, and an indicator level for EC level 10.2) and use the F-test for model utility from chapter 11. Note: you do not need to do both the ANOVA test (from Chapter 9) and the F-test for model utility (from Chapter 11); you can use just one of them to answer the question. b) Now, fit a simple linear regression model to the 18 observations where y, the dependent variable, is tomato yield, and x, the independent variable, is the actual EC level value (1.6 nmhos/cm, ). Test the null hypothesis (at -0.05) that the coefficient of EC level (B) is 0 versus the alternative that it is not 0. Interpret your results in the context of the problem. c) Compare your results from parts a) and b). Note that in a) you treated salinity as a categorical variable, and in b) you treated salinity as a continuous variable. Recall in the beginning of the quarter when we discussed type of variables (continuous and categorical/discrete), we noted that a variable could sometimes be considered continuous and sometimes categorical/discrete and the decision to treat the variable as continuous or categorical could affect interpretation of results.Explanation / Answer
Result:
a).
One-way ANOVA: yield versus ec
Method
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor
Levels
Values
ec
4
1.6, 3.8, 6.0, 10.2
Analysis of Variance
Source
DF
Adj SS
Adj MS
F-Value
P-Value
ec
3
456.5
152.168
17.11
0.000
Error
14
124.5
8.893
Total
17
581.0
Model Summary
S
R-sq
R-sq(adj)
R-sq(pred)
2.98207
78.57%
73.98%
65.08%
Means
ec
N
Mean
StDev
95% CI
1.6
5
58.28
3.60
(55.42, 61.14)
3.8
4
55.40
2.66
(52.20, 58.60)
6.0
4
50.85
2.43
(47.65, 54.05)
10.2
5
45.50
2.90
(42.64, 48.36)
Pooled StDev = 2.98207
Calculated F=17.11, P=0.000 which is < 0.05 level.
We conclude that mean tomato yield is different among atleast 2 of salinity levels.
b).
Regression Analysis: yield versus ec
Analysis of Variance
Source
DF
Adj SS
Adj MS
F-Value
P-Value
Regression
1
453.047
453.047
56.65
0.000
ec
1
453.047
453.047
56.65
0.000
Error
16
127.956
7.997
Lack-of-Fit
2
3.458
1.729
0.19
0.825
Pure Error
14
124.498
8.893
Total
17
581.003
Model Summary
S
R-sq
R-sq(adj)
R-sq(pred)
2.82794
77.98%
76.60%
71.48%
Coefficients
Term
Coef
SE Coef
T-Value
P-Value
VIF
Constant
60.67
1.28
47.37
0.000
ec
-1.509
0.200
-7.53
0.000
1.00
Regression Equation
yield
=
60.67 - 1.509 ec
To test the significance of the model, Calculated F=56.65, P=0.000 which is < 0.05 level.
We conclude that the regression model is useful in predicting tomato yield .
c).
Both in part a and part b gave similar result.
Sometimes considering categorical variable loose some information that could affect the interpretation of the results.
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
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