1. Problem and Data Set

The researcher is conducting an experiment to understand the joint effects of temperature and humidity on the growth of a specific plant species. The temperature has two levels: Low (20°C) and High (30°C), while humidity also has two levels: Low (40%) and High (80%). Each combination of temperature and humidity will be tested on 16 plant samples. For each combination of temperature and humidity, the researcher measures the height of 12 randomly selected plants after two weeks from the 3 different lots. The growth in centimeters were recorded as follows.

Temperature Humidity Lot Plant Growth
Low Low 1 12.5
Low High 1 15.2
High Low 1 14.8
High High 1 18.3
Low Low 2 12.8
Low High 2 16.3
High Low 2 13.4
High High 2 17.9
Low Low 3 13.0
Low High 3 14.5
High Low 3 14.0
High High 3 16.5

2. Write your experimental question.

Is there significant effect of the temperature, humidity and joint effects of temperature and humidity on the growth of a specific plant species at \(α=0.5\)?

3. Construct your null and alternative hypotheses.

Temperature:

Ha: There is a main effect of temperature on the growth of a specific plant species.
Ho: There is no main effect of temperature on the growth of a specific plant species.

Humidity

Ha: There is a main effect of humidity on the growth of a specific plant species.
H0: There is no main effect of humidity on the growth of a specific plant species.

Interaction:

Ha: There is a significant interaction effect between temperature and humidity.
H0: There is no significant interaction effect between temperature and humidity.

4. Fit a full factorial model with interaction to the data.

   Temperature Humidity Lot Plantgrowth
1          Low      Low   1        12.5
2          Low      Low   2        12.8
3          Low      Low   3        13.0
4          Low     High   1        15.2
5          Low     High   2        16.3
6          Low     High   3        14.5
7         High      Low   1        14.8
8         High      Low   2        13.4
9         High      Low   3        14.0
10        High     High   1        18.3
11        High     High   2        17.9
12        High     High   3        16.5

Call:
lm(formula = Plantgrowth ~ Temperature + Humidity + Temperature * 
    Humidity, data = lab5)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.06667 -0.36667 -0.01667  0.43333  0.96667 

Coefficients:
                           Estimate Std. Error t value Pr(>|t|)    
(Intercept)                 17.5667     0.4353  40.360 1.56e-10 ***
TemperatureLow              -2.2333     0.6155  -3.628 0.006702 ** 
HumidityLow                 -3.5000     0.6155  -5.686 0.000462 ***
TemperatureLow:HumidityLow   0.9333     0.8705   1.072 0.314917    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.7539 on 8 degrees of freedom
Multiple R-squared:  0.8922,    Adjusted R-squared:  0.8517 
F-statistic: 22.06 on 3 and 8 DF,  p-value: 0.000318

5. Is there a significant interaction effect between temperature and humidity?

                     Df Sum Sq Mean Sq F value   Pr(>F)    
Temperature           1  9.363   9.363   16.48 0.003639 ** 
Humidity              1 27.603  27.603   48.57 0.000116 ***
Temperature:Humidity  1  0.653   0.653    1.15 0.314917    
Residuals             8  4.547   0.568                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Base on the data above, the interactive between temperature and humidity has p-value = 0.314917 which is clearly greater than 0.05. Hence, there is no significant interaction effect between temperature and humidity

6. Is there evidence of confounding between main effects and lot effects?


Call:
lm(formula = Plantgrowth ~ Temperature + Humidity + Lot, data = lab5)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.9000 -0.5417  0.1000  0.5792  0.8167 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)     18.0333     0.6290  28.670 2.37e-09 ***
TemperatureLow  -1.7667     0.4193  -4.213  0.00294 ** 
HumidityLow     -3.0333     0.4193  -7.234 8.94e-05 ***
Lot             -0.3500     0.2568  -1.363  0.21000    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.7263 on 8 degrees of freedom
Multiple R-squared:  0.8999,    Adjusted R-squared:  0.8624 
F-statistic: 23.98 on 3 and 8 DF,  p-value: 0.0002368

Base on the data above, the p-value of lot is 0.21 which is clearly greater than 0.05. Thus, there is no significant evidence of confounding between lot effects and the main effects.

7. How would you modify the design to remove confounding? To minimize the impact of confounding in a study’s design, various strategies can be utilized to strengthen the accuracy of causal conclusions. One of these are organizing experimental units based on shared characteristics into groups, and random assignment of treatments within each specific group or block. But in this case, there is no sufficient evidence of confounding between main effects and lot effects thus, there is no need to modify the design.

8. Compare the precision of estimating main effects under the current vs modified design. Since we’ve mentioned that there’s no need to modify the design, we will just discuss the precision of estimating main effects under the current. Note that the main effects exhibit a small standard error, signifying increased accuracy in estimating these effects. Furthermore, it is evident that the model displays a diminutive residual standard error. A decreased residual standard error implies that the model’s predictions closely align with the observed values, indicating an enhanced level of fit.

9. Provide interpretations of your findings. Based on the ANOVA table, the p-value of temperature is equal to 0.003639 less than 0.05 and F-value of 16.4751 and 48.5689 which is greater than the tabulated F=5.32. Thus, we reject the null hypothesis. This means that the main effects of Temperature and Humidity is statistically significant. Hence, we fail to reject the null hypothesis and this suggests that the interaction effect between temperature and humidity is not statistically significant.

10. Give details on the syntax used to produce your answer.

This code is used to insert the picture of the given data

knitr::include_graphics("EX1.png", error = FALSE)

The corresponding data converted to excel

library(readxl)
LAB5<-read_xlsx("D:/stat//lab5.xlsx")
library(readxl)
library(kableExtra)
A<-read_xlsx("D:/stat//LAB5.xlsx")
kable(A, format = "html") %>%
  kable_styling(full_width = FALSE) %>%
  row_spec(0, bold = TRUE, color = "black", background = "lightgray") %>%
  row_spec(1:4, background = "white")
Temperature Humidity Lot Plant Growth
Low Low 1 12.5
Low High 1 15.2
High Low 1 14.8
High High 1 18.3
Low Low 2 12.8
Low High 2 16.3
High Low 2 13.4
High High 2 17.9
Low Low 3 13.0
Low High 3 14.5
High Low 3 14.0
High High 3 16.5

For convenience, we made this data.frame.

lab5 <- data.frame(
  Temperature = rep(c("Low", "Low", "High", "High"), each = 3),  
  Humidity = rep(c("Low", "High","Low", "High"), each = 3),   
  Lot = rep(1:3, times = 4),            
  Plantgrowth = c(12.5,12.8,13.0,15.2,16.3,14.5,14.8,13.4,14.0,18.3,17.9,16.5)  
)
lab5
   Temperature Humidity Lot Plantgrowth
1          Low      Low   1        12.5
2          Low      Low   2        12.8
3          Low      Low   3        13.0
4          Low     High   1        15.2
5          Low     High   2        16.3
6          Low     High   3        14.5
7         High      Low   1        14.8
8         High      Low   2        13.4
9         High      Low   3        14.0
10        High     High   1        18.3
11        High     High   2        17.9
12        High     High   3        16.5

We use lm to fit the model.

C<-lm(Plantgrowth ~ Temperature + Humidity + Temperature*Humidity, lab5)
summary(C)

Call:
lm(formula = Plantgrowth ~ Temperature + Humidity + Temperature * 
    Humidity, data = lab5)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.06667 -0.36667 -0.01667  0.43333  0.96667 

Coefficients:
                           Estimate Std. Error t value Pr(>|t|)    
(Intercept)                 17.5667     0.4353  40.360 1.56e-10 ***
TemperatureLow              -2.2333     0.6155  -3.628 0.006702 ** 
HumidityLow                 -3.5000     0.6155  -5.686 0.000462 ***
TemperatureLow:HumidityLow   0.9333     0.8705   1.072 0.314917    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.7539 on 8 degrees of freedom
Multiple R-squared:  0.8922,    Adjusted R-squared:  0.8517 
F-statistic: 22.06 on 3 and 8 DF,  p-value: 0.000318

We use this code to get the ANOVA table.

anova<-aov(C)
summary(anova)
                     Df Sum Sq Mean Sq F value   Pr(>F)    
Temperature           1  9.363   9.363   16.48 0.003639 ** 
Humidity              1 27.603  27.603   48.57 0.000116 ***
Temperature:Humidity  1  0.653   0.653    1.15 0.314917    
Residuals             8  4.547   0.568                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

We use lm to fit the model.

B <- lm(Plantgrowth ~ Temperature + Humidity + Lot, data = lab5)
summary(B)

Call:
lm(formula = Plantgrowth ~ Temperature + Humidity + Lot, data = lab5)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.9000 -0.5417  0.1000  0.5792  0.8167 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)     18.0333     0.6290  28.670 2.37e-09 ***
TemperatureLow  -1.7667     0.4193  -4.213  0.00294 ** 
HumidityLow     -3.0333     0.4193  -7.234 8.94e-05 ***
Lot             -0.3500     0.2568  -1.363  0.21000    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.7263 on 8 degrees of freedom
Multiple R-squared:  0.8999,    Adjusted R-squared:  0.8624 
F-statistic: 23.98 on 3 and 8 DF,  p-value: 0.0002368