In: Statistics and Probability
Four different types of insecticides are used on strawberry plants. The number of strawberries on each randomly selected plant is given below. Find the test statistic F to test the hypothesis that the type of insecticide makes no difference in the mean number of strawberries per plant. Use α = 0.01.
Insecticide 1 | Insectiside 2 | Insecticide 3 | Insecticide 4 |
6 | 5 | 6 | 3 |
5 | 8 | 3 | 5 |
6 | 5 | 4 | 3 |
7 | 5 | 2 | 4 |
7 | 5 | 3 | 4 |
6 | 6 | 3 | 5 |
STEP 1: Hypothesis: Ho:______________________________________ H1:______________________________________
STEP 2: Restate the level of significance: _______________________
STEP 4: p-value __________________________ from appropriate test on calculator.
STEP 5: Conclusion.
Here we will perform the one way anova, we use the R software for performing anova which is free software. The R-codes for preforming the analysis is given below along with output
>data=read.csv(file.choose())
>attach(data)
>owanova<-aov(strawaberries~as.factor(insecticide), qr=TRUE)
One-Way ANOVA | |||||
---|---|---|---|---|---|
df | Sum of Squares | Mean Squares | F | Pr(>F) | |
insecticide | 3 | 29.66666667 | 9.88888889 | 8.35680751 | 0.00084575 |
Residuals | 20 | 23.66666667 | 1.18333333 | NA | NA |
STEP 1: Hypothesis: Ho: The type of insecticide makes no difference in the mean number of strawberries per plant.
H1: The type of insecticide makes difference in the mean number of strawberries per plant.
STEP 2: Restate the level of significance: 0.05, i.e., 5%
STEP 4: p-value 0.0008457 from appropriate test on calculator.
STEP 5: Conclusion: here the p-value is less than 5% level of significance. Therefore we may reject the null hypothesis at 5% level of significance. And the conclude that, the type of insecticide may be makes difference in the mean number of strawberries per plant.