In: Statistics and Probability
This dataset contains two variables: fertilizer type (1, 2, or 3) and crop yield per square foot. Using ANOVA, test the null hypothesis that the three fertilizers are equally effective. Hint: Notice that the data are arranged in what's called a "long" format (one long column of data). To you ANOVA, you first need to rearrange the data into three columns.
1. What is the value of the F-statistic? Answer
2. What is the p-value that the three means are the same?
Answer
3. At α=.05, should the null hypothesis be rejected?
AnswerRejectDon’t reject
4. Which fertilizer should you choose based on these results (be careful here!)?
Data Set
| fertilizer | yield | 
| 1 | 177.228692 | 
| 1 | 177.550041 | 
| 1 | 176.408462 | 
| 1 | 177.703625 | 
| 1 | 177.125486 | 
| 1 | 176.778342 | 
| 1 | 176.746302 | 
| 1 | 177.061164 | 
| 1 | 176.274949 | 
| 1 | 177.967203 | 
| 1 | 176.6013 | 
| 1 | 177.030543 | 
| 1 | 177.479507 | 
| 1 | 176.87413 | 
| 1 | 176.114388 | 
| 1 | 176.008395 | 
| 1 | 176.108313 | 
| 1 | 178.357441 | 
| 1 | 177.262445 | 
| 1 | 176.918845 | 
| 1 | 176.239016 | 
| 1 | 176.57307 | 
| 1 | 176.039298 | 
| 1 | 176.817922 | 
| 1 | 176.160587 | 
| 1 | 177.226424 | 
| 1 | 175.938533 | 
| 1 | 177.164937 | 
| 1 | 175.36084 | 
| 1 | 177.276996 | 
| 1 | 175.945444 | 
| 1 | 175.88278 | 
| 2 | 176.479341 | 
| 2 | 176.044342 | 
| 2 | 177.412462 | 
| 2 | 177.360818 | 
| 2 | 177.385499 | 
| 2 | 176.975808 | 
| 2 | 177.379779 | 
| 2 | 177.997995 | 
| 2 | 176.434863 | 
| 2 | 176.933265 | 
| 2 | 175.98348 | 
| 2 | 177.034093 | 
| 2 | 176.436762 | 
| 2 | 176.067745 | 
| 2 | 177.121049 | 
| 2 | 177.197721 | 
| 2 | 176.603724 | 
| 2 | 177.208171 | 
| 2 | 177.148829 | 
| 2 | 176.819077 | 
| 2 | 176.999067 | 
| 2 | 178.134605 | 
| 2 | 176.429156 | 
| 2 | 176.668323 | 
| 3 | 177.104186 | 
| 3 | 178.079635 | 
| 3 | 176.903422 | 
| 3 | 177.540284 | 
| 3 | 177.03271 | 
| 3 | 178.286042 | 
| 3 | 176.40541 | 
| 3 | 176.43083 | 
| 3 | 177.396331 | 
| 3 | 176.925576 | 
| 3 | 177.055046 | 
| 3 | 177.344164 | 
| 3 | 177.128368 | 
| 3 | 177.168302 | 
| 3 | 176.353941 | 
| 3 | 179.060899 | 
| 3 | 176.300517 | 
| 3 | 177.593352 | 
| 3 | 177.115245 | 
| 3 | 177.794457 | 
Solution:
We can use the excel "ANOVA:Single Factor" data analysis tool to find the answer to the given questions. The excel output is given below:
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
| Groups | Count | Sum | Average | Variance | ||
| 1 | 32 | 5656.22542 | 176.7570444 | 0.469119716 | ||
| 2 | 24 | 4246.255974 | 176.9273323 | 0.31409632 | ||
| 3 | 20 | 3545.018717 | 177.2509359 | 0.475272296 | ||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit | 
| Between Groups | 3.008554181 | 2 | 1.50427709 | 3.565667772 | 0.033304083 | 3.122102932 | 
| Within Groups | 30.79710019 | 73 | 0.421878085 | |||
| Total | 33.80565437 | 75 | 
1. What is the value of the F-statistic?
Answer: F = 3.566
2. What is the p-value that the three means are the same?
Answer: p-value = 0.0333
3. At α=.05, should the null hypothesis be rejected?
Answer: Reject
4. Which fertilizer should you choose based on these results (be careful here!)?
Answer: We should choose the fertilizer type 2