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b. [2 pt] What statistical method can you use to determine which fertilizer is different and why is this method better than conducting many pairwise

b. [2 pt] What statistical method can you use to determine which fertilizer is different and why is this method better than conducting many pairwise tests? (hint: this can be easily done by adding a short statement to the above code)c. [1 pt] After using the method from part b, you find that fertilizer a, labeled fert_a in dataset, has a significantly different crop yield than the other fertilizers. You subset your data into crop_subset to include the variable fert_a which is the amount of fertilizer a (in ounces) and the corresponding crop_yield. You want to test whether there is a linear relationship between these 2 variables. Write the null and alternative hypotheses in the context of this problem.H0:Ha:d. [2 pts] You use R to fit a linear model of your data and assign it to an object called fit and find an intercept of 3.72 ounces and a slope of 1.696. You are given another dataframe, newdata = data.frame(fert_a = 150), to predict the crop yield for a fert_a value of 150 ounces. Calculate this crop yield by hand using your linear model and write the line of code that outputs the prediction interval.Prediction:Code:

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Question 5 [7 pts total] You are working in the Department of Agriculture and are given data about the effects of 3 different types of fertilizer on crop yield. You are tasked with analyzing whether any of the fertilizers produce a significantly better average crop yield. The dataset is called crop_data and the variables of Interest are crop yield (number of crops/ounce of fertilizer) and fertilizer (type of fertilizer). a. [2 pts] Write the 2 lines of code that give you the output below, assigning your first line of code to an object called crop. What can you conclude about the differences between the fertilizers and their crop yielda based on these results? ## # A tibble: 2 x 6 ## term df sumsq meansq statistic p. value # # ## 1 fertilizer 2 6.07 3.03 7. 86 0.000701 ## 2 Residuals 93 35.9 0.386 NA NA Code: Conclusions

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