In an agricultural experiment, the effects of two fertilizers on the production

Question

In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Sixteen randomly selected plots of land were treated with fertilizer A, and 12 randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results: Directions: For this assignment, complete the following: What is the goal of this study? Define the main variable. What constitutes the population of this study? What is the sample size? Compute the mean, median, and standard deviation of each data set. Draw appropriate graphs for each data set (boxplots are recommended). Include a graph comparing the means and standard deviations of the two data sets. Use EXCEL to generate these graphs. Write a paragraph describing the similarities and differences between the two data sets, using appropriate descriptive statistics such as means, median, standard deviation, etc. Perform a hypothesis test to determine if there is a significant difference between the mean yields for the two types of fertilizers? Use alpha = 0.05 and let: mu_1 mean yield for plot using fertilizer A mu_2 mean yield for plot using fertilizer B

Explanation / Answer

1. The goal or purpose of the study is to study the effect of fertilizers on the production of oranges.

2. The main variable of the study (the outcome or response variable) is the number of pounds of harvested fruits.

3. Population refer to entire group of individuals or instances about whom one eants to learn. The population of study is the plot smeant for harvesting oranges and had been treated with fertilizer A and B. Sample refer to a subset of the population, examined in order to learn about the population. The sample cosntitutes 16 plot streated with fertilizer A and 12 plot streated with fertilizer B.

4. Mean is the arithmetic average of the data. The formula for mean is: xbar=sigma x/n, where, x denote number of pounds of harvested oranges, and n refer to sample size.

Thus, x1bar=(445+523+...+506)/16=457.313

Similarly, x2bar=(362+414+...+384)/12=394.167

Median refer to the point in a distribution of scores above and below wich exactly half of cases fall. For fertilizer A, the data is even, therefore, median is average of the two middle most values of the sorted data that is: (448+457)/2=452.5

Similarly, median for fertilizer B is 390 [(387+393)/2].

Standard deviation is the measure of dispersion and the statistic is computed by summing the sqaured deviations of the scores around the mean, dividing by N-1 and taking square root of it.

s=sqrt[1/N-1 sigma (x-xbar)^2]

s1=sqrt[1/16-1 {(400-457.313)^2+...+(506-457.313)^2}]=38.482

s2=sqrt[1/12-1 {(362-394.167)^2+...+(437-394.167)^2}]=22.954

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.