Question

Plot	Fertilizer A	Fertilizer B	Fertilizer C
1	563	588	575
2	593	624	593
3	542	576	564
4	649	672	653
5	565	583	556
6	587	612	590
7	595	617	607
8	429	446	423
9	500	515	483
10	610	641	626
11	524	547	523
12	559	586	568
13	546	582	551
14	503	530	502
15	550	573	567
16	492	518	495
17	497	529	513
18	619	643	626
19	473	497	479
20	533	556	540

Fundamental to improving crop yield is the necessity to quantify effects of such factors as crop varieties, nutrients, and fertilizers on crop yields. Statisticians have been integral to this effort for a century. Ronald Fisher, one of statistics’ most important historical figures, developed his principles of experimental design in the context of agricultural field trials in the 1910s and 20s. Fisher invented statistical models to account for confounding factors affecting crop yields such as soil type or weather. These principles, along with new statistical methods, continue to have an important role in agricultural research. For example, developments in spatial statistical models have contributed to precision agriculture, a farm management strategy that uses global positioning systems to target interventions to the needs of a crop at a specific location in a field. One of the first uses of some techniques in statistics goes back to 1920s for the purpose of testing the effectiveness of fertilizers on plots of lands. The objective of this exercise was to see if different amounts of fertilizers are yielding different amounts of crops. Recently, a scientist at one of the top agricultural colleges in the U.S. decided to conduct this experiment on some new fertilizers and was interested to see the impact of them on crop yields. Accordingly, she selected three types of fertilizers A, B, and C and applied fertilizer A to 20 1- acre plots of land, fertilizer B to another 20 plots, and fertilizer C to yet another plots of land. At the end of growing season, the crop yields were recorded and is summarized in the data file.

f)Calculate a 90% confidence interval estimate of applying fertilizer B data for crop yield in all plots of lands and interpret your results.

g) Assuming population mean of crop yield is 570 bushels for all fertilizers with standard deviation of 40 bushels for all. Formulate a test hypothesis that crop yield by applying fertilizer C differs from the population crop yield for all fertilizers. Conduct the hypothesis test, conclude your analysis and explain your answer. Use both critical value and p-value approach with alpha=0.05

h) Calculate a 90% confidence interval estimate of the difference between the population mean yield of fertilizers B and A. Can we conclude at 0.05 level of significance, that the crop yield using fertilizer B is greater than the crop yield using fertilizer A? (hint: you can use the template in chapter 10 to calculate degrees of freedom and the standard error)

i) If we assume that observations are now plots of lands, can the scientist infer that there are differences between the three types of fertilizers?

Answer 1

f)Calculate a 90% confidence interval estimate of applying fertilizer B data for crop yield in all plots of lands and interpret your results.

The 90% confidence interval estimate of applying fertilizer B data for crop yield in all plots of lands is between 550.208 and 593.292.

We are 90% confident that the true estimate of applying fertilizer B data for crop yield in all plots of lands is between 550.208 and 593.292.

550.208	confidence interval 90.% lower
593.292	confidence interval 90.% upper
21.542	margin of error

g) Assuming population mean of crop yield is 570 bushels for all fertilizers with standard deviation of 40 bushels for all. Formulate a test hypothesis that crop yield by applying fertilizer C differs from the population crop yield for all fertilizers. Conduct the hypothesis test, conclude your analysis and explain your answer. Use both critical value and p-value approach with alpha=0.05

The hypothesis being tested is:

H0: µ = 570

Ha: µ ≠ 570

570.000	hypothesized value
551.700	mean Fertilizer C
40.000	std. dev.
8.944	std. error
20	n
-2.05	z
.0408	p-value (two-tailed)

The p-value is 0.0408.

Since the p-value (0.0408) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that µ ≠ 570.

h) Calculate a 90% confidence interval estimate of the difference between the population mean yield of fertilizers B and A. Can we conclude at 0.05 level of significance, that the crop yield using fertilizer B is greater than the crop yield using fertilizer A? (hint: you can use the template in chapter 10 to calculate degrees of freedom and the standard error)

-54.643	confidence interval 90.% lower
4.043	confidence interval 90.% upper
29.343	margin of error

The 90% confidence interval estimate of the difference between the population mean yield of fertilizers B and A is between -54.643 and 4.043.

We cannot conclude that the crop yield using fertilizer B is greater than the crop yield using fertilizer A.

i) If we assume that observations are now plots of lands, can the scientist infer that there are differences between the three types of fertilizers?

	Mean	n	Std. Dev
	546.5	20	54.35	Fertilizer A
	571.8	20	55.72	Fertilizer B
	551.7	20	57.79	Fertilizer C
	556.6	60	56.10	Total
ANOVA table
Source	SS	df	MS	F	p-value
Treatment	7,131.03	2	3,565.517	1.14	.3276
Error	1,78,562.90	57	3,132.682
Total	1,85,693.93	59

No, there are no differences between the three types of fertilizers.

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