Question

Students planted alfalfa seeds in 15 cups and randomly chose 5 to get plain water, 5 to get a moderate amount of acid, and 5 to get a stronger acid solution. They propsed that the distance from the window might affect the growth rates as it was the main source of light

The cups were arranged in 5 rows of 3 with a cup of each acid level in each row. They were arranged as follows in rows a,b,c,d,e with a being furthest from the window and e being the closest. Each cup was an experimental unit and the response variable was the average height of the alfslfa sprout n each cup after 4 days. The data is shown below.

Ht4	Acid	Row
1.45	water	a
2.79	water	b
1.93	water	c
2.33	water	d
4.85	water	e
1	1.5HCl	a
0.7	1.5HCl	b
1.37	1.5HCl	c
2.8	1.5HCl	d
1.46	1.5HCl	e
1.03	3.0HCl	a
1.22	3.0HCl	b
0.45	3.0HCl	c
1.65	3.0HCl	d
1.07	3.0HCl	e

a. Find the means for each row of cups(a,b..,e) and each treatment(water,1.5HCL, 3.0 HCL) Also find the average and standard deviation for the growth in all 15 cups.

b. Construct a two-way main effects ANOVA table for testing for difference in average growth due to acid treatments using the rows as a blocking variable.

c. Based on the ANOVA would you conclude that there is significanht difference in average growth due to the treatments? Why or why not?

d. Based on the ANOVA, would you conclude that there is a significant difference in average growth due to the distance from the window? why or why not?

Answer 1

a)

b)

c) The estimated p-value of treatment (Acid) from the ANOVA table is 0.049 and less than 0.05 level of significance. Hence, we can conclude that there is the significant difference in average growth due to the treatments at 0.05 level of significance.

d) The estimated p-value of Row from the ANOVA table is 0.324 and more than 0.05 level of significance. Hence, we can conclude that there is an insignificant difference in average growth due to the distance from the window at 0.05 level of significance.