Question

Two different plant fertilizers are available. Set up an experiment to determine which will grow the largest plant by answering the following questions.

The yield data (kg/plot) were collected from 20 plots. Each fertilizer (A or B) was randomly assigned to 10 plots.

Fertilizer A	17.5	17.9	18.2	16.9	17.2	17.5	17.3	18.9	16.8	17.1
Fertilizer B	18.4	17.9	18.2	17.6	17.6	18.8	18.1	17.4	17.8	17.6

f. Conduct the appropriate hypothesis testing. Report the test-statistics and p-values.

g. Make a conclusion based on a 10% significance level. Briefly explain how do you make this conclusion.

Answer 1

f.

R code:

Fertilizer_A=c(17.5,17.9,18.2,16.9,17.2,17.5,17.3,18.9,16.8,17.1)
Fertilizer_B=c(18.4,17.9,18.2,17.6,17.6,18.8,18.1,17.4,17.8,17.6)
shapiro.test(Fertilizer_A)
shapiro.test(Fertilizer_B)

Output:

> shapiro.test(Fertilizer_A)

Shapiro-Wilk normality test

data: Fertilizer_A
W = 0.91367, p-value = 0.3072

> shapiro.test(Fertilizer_B)

Shapiro-Wilk normality test

data: Fertilizer_B
W = 0.93227, p-value = 0.4706
Since p-values for both samples are greater than 0.10 so we can assume these two samples come from two independent normal populations.

R code:

Fertilizer_A=c(17.5,17.9,18.2,16.9,17.2,17.5,17.3,18.9,16.8,17.1)
Fertilizer_B=c(18.4,17.9,18.2,17.6,17.6,18.8,18.1,17.4,17.8,17.6)
var.test(Fertilizer_A, Fertilizer_B, alternative = "two.sided")

Output:

F test to compare two variances

data: Fertilizer_A and Fertilizer_B
F = 2.1954, num df = 9, denom df = 9, p-value = 0.257
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.5453119 8.8387583
sample estimates:
ratio of variances
2.195423
Since p-value = 0.257>0.1 so we can assume two variances are equal.

R code:

Fertilizer_A=c(17.5,17.9,18.2,16.9,17.2,17.5,17.3,18.9,16.8,17.1)
Fertilizer_B=c(18.4,17.9,18.2,17.6,17.6,18.8,18.1,17.4,17.8,17.6)
t.test(Fertilizer_A,Fertilizer_B,, var.equal=TRUE,alternative ="two.sided",conf.level = 0.90)

Output:

Two Sample t-test

data: Fertilizer_A and Fertilizer_B
t = -1.6669, df = 18, p-value = 0.1128
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-0.83652325 0.01652325
sample estimates:
mean of x mean of y
17.53 17.94

Value of test statistic=-1.6669, p-value=0.1128.

g. Since p-value>0.10 so there is insignificant difference between Fertilizer A and Fertilizer B.