Question

Use R studio to answer the following

Organic Inorganic 570 482 592 612 630 642 511 582 634 444 513 422 558 460 599

We are interested in testing for significant difference in the mean biomass of pine seedlings grown with organic versus inorganic fertilizer. What case are we in?  In other words, can we assume that s²₁ = s²₂. (s_org=50 and s_in=87 with samples of sizes of n_org = 7 and n_in = 8. Use a = 0.10.

1. Complete the test with the organic fertilizer assigned as the first (numerator) population. (Hint: read in the data as vectors, or copy into a .csv file to enter into R)
2. Complete the test with the inorganic fertilizer assigned as the first (numerator) population.

Answer 1

Ans a ) the test with the organic fertilizer assigned as the first (numerator) population

using R

> organic=c(570,592,630,511,634,513,558)
> inorganic=c(482,612,642,582,444,422,460,599)
> var.test(organic, inorganic, ratio = 1, alternative = c("two.sided"))

   F test to compare two variances

data: organic and inorganic
F = 0.33038, num df = 6, denom df = 7, p-value = 0.1984
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.06454586 1.88169348
sample estimates:
ratio of variances
0.3303842

since p value of F stat is 0.1984 which is greater tha 0.05 so we can assume variance are equal .

b )  the test with the inorganic fertilizer assigned as the first (numerator) population.

using R

> organic=c(570,592,630,511,634,513,558)
> inorganic=c(482,612,642,582,444,422,460,599)
> var.test(inorganic, organic, ratio = 1, alternative = c("two.sided"))

   F test to compare two variances

data: inorganic and organic
F = 3.0268, num df = 7, denom df = 6, p-value = 0.1984
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.5314362 15.4928612
sample estimates:
ratio of variances
3.026779

since p value of F stat is 0.1984 which is greater tha 0.05 so we can assume variance are equal .