Question

A group of students wish to examine how pesticides affect seedling growth. Seeds are randomly assigned to be planted in pots with soil treated with pesticide (treatment group, group 1) or in pots with untreated soil (control group, group 2). Seedling growth (in mm) is recorded after 2 weeks. The data for each group is given below.

Pesticide Group (Group 1)		Untreated Group (Group 2)
24	71	42	26
61	49	33	54
59	54	46	60
46	43	37	28
43	53	43	62
44	57	41	20
52	49	10	53
43	56	42	48
58	33	55	37
67		19	85
62		17	42
57		55

Is there evidence that the average growth in mm differs between the pesticide and untreated groups?

What is the value of the test statistic?

The p-value?

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

This is a two tailed test.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 51.47619

X2bar = 41.52174

S1 = 11.00736

S2 = 17.14873

n1 = 21

n2 = 23

df = n1 + n2 – 2 = 21 + 23 – 2 = 42

We assume α = 0.05

Critical value = - 2.0181 and 2.0181

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(21 – 1)* 11.00736^2 + (23 – 1)* 17.14873^2]/(21 + 23 – 2)

S_p² = 211.7376

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (51.47619 – 41.52174) / sqrt[211.7376*((1/21)+(1/23))]

t = 2.2666

Test statistic = 2.2666

P-value = 0.0286

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the average growth in mm differs between the pesticide and untreated groups.