Question

1. A researcher is investigating the health of some plants that have been treated with a chemical to increase their rate of growth. In the problems below, data is given about levels of two different elements in the plants. The data for each element show the levels of the element in the treated plants, and the levels of the element in untreated plants.

A.) The nitrogen levels of the treated and untreated plants are given in parts per million in the chart below.

b. State the null hypothesis about nitrogen for the experiment.

c. Compare the results for the control group and the treatment group. Do you think that the researcher has enough evidence to reject the null hypothesis?

D. The potassium levels of the treated and untreated plants are given in parts per million in the chart below.

e. State the null hypothesis about potassium for the experiment.

f. Compare the results for the control group and the treatment group. Do you think that the researcher has enough evidence to reject the null hypothesis?

Untreated plants Treated plants 140 100 120 110130 100 130 150 120 120 180 130 150 220 150170 160 130 Untreated plants Treated plants 200 250 320 310 340 310 270 340 370 310 260 280 310 220 260290 220 250

Answer 1

a) , b) and c)

Untreated	treated
140	120
100	180
120	130
110	150
130	220
100	150
130	170
150	160
120	130

Ho : μ1 = μ2

$Ha: \mu_1 \neq \mu_2$

t-Test: Two-Sample Assuming Equal Variances
	Untreated	treated
Mean	122.2222222	156.6666667
Variance	294.4444444	950
Observations	9	9
Pooled Variance	622.2222222
Hypothesized Mean Difference	0
df	16
t Stat	-2.929224666
P(T<=t) one-tail	0.004913031
t Critical one-tail	1.745883676
P(T<=t) two-tail	0.009826061
t Critical two-tail	2.119905299

p-value = 0.0098 < alpha

hence we reject the null hypothesis

we conclude that the researcher has enough evidence to reject the null hypothesis

d) ,e) and f)

data

Untreated	treated
200	310
250	260
320	280
310	310
340	220
310	260
270	290
340	220
370	250

t-Test: Two-Sample Assuming Equal Variances
	Untreated	treated
Mean	301.1111111	266.6666667
Variance	2761.111111	1150
Observations	9	9
Pooled Variance	1955.555556
Hypothesized Mean Difference	0
df	16
t Stat	1.652305552
P(T<=t) one-tail	0.058978919
t Critical one-tail	1.745883676
P(T<=t) two-tail	0.117957838
t Critical two-tail	2.119905299

p-value =0.1180 > alpha

we conclude that the researcher has not enough evidence to reject the null hypothesis