Question

Here is the dataset containing plant growth measurements of plants grown in solutions of commonly-found chemicals in roadway runoff. Researchers wish to determine roadway runoff with different compositions has a different effect on plant growth.

Phragmites australis, a fast-growing non-native grass common to roadsides and disturbed wetlands of Tidewater Virginia, was grown in a greenhouse and watered with one of the following treatments:

• Distilled water (control);
• A weak petroleum solution (representing standard roadway runoff);
• Sodium chloride solution;
• Magnesium chloride solution;
• De-icing brine (50% sodium chloride and 50% magnesium chloride).

Twenty grass preparations were used for each solution, and total growth (in cm) was recorded after watering every other day for 40 days.

1.) Perform the correct statistical test to determine the p-value.

• Report your answer in scientific notation
  • e.g. 1.00E-04

Distilled H2O	Petro	NaCl	MgCl	NaCl + MgCl
19.93	19.85	19.87	19.91	19.73
19.91	20.06	19.88	19.92	19.77
20.08	19.99	20.04	19.84	19.75
19.99	19.88	20.05	19.98	19.93
19.9	19.98	20.06	19.82	19.94
19.98	20.08	19.83	19.92	19.79
19.92	20.1	19.9	20.09	19.84
20.01	19.82	19.83	20.1	19.94
19.96	20.01	19.85	20.04	19.89
20.13	20.1	19.87	20.04	19.72
20.15	19.84	19.85	19.87	19.88
20.04	20.03	19.93	19.89	20
19.98	20.01	19.82	19.77	19.74
20.03	19.96	19.85	19.97	19.95
20.13	19.91	20.06	19.84	19.79
20	20.03	20.04	20.07	19.85
20.07	19.92	20	19.83	19.74
19.98	19.94	19.9	19.9	19.78
20.02	20.01	19.94	19.95	19.88
19.94	19.8	20.05	19.78	19.83

2.) Based on the p-value from your Phragmites dataset analysis, select the options that are TRUE.

		The p-value is greater than α(0.05).
		The calculated F value was less than the critical F value.
		A post-hoc test is necessary to determine statistically-significant difference(s).
		The p-value is less than α(0.05).
		The calculated F value was greater than the critical F value.

3.) Which is / are (an) appropriate evaluation(s) of the results of your Phragmites data analysis?

		I would fail to reject the null hypothesis.
		There is a significant difference in growth between the five groups of Phragmites plants.
		The control plants' growth rate was greater than the contaminated plants.
		I would reject the null hypothesis.
		There is no significant difference in growth between the five groups of Phragmites plants.

Answer 1

using Excel

data -> data analysis -> Anova Single Factor

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Distilled H2O	20	400.15	20.0075	0.0055
Petro	20	399.32	19.9660	0.0086
NaCl	20	398.62	19.9310	0.0081
MgCl	20	398.53	19.9265	0.0103
NaCl + MgCl	20	396.74	19.8370	0.0073
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.3180	4	0.0795	9.9463	0.0000	2.4675
Within Groups	0.7593	95	0.0080
Total	1.0773	99

since p-value = 0.0000 < alpha

we reject the null hypothesis

we conclude that there is significant difference in means

2)

option C),D) and E) are correct

it wrong then remove option C)

3)

There is a significant difference in growth between the five groups of Phragmites plants.

I would reject the null hypothesis.

option )A and C) are correct