Consider the following output of a single-factor ANOVA performed on a continuous variable x grouped by categories A, B, C, D, E and F

SUMMARY
Groups	Count	Sum	Average	Variance
A	25	1203.5	48.1	188.9
B	23	1182.5	51.4	50.1
C	12	646	53.8	82.7
D	27	1385	51.3	87.3
E	23	1121	48.7	76.2
F	22	1226	55.7	59.6

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	908.6	5	181.7	1.949	0.091	2.286
Within Groups	11744.7	126	93.2
Total	12653.2	131

Take p=0.05 to be significant.

Select the correct statement:

Question 1 options:

	We do not have enough evidence to reject the null hypothesis. We should not perform POST-HOC analysis to test for the difference between individual means
	We do not have enough evidence to reject the null hypothesis. We can now perform POST-HOC analysis to test for the difference between individual means.
	We have enough evidence to reject the null hypothesis. We should not perform POST-HOC analysis to test for the difference between individual means
	We have enough evidence to reject the null hypothesis. We can now perform POST-HOC analysis to test for the difference between individual means.

Eucalyptus

The next 4 questions relate to Task 1 - Comparing growth of Eucalyptus seedlings under different treatments. This material was covered in the week 11 computing lab on ANOVA.

As part of a project to regenerate native vegetation on a previously farmed land, a random selection of 64Eucalyptus Obliqua seedlings were planted on Tasmanian Peninsula. The seedlings' height was measured (in centimetres) and then they were randomly allocated to one of the four treatments:

1. Herbicide applied (H);
2. Scalping performed (S);
3. Applying herbicide and scalping (H+S);
4. no treatment (the control group) (C).

You should use JASP (or SPSS, or Excel) output you gathered to answer the questions.

Question 2 (1 point)

For the ANOVA analysis comparing "H_2yr"test scores for the four treatments the p-value was 0.103. This means:

Question 2 options:

	We cannot reject the null hypothesis that the means for H_2yr for the four treatment groups are the same.
	The p-value is significant so the mean test scores for the the four treatments are most likely all the same.
	The p-value is significant so at least one of the mean test scores is different from the others.
	The p-value is quite large so the means are most likely all different from each other.