Question

As part of a study of wheat maturation, an agronomist selected a sample of wheat plants at random from a field plot. For each plant, the agronomist measured the moisture content from two locations: one from the central portion and one from the top portion of the wheat head. The agronomist hypothesizes that the central portion of the wheat head has more moisture than the top portion. What can the agronomist conclude with α = 0.01? The moisture content data are below.

central	top
62.7 63.6 60.9 63.1 62.7 63.7 62.5	61.7 63.6 60.2 62.9 61.6 62.8 62.3

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- wheat maturation top portion moisture content wheat head central portion
Condition 2:
---Select--- wheat maturation top portion moisture content wheat head central portion

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

The central portion of the wheat head had significantly more moisture than the top portion.The central portion of the wheat head had significantly less moisture than the top portion.    There was no significant moisture difference between the central and top portion of the wheat head.

Answer 1

Observations	Central	Top	d=Cental - Top	d2
1	62.7	61.7	1	1
2	63.6	63.6	0	0
3	60.9	60.2	0.7	0.49
4	63.1	62.9	0.2	0.04
5	62.7	61.6	1.1	1.21
6	63.7	62.8	0.9	0.81
7	62.5	62.3	0.2	0.04
SUM =			4.1	3.59