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 62.7 60.2 62.5 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--- top portion wheat maturation moisture content wheat head central portion
Condition 2:
---Select--- top portion wheat maturation moisture content wheat head central portion

c) Compute the appropriate test statistic(s) to make a decision about H₀.

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. (Please say if "na")
[  ,  ]

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. (select one)

-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

Central	Top	Difference
62.7	61.7	1
63.6	62.7	0.9
60.9	60.2	0.7
63.1	62.5	0.6
62.7	61.6	1.1
63.7	62.8	0.9
62.5	62.3	0.2

Sample mean of the difference using excel function AVERAGE(), x̅d = 0.7714

Sample standard deviation of the difference using excel function STDEV.S(), sd = 0.3039

Sample size, n = 7

a) appropriate test statistic : Related-Samples t-test

c) Null and Alternative hypothesis:

Ho : µd ≤ 0 ; H1 : µd > 0

Test statistic:

t = (x̅d)/(sd/√n) = (0.7714)/(0.3039/√7) = 6.7151

df = n-1 = 6

Critical value :

Right tailed critical value, t-crit = ABS(T.INV(0.01, 6)) = 3.143

Conclusion:

Reject the null hypothesis

d) 99% Confidence interval :

At α = 0.01 and df = n-1 = 6, two tailed critical value, t-crit = T.INV.2T(0.01, 6) = 3.707

Lower Bound = x̅d - t-crit*sd/√n = 0.7714 - 3.707 * 0.3039/√7 = 0.3455

Upper Bound = x̅d + t-crit*sd/√n = 0.7714 + 3.707 * 0.3039/√7 = 1.1973

0.3455 < µd < 1.1973

e) Cohen's d = (x̅_d- µ_d)/s = 2.538

r² = d²/(d²+4) = 0.6169

f) Make an interpretation based on the results. (select one)

-The central portion of the wheat head had significantly more moisture than the top portion.