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 less moisture than the top portion. What can the agronomist conclude with α = 0.05? The moisture content data are below.

central	top
61.7 63.6 60.2 62.5 61.6 62.8 62.3	62.7 63.6 60.1 63.1 62.7 63.7 62.5

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

For the score differences we have

D¯=−0.529 ,sD​=0.496

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ < 0

This corresponds to a left-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=6.

Hence, it is found that the critical value for this left-tailed test is tc​=−1.943, for α=0.05 and df=6.

The rejection region for this left-tailed test is R={t:t<−1.943}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

t=​D¯/sD​/sqrt(n)​ =​−0.529/0.496/ sqrt(7) ​=−2.821

(4) Decision about the null hypothesis

Since it is observed that t=−2.821<tc​=−1.943, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0152, and since p=0.0152<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is less than μ2​, at the 0.05 significance level.

Confidence Interval: The 95% confidence interval is −0.987<μD​<−0.07.

e) Cohen's d = (62.629 - 62.1) ⁄ 1.143848 = 0.462474. MEDIUM EFFECT

( f) An interpretation based on the results:

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