To study the effects of row spacing on the productivity of a plant species, plan

Question

To study the effects of row spacing on the productivity of a plant species, plants are to be grown at 30 cm and 50 cm spacing. The observed productivity values (in terms of growth) are given below.

30 cm58 53 59 45 69 72 55 56 58.375 8.651

50 cm52 49 55 51 53 61 50 52 52.875 3.758

The data are presented to you and the researchers want to know whether increasing row spacing reduces productivity. As a statistical consultant, before you choose the appropriate test, you would like to know more on how those data are collected. You approach the researchers who carried out the experiments. The project was directed by a faculty member and a technician. Unfortunately, both of their descriptions on how the experiment was conducted do not agree to each other. Here are their descriptions:

Faculty: The design used sixteen comparable fields at the experimental station. Once the fields had been identified, it was determined at random which eight fields were planted at 30 cm and the other fields were planted at 50 cm spacing. Once the plants had reached maturity, the productivity was determined. Technician: At the time of planting, only eight fields were available at the experimental station. They were furthermore not comparable in soil types as the faculty member assumed. Since the plant species grows quickly and can reach maturity between spring and fall more than once, the following was done: Plant each of the eight available fields twice during the season. After tilling the fields, start on some fields with 30 cm spacing. After the plants have reached maturity, harvest, till the soil again and plant at 50 cm row spacing. On other fields, start with 50 cm spacing and then switch to 30 cm spacing after harvest. Which spacing goes first on each of the eight fields is determined by flipping a coin.

Using all the information above, answer the following questions.

(a) Test whether increasing row spacing reduces productivity if the experiment were conducted as the faculty member had planned, at = 0.05 level (list your hypotheses, critical region(s), and conclusion). Assume normality on the data. If any necessary test(s) is (are) needed before completing your task, do it (them) at level 0.10.

(b) Test the same hypotheses as in part (a) at level = 0.05 using the technician’s scenario. The data in the table is organized such that the values in a column represent observations from the same field.

(c) Comment on the results by both tests in parts (a) and (b). Was the redesign by the technician reasonable?

Explanation / Answer

a.
Given that,
mean(x)=58.375
standard deviation , s.d1=8.651
number(n1)=8
y(mean)=52.875
standard deviation, s.d2 =3.758
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =58.375-52.875/sqrt((74.8398/8)+(14.12256/8))
to =1.649
| to | =1.649
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.895
we got |to| = 1.64932 & | t | = 1.895
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.6493 ) = 0.07154
hence value of p0.05 < 0.07154,here we do not reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.649
critical value: 1.895
decision: do not reject Ho
p-value: 0.07154
we do not have enough evidence to support the claim

Given that,
mean(x)=58.375
standard deviation , s.d1=8.651
number(n1)=8
y(mean)=52.875
standard deviation, s.d2 =3.758
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.415
since our test is right-tailed
reject Ho, if to > 1.415
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =58.375-52.875/sqrt((74.8398/8)+(14.12256/8))
to =1.649
| to | =1.649
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.415
we got |to| = 1.64932 & | t | = 1.415
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 1.6493 ) = 0.07154
hence value of p0.1 > 0.07154,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.649
critical value: 1.415
decision: reject Ho
p-value: 0.07154
we have enough evidence that increasing row spacing reduces productivity if the experiment were conducted as the faculty member had planned

b.

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 5.5
We have d = 5.5
pooled variance = calculate value of Sd= S^2 = sqrt [ 522-(44^2/8 ] / 7 = 6.325
to = d/ (S/n) = 2.46
critical Value
the value of |t | with n-1 = 7 d.f is 1.895
we got |t o| = 2.46 & |t | =1.895
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.4597 ) = 0.02174
hence value of p0.05 > 0.02174,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: 2.46
critical value: reject Ho, if to > 1.895
decision: Reject Ho
p-value: 0.02174

c.

problem (a) we changes the level of significance at 0.05 and 0.10 then the solution is at 0.05 level failed to reject the null hypothesis and at level 0.10 reject null hypothesis
problem (b) at 0.05 level of significance reject the null hypothesis
so that the redesign by the technician reasonable.

X Y X-Y (X-Y)^2 58 52 6 36 53 49 4 16 59 55 4 16 45 51 -6 36 69 53 16 256 72 61 11 121 55 50 5 25 56 52 4 16 0 0 0 0 0 0 0 0 44 522

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