Using the information above: 1) use a T-Test to determine if there is a signific

Question

Using the information above:

1) use a T-Test to determine if there is a significant difference between the two species in terms of days of germination

2) use a X^2 (x-square) test to determine if there is a difference in germination rate.

Please show the work to questions.

thank you;

In a germination experiment the germination success and timing of wild radish (Raphanus raphanistrum) and crop radish (Raphanus sativus) were measured. The seeds are from the two plant species that are closely related and differ in the level of domestication. As an evolutionary botanist, it is suspect that domestication could have selected for differences in a host of traits, including germination success (Percent germination) as well as timing of germination (Rate of germination). We want to know if species potentially influences these two traits. The following data was collected orn the 20 seeds that were originally planted for each species. Species Wild Radish11 | Days to Germinate 5,3,1,4,9,9,7,14,7,8,13 3.4,1.4,5.5.2.3.1.1.44,5.4.4,5.2.1. # Germinated

Explanation / Answer

a.
Given that,
mean(x)=7.2727
standard deviation , s.d1=3.9771
number(n1)=11
y(mean)=3.2222
standard deviation, s.d2 =1.5168
number(n2)=18
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.228
since our test is two-tailed
reject Ho, if to < -2.228 OR if to > 2.228
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =7.2727-3.2222/sqrt((15.81732/11)+(2.30068/18))
to =3.237
| to | =3.237
critical value
the value of |t | with min (n1-1, n2-1) i.e 10 d.f is 2.228
we got |to| = 3.23703 & | t | = 2.228
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.237 ) = 0.009
hence value of p0.05 > 0.009,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.237
critical value: -2.228 , 2.228
decision: reject Ho
p-value: 0.009

we have enough evidence to support there is a significant difference between the two species in terms of days of germination

b.

Given that,
population variance (^2) =15.8173
sample size (n) = 11
sample variance (s^2)=2.3006
null, Ho: ^2 =15.8173
alternate, H1 : ^2 !=15.8173
level of significance, = 0.05
from standard normal table, two tailed ^2 /2 =18.307
since our test is two-tailed
reject Ho, if ^2 o < - OR if ^2 o > 18.307
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(11 - 1 ) * 2.3006 / 15.8173 = 10*2.3006/15.8173 = 1.45
| ^2 cal | =1.45
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=10 is 18.307
we got | ^2| =1.45 & | ^2 | =18.307
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.9991
ANSWERS
---------------
null, Ho: ^2 =15.8173
alternate, H1 : ^2 !=15.8173
test statistic: 1.45
critical value: -18.307 , 18.307
p-value:0.9991
decision: do not reject Ho

we do not have enough evidence to support there is a difference in germination rate

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.