Determine the Test Statistic: A) 3.9905 B) 5.1582 C) 5.5327 D) 7.1427 Please sho

Question

Determine the Test Statistic:

A) 3.9905

B) 5.1582

C) 5.5327

D) 7.1427

Please show your steps so that I can understand these concepts.

In a study performed by Pierce (1948), the frequency (the number of wing vibrations per second) of chirps made by a striped ground cricket, was recorded at various ground temperatures. Since crickets are ectotherms (cold-blooded), the rate of their physiological processes and their overall metabolism are influenced by temperature Consequently, there is reason to believe that temperature would have a profound effect on aspects of their behavior, such as chirp frequency. Is there a linear relation- ship between chirp frequency and temperature? The data below correspond to a small sample of observations Subject Temperature (o F) Frequency (Chirps/Sec) 20 88.6 18.4 84.3 16.3 83.3 15.5 75.2 14.4 76.3 16.92 81.54 Mean 2.26 5.66 SD 7173

Explanation / Answer

Given that,
value of r =0.9173
number (n)=5
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.05
from standard normal table, two tailed t /2 =3.182
since our test is two-tailed
reject Ho, if to < -3.182 OR if to > 3.182
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.9173/(sqrt( ( 1-0.9173^2 )/(5-2) )
to =3.99
|to | =3.99
critical value
the value of |t | at los 0.05% is 3.182
we got |to| =3.99 & | t | =3.182
make decision
hence value of | to | > | t | and here we reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: A) 3.9905
critical value: -3.182 , 3.182
decision: reject Ho

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