B) compute the appropriate statistic and state your conclusion 2. Since muscle t

Question

B) compute the appropriate statistic and state your conclusion 2. Since muscle tension in the head region has been associated with tension headaches, you reason that if the produced. You design an experiment in which nine subjects with tension headaches participate. The subjects keep daily logs of the number of headaches they experience during a 2-week base-line period. Then you train them to lower their muscle tension in the head region, using a biofeedback device. For this experiment, the biofeedback device is connected to the frontalis muscle, a muscle in the forehead region. The device tells the subject the amount of tension in the muscle to which it is attached (in this case, frontalis) and helps them achieve low tension levels. After 6 weeks of training, during which the subjects have become successful at maintaining low frontalis muscle tension, they again keep a 2-week log of the number of headaches experienced. The following are the number of headaches recorded during each 2-week period. 0.05 (2 tail) muscle tension could be changed, an effect on headaches would be Baseline No. of After training Headaches Subject No. No. of Headaches 17 13 10 a. State the null and alternative hypotheses

Explanation / Answer

2.

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 5.67
We have d = 5.67
pooled variance = calculate value of Sd= S^2 = sqrt [ 427-(51^2/9 ] / 8 = 4.15
to = d/ (S/n) = 4.1
critical Value
the value of |t | with n-1 = 8 d.f is 2.306
we got |t o| = 4.1 & |t | =2.306
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 4.0955 ) = 0.0035
hence value of p0.05 > 0.0035,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: 4.1
critical value: reject Ho, if to < -2.306 OR if to > 2.306
decision: Reject Ho
p-value: 0.0035

we have enough evidence to support the claim

X Y X-Y (X-Y)^2 17 3 14 196 13 7 6 36 6 2 4 16 5 3 2 4 5 6 -1 1 10 2 8 64 8 1 7 49 6 0 6 36 7 2 5 25

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