\"Tennis elbow\" is an orthopedic condition thought to be related to the force e
ID: 3310225 • Letter: #
Question
"Tennis elbow" is an orthopedic condition thought to be related to the force experienced when hitting the ball. A recent article in the International Journal of Sport Biomechanics presented the force (in newtons) on the hand after impact for six advanced and eight intermediate players. Using the data below, test ( = .05) to see if advanced and intermediate players experience the same force. Assume relevant distributions are normal 4. Force Experienced (N) Advanced Intermediate 44.7 15.58 26.31 19.16 55.75 24.13 28.54 10.56 46.9932.88 39.46 21.47 14.32 33.09 Mean 40.3 21.4 Std Dev 11.38.3 a. State the hypothesis late test statistic and the level of sig c. State the decision rules, label/complete the sketch to show rejection regionExplanation / Answer
Given that,
mean(x)=40.3
standard deviation , s.d1=11.3
number(n1)=6
y(mean)=21.4
standard deviation, s.d2 =8.3
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =40.3-21.4/sqrt((127.69/6)+(68.89/8))
to =3.4568
| to | =3.4568
critical value
the value of |t | with min (n1-1, n2-1) i.e 5 d.f is 2.571
we got |to| = 3.45683 & | t | = 2.571
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.4568 ) = 0.018
hence value of p0.05 > 0.018,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.4568
critical value: -2.571 , 2.571
decision: reject Ho
p-value: 0.018
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