In: Statistics and Probability
How can I apply the data that is in the box to these questions? Specifically, part F. Thanks!
Grunt | No grunt |
m=490 ms | M=483ms |
SS=3100 | SS=2511 |
N=32 | N=32 |
-First, calculate the pooled variance.
Note: the sum of squares for each sample has already been calculated and is provided in the data box above (for the grunt sample, SS = 3100, and for the no grunt sample, SS = 2511)
-Second, calculate the t-statistic.
Given that,
mean(x)=490
standard deviation , s.d1=10
number(n1)=32
y(mean)=483
standard deviation, s.d2 =9
number(n2)=32
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.67
since our test is right-tailed
reject Ho, if to > 1.67
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2
)/(n1+n2-2)
s^2 = (31*100 + 31*81) / (64- 2 )
s^2 = 90.5
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=490-483/sqrt((90.5( 1 /32+ 1/32 ))
to=7/2.378
to=2.943
| to | =2.943
critical value
the value of |t α| with (n1+n2-2) i.e 62 d.f is 1.67
we got |to| = 2.943 & | t α | = 1.67
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: right tail -ha : ( p > 2.9433 ) = 0.00228
hence value of p0.05 > 0.00228,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
T test for difference of means with equal variance
d.
test statistic: 2.943
c.
critical value: 1.67
e.
pooled variance = 90.5
decision: reject Ho
p-value: 0.00228
f.
we have enough evidence to support the claim that if grunting
negatively impacts the opponent’s ability to judge the trajectory
of the ball, then response times should be longer for the grunt
group than for the no grunt group.