80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in
ID: 3208352 • Letter: 8
Question
80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis.
Relaxed time (seconds) Jumping time (seconds)
26 21
47 40
30 28
22 21
23 25
45 43
37 35
29 32
Explanation / Answer
Given that,
mean(x)=32.375
standard deviation , s.d1=9.6205
number(n1)=8
y(mean)=30.625
standard deviation, s.d2 =8.3313
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =32.375-30.625/sqrt((92.55402/8)+(69.41056/8))
to =0.389
| to | =0.389
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.365
we got |to| = 0.38893 & | t | = 2.365
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.3889 ) = 0.709
hence value of p0.05 < 0.709,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.389
critical value: -2.365 , 2.365
decision: do not reject Ho
p-value: 0.709
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