Given that,
mean(x)=108
standard deviation , s.d1=7.2
number(n1)=10
y(mean)=103
standard deviation, s.d2 =8.1
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.734
since our test is right-tailed
reject Ho, if to > 1.734
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2
)/(n1+n2-2)
s^2 = (9*51.84 + 9*65.61) / (20- 2 )
s^2 = 58.725
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=108-103/sqrt((58.725( 1 /10+ 1/10 ))
to=5/3.427
to=1.459
| to | =1.459
critical value
the value of |t α| with (n1+n2-2) i.e 18 d.f is 1.734
we got |to| = 1.459 & | t α | = 1.734
make decision
hence value of |to | < | t α | and here we do not reject
Ho
p-value: right tail -ha : ( p > 1.459 ) = 0.0809
hence value of p0.05 < 0.0809,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.459
critical value: 1.734
decision: do not reject Ho
p-value: 0.0809
we do not have enough evidence to support the claim that if the
first born twins have a higher IQ than second born twins.
option:A
Type 2 error is possible when its fails to reject the null
hypothesis.
A research was done to see if the firs- born twins have a higher IQ than...
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