In: Statistics and Probability
Assumed data,because not given in the data
sample mean, x =22
standard deviation, s =6
Given that,
population mean(u)=23.4
number (n)=252
null, Ho: μ=23.4
alternate, H1: μ<23.4
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.651
since our test is left-tailed
reject Ho, if to < -1.651
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =22-23.4/(6/sqrt(252))
to =-3.7041
| to | =3.7041
critical value
the value of |t α| with n-1 = 251 d.f is 1.651
we got |to| =3.7041 & | t α | =1.651
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -3.7041 ) = 0.00013
hence value of p0.05 > 0.00013,here we reject Ho
ANSWERS
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null, Ho: μ=23.4
alternate, H1: μ<23.4
test statistic: -3.7041
critical value: -1.651
decision: reject Ho
p-value: 0.00013
we have enough evidence to support the claim that the average
bodyfat for Crossfit Athletes is less than the average bodyfat for
the American population.