A researcher at a medical college conducted a study of 60 randomly selected male

Question

A researcher at a medical college conducted a study of 60 randomly selected male soccer players and concluded that frequently "heading" the ball in soccer lowers players' IQs. The soccer players were divided into two groups, based on whether they averaged 10 or more headers per game. Mean IQs were reported in the article, but the sample sizes and standard deviations were not given. Suppose that these values were as given in the accompanying table. (Use a statistical computer package to calculate the P- value. Use mu_fewer than to headers - mu_10 headers- Round your test statistic to two decimal places, your of down to the nearest whole number, and the P-value to three decimal places.)

Explanation / Answer

Given that,
mean(x)=116
standard deviation , s.d1=10
number(n1)=33
y(mean)=107
standard deviation, s.d2 =8
number(n2)=27
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.663
since our test is two-tailed
reject Ho, if to < -2.663 OR if to > 2.663
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (32*100 + 26*64) / (60- 2 )
s^2 = 83.8621
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=116-107/sqrt((83.8621( 1 /33+ 1/27 ))
to=9/2.3764
to=3.7872
| to | =3.7872
critical value
the value of |t | with (n1+n2-2) i.e 58 d.f is 2.663
we got |to| = 3.7872 & | t | = 2.663
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 3.7872 ) = 0.0004
hence value of p0.01 > 0.0004,here we reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.7872
critical value: -2.663 , 2.663
decision: reject Ho
p-value: 0.0004

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.