#4. In a study of cartilage thickness, a random sample of 11 men had an average
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#4. In a study of cartilage thickness, a random sample of 11 men had an average volume of femoral cartilage of 18.7 cm with a S.D. of 3.3 cm2, while the average femoral cartilage volume of a sample of 9 women was 11.2 cm with S.D. of 2.4 cm2. Can you conclude that the average femoral cartilage volume is different smaller for women than for men? (The appropriate df is 20) (a)Give appropriate null and alternative hypotheses. Define your parameters exactly (b)Give a p-value and conclusion ( c)If there were reason to believe that the varia nce in femoral cartilage volume is the same for men and women then what would be the best estimate of that common variance?Explanation / Answer
4.
Given that,
mean(x)=18.7
standard deviation , s.d1=3.3
number(n1)=11
y(mean)=11.2
standard deviation, s.d2 =2.4
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /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 = (10*10.89 + 8*5.76) / (20- 2 )
s^2 = 8.61
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=18.7-11.2/sqrt((8.61( 1 /11+ 1/9 ))
to=7.5/1.319
to=5.687
| to | =5.687
critical value
the value of |t | with (n1+n2-2) i.e 18 d.f is 1.734
we got |to| = 5.687 & | t | = 1.734
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 5.6867 ) = 0.00001
hence value of p0.05 > 0.00001,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 5.687
critical value: 1.734
decision: reject Ho
b.
p-value: 0.00001
c.
common variance for both men and women s^2 = 8.61
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