A study was conducted to determine standard reference values for musculoskeletal

Question

A study was conducted to determine standard reference values for musculoskeletal ultrasonography in healthy adults. Independed random samples of men and women were obtained and the sagittal diameter (in mm) of the biceps tendon was table measured in each subject. The resulting summary statistics are given in the following table. Assume the underlying populations are normal, with equal variances. a. Is there any evidence to suggest that the population mean sagittal diameter of women's biceps tendons is different from that of men's biceps tendons? Use alpha = 0.01. b. Construct a 95% confidence interval for the difference in population mean sagittal diameters, mu_1 = mu_2.

Explanation / Answer

a.

Given that,
mean(x)=2.5
standard deviation , s.d1=0.49
number(n1)=54
y(mean)=2.8
standard deviation, s.d2 =0.49
number(n2)=48
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.626
since our test is two-tailed
reject Ho, if to < -2.626 OR if to > 2.626
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (53*0.2401 + 47*0.2401) / (102- 2 )
s^2 = 0.2401
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=2.5-2.8/sqrt((0.2401( 1 /54+ 1/48 ))
to=-0.3/0.0972
to=-3.0863
| to | =3.0863
critical value
the value of |t | with (n1+n2-2) i.e 100 d.f is 2.626
we got |to| = 3.0863 & | t | = 2.626
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -3.0863 ) = 0.0026
hence value of p0.01 > 0.0026,here we reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.0863
critical value: -2.626 , 2.626
decision: reject Ho
p-value: 0.0026

b.
CI = x1 - x2 ± t a/2 * Sqrt(S^2(1/n1+1/n2))
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=2.5
Standard deviation( sd1 )=0.49
Sample Size(n1)=54
Mean(x2)=2.8
Standard deviation( sd2 )=0.49
Sample Size(n2)=48
S^2 = (53*0.2401 + 47*0.2401) / (102- 2 )
S^2 = 0.2401
CI = [ ( 2.5-2.8) ± t a/2 * 0.49 Sqrt( 1/54+1/48)]
= [ (-0.3) ± t a/2 * 0.49 * Sqrt( 0.03935185) ]
= [ (-0.3) ± 2.00099538 * 0.49 * Sqrt( 0.03935185) ]
= [ (-0.49450231 , -0.10549769 ]

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