A study was conducted to compare musculoskeletal ultrasonography in healthy adul

Question

A study was conducted to compare musculoskeletal ultrasonography in healthy adults and adults with osteoporosis For an independent ran- dom sample of 18 healthy women, sagittal diameter (in mm) of the biceps tendon was taken and resulted in sample statistics 2.45 and sr = 0.492. For an independent random sample of 34 women with osteoporosis, the equivalent sample statistics were y = 2.18 and Sy = 0.318. (a) Ignore the sample standard deviations and assume it is known that = 0.25 and -0.1. Assuming both samples are obtained from normal populations, construct a 90% confidence interval for the difference in mean sagittal diameter. (b) Assume it is known that the samples are obtained from normal populations and have a common (but unknown) variance. Using the sample standard deviations, calculate a pooled estimate of variance for the common population variance and construct a 99% confidence interval for the difference in mean sagittal diameter.

Explanation / Answer

PART A.
given that,
mean(x)=2.45
standard deviation , 1 =0.5
population size(n1)=18
y(mean)=2.18
standard deviation, 2 =0.3162
population size(n2)=34
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.25/18)+(0.09998/34))
= 0.1297
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.1297
= 0.2134
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (2.45-2.18) ± 0.2134 ]
= [0.0566 , 0.4834]
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interpretations:
1. we are 90% sure that the interval [0.0566 , 0.4834] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.1 true mean
difference is zero

PART B.
given that,
mean(x)=2.45
standard deviation , s.d1=0.492
number(n1)=18
y(mean)=2.18
standard deviation, s.d2 =0.318
number(n2)=34
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (17*0.242 + 33*0.101) / (52- 2 )
s^2 = 0.149
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 0.149 * (1/18+1/34) )
=0.113
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.01
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 50 d.f is 2.678
margin of error = 2.678 * 0.113
= 0.301
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (2.45-2.18) ± 0.301 ]
= [-0.031 , 0.571]

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