2. A study was conducted to compare musculoskeletal ultrasonography in healthy a

Question

2. 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 x = 2.45 and sx-0492. For an independent random sample of 34 women with osteoporosis, the equivalent sample statistics were 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
number(n1)=18
y(mean)=2.18
standard deviation, 2 =0.3162
number(n2)=34
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 2.45-2.18) ±Z a/2 * Sqrt( 0.25/18+0.09998244/34)]
= [ (0.27) ± Z a/2 * Sqrt( 0.0168) ]
= [ (0.27) ± 1.645 * Sqrt( 0.0168) ]
= [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
sample size, n1=18
y(mean)=2.18
standard deviation, s.d2 =0.318
sample size,n2 =34
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 2.45-2.18) ± t a/2 * sqrt( 0.149 * (1/18+1/34) ]
= [ (0.27) ± 0.301 ]
= [-0.031 , 0.571]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-0.031 , 0.571]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

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