012 point Dual energy X ay absorptiometry DXA is technique or measuring bone hea
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012 point Dual energy X ay absorptiometry DXA is technique or measuring bone health One of the most common meas es is total body bonc mineral content TBB, C). A highly skilled operator is equired to take the measurements. Recently, a new DXA machine as purchased by research lab and bo operators were trained to take the measurements. T BMC for eight subjects w as measured by both operators The units 9 ams 9). A co o son f the means for the bo operators provides check on tht train ng they receive and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Herc are the dato; Subject Operator 1 1.328 1.336 1.078 1.228 0.939 1.009 1.178 1.289 1.323 1.322 1.073 1.233 0.934 1.019 1.184 1.304 (a) Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. (Usc Operator 1 minus Operator 2. Round your answers to four decimal places.) Describe the distribution ofthese differences u words. The sarnple is too small to make judgments about skewness or symmetry. The distribution is Icft skewed. The distribution is Normal The distribution is right skewed The distribution is uniform. (b) Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.) Give the degrees of freedom Give the Pivalue- (Round your enswer to four decimal places.) Give We can reject Ho based on this sample. We cannot reject Ho based an this sample (c Th¢ sample here is rather small, so ve may not haver uch po er to detect differences 4f interest. Use 95% confidence interva to provide ar ge of differences that re compottle with these data. Round your answers to fur deci al places (d)The eight subjects used for this comparison were not a random sample. In fact, they were friends of the researchers whose ages and weights were similar to the types of people who would be measured with this DXA. Comment on the appropriateness of this procedure for selecting a sample, and discuss any consequences regarding the interpretation of the significance testing and confidence interval results The subjects from this sample, test results, and confidence interval are representative of future subjects. The subjects from this sampie may be repreentative of future subjects, but the tet results and confidence interval are suspect because this is not a random samplcExplanation / Answer
a.
mean(x)=1.173
standard deviation , s.d1=0.1503
number(n1)=8
y(mean)=1.174
standard deviation, s.d2 =0.1495
number(n2)=8
distribution is normal
b.
Given that,
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.173-1.174/sqrt((0.02259/8)+(0.02235/8))
to =-0.0133
| to | =0.0133
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.895
we got |to| = 0.01334 & | t | = 1.895
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.0133 ) = 0.50514
hence value of p0.05 < 0.50514,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -0.0133
critical value: 1.895
decision: do not reject Ho
p-value: 0.50514
c.
TRADITIONAL METHOD
given that,
mean(x)=1.173
standard deviation , s.d1=0.1503
number(n1)=8
y(mean)=1174
standard deviation, s.d2 =0.1495
number(n2)=8
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((0.0226/8)+(0.0224/8))
= 0.075
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.3646
margin of error = 2.365 * 0.075
= 0.1773
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1.173-1174) ± 0.1773 ]
= [-1173.0043 , -1172.6497]
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DIRECT METHOD
given that,
mean(x)=1.173
standard deviation , s.d1=0.1503
sample size, n1=8
y(mean)=1174
standard deviation, s.d2 =0.1495
sample size,n2 =8
CI = x1 - x2 ± t 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)
ta/2 = t-table value
CI = confidence interval
CI = [( 1.173-1174) ± t a/2 * sqrt((0.0226/8)+(0.0224/8)]
= [ (-1172.827) ± t a/2 * 0.075]
= [-1173.0043 , -1172.6497]
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d.
interpretations:
1. we are 95% sure that the interval [-1173.0043 , -1172.6497] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
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