Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health.

Question

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body bone mineral content (TBBMC). A highly skilled operator is required to take the measurements. Recently, a new DXA machine was purchased by a research lab and two operators were trained to take the measurements. TBBMC for eight subjects was measured by both operators. The units are grams (g). A comparison of the means for the two operators provides a check on the training they received and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Here are the data:

(a) Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. (Use Operator 1 minus Operator 2. Round your answers to four decimal places.)

x bar =

s =

Describe the distribution of these differences using words (choose one):

a. The sample is too small to make judgments about skewness or symmetry.

b. The distribution is right skewed.

c. The distribution is uniform.

d. The distribution is left skewed.

e. The distribution is Normal.

(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.)

t =

df =

P-value =

Give your conclusion:

a. We can reject H0 based on this sample.

b. We cannot reject H0 based on this sample.

(c) The sample here is rather small, so we may not have much power to detect differences of interest. Use a 95% confidence interval to provide a range of differences that are compatible with these data. (Round your answers to four decimal 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:

a. The subjects from this sample may be representative of future subjects, but the test results and confidence interval are suspect because this is not a random sample.

b. The subjects from this sample, test results, and confidence interval are representative of future subjects.

Subject Operator 1 2 3 4 5 6 7 8 1 1.326 1.335 1.076 1.225 0.938 1.006 1.179 1.289 2 1.323 1.322 1.073 1.233 0.934 1.019 1.184 1.304

Explanation / Answer

Given that,
mean(x)=1.1718
standard deviation , s.d1=0.1504
number(n1)=8
y(mean)=1.174
standard deviation, s.d2 =0.1495
number(n2)=8
null, Ho: u1 > u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.895
since our test is left-tailed
reject Ho, if to < -1.895
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.1718-1.174/sqrt((0.02262/8)+(0.02235/8))
to =-0.029
| to | =0.029
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.895
we got |to| = 0.02934 & | t | = 1.895
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:left tail - Ha : ( p < -0.0293 ) = 0.48871
hence value of p0.05 < 0.48871,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 > u2
alternate, H1: u1 < u2
test statistic: -0.029
critical value: -1.895
decision: do not reject Ho
p-value: 0.48871
c.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=1.1718
Standard deviation( sd1 )=0.1504
Sample Size(n1)=8
Mean(x2)=1.174
Standard deviation( sd2 )=0.1495
Sample Size(n2)=8
CI = [ ( 1.1718-1.174) ±t a/2 * Sqrt( 0.02262016/8+0.02235025/8)]
= [ (-0.0022) ± t a/2 * Sqrt( 0.0056) ]
= [ (-0.0022) ± 2.365 * Sqrt( 0.0056) ]
= [-0.1795 , 0.1751]

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