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.

	Subject
Operator	1	2	3	4	5	6	7	8
1	1.329	1.339	1.076	1.228	0.938	1.005	1.180	1.289
2	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. (Use Operator 1 minus Operator 2. Round your answers to four decimal places.)

x=

s=

Describe the distribution of these differences using words.

The distribution is left skewed.

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

The distribution is uniform.

The distribution is right skewed.

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 =

Give the degrees of freedom.

Give the P-value. (Round your answer to four decimal places.)

Give your conclusion. (Use the significance level of 5%.)   

(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.)

,

Answer 1

a)

S. No	Operator 1	Operator 2	diff:(d)=x1-x2	d2
1	1.329	1.323	0.006	0.00
2	1.339	1.322	0.017	0.00
3	1.076	1.073	0.003	0.00
4	1.228	1.233	-0.005	0.00
5	0.938	0.934	0.004	0.00
6	1.005	1.019	-0.014	0.00
7	1.18	1.184	-0.004	0.00
8	1.289	1.304	-0.015	0.00
	total	=	Σd=-0.008	Σd2=0.0008
	mean dbar=	d̅     =	-0.0010
	degree of freedom =n-1                            =		7
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		0.010717

from above:

x =-0.0010

s =0.0107

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

b)

std error=Se=SD/√n=		0.0038
test statistic            =	(d̅-μd)/Se         =	-0.264

degrees of freedom =7

p value

=

0.3997

from excel: tdist(0.264,7,1)

since test statistic does not falls in rejection region we fail to reject null hypothesis

we do not have have sufficient evidence to conclude that one of the operators is producing measurements that are consistently higher than the other.

c)

for 95% CI; and 7 degree of freedom, value of t=			2.365
therefore confidence interval=sample mean -/+ t*std error
margin of errror          =t*std error=		0.008961
lower confidence limit                     =		-0.0100
upper confidence limit                    =		0.0080
from above 95% confidence interval for population mean =(-0.0100 ,0.0080)