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	1.328	1.340	1.075	1.228	0.937	1.008	1.180	1.288
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 right skewed. The distribution is Normal.     The distribution is left skewed. The distribution is uniform. The sample is too small to make judgments about skewness or symmetry.

(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%.)

We can reject H₀ based on this sample. 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.)

Answer 1

(a)

Following table shows the calculations:

Operator 1	Operator 2	d=operator1-operator 2	(d-mean)^2
1.328	1.323	0.005	3.6E-05
1.34	1.322	0.018	0.000361
1.075	1.073	0.002	9E-06
1.228	1.233	-0.005	1.6E-05
0.937	0.934	0.003	0.000016
1.008	1.019	-0.011	1E-04
1.18	1.184	-0.004	9E-06
1.288	1.304	-0.016	0.000225
Total		-0.008	0.000772

Sample size: n=8

Now,

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

(b)

Hypotheses are:

The test statistics will be

Degree of freedom: df=n-1=7

The P-value of the test is: 0.7949

We cannot reject H0 based on this sample.

(c)

For 95% confidence interval t critical value for df=7 is 2.4314. So required confidence interval will be

So required confidence interval is (-0.0100, 0.0080).