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	1.337	1.076	1.226	0.938	1.006	1.178	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=

(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.)

(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

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.

We have given the dataset of 8 observations.

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.)

Difference = Operator1 - Operator2

And we have to find sample mean and sample standard deviation of the difference.

We can do this in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> Display descriptive statistics --> Variables : Operator1-Operator2 --> Statistics --> Select mean and standard deviation --> ok --> ok

Descriptive Statistics: operator1-operator2

Variable Mean StDev
operator1-operator2 -0.00175 0.01021

x= -0.00175

s=0.01021

(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.)

Here we have to test the hypothesis that,

H0 : mu1 = mu2 Vs H1 : mu1 not= mu2

where mu1 : population mean for operator1

mu2 is population mean for operator2.

Assume alpha = level of significance = 0.05

Here sample size is of 8 which is too small.

So we will use Two sample t-test assuming equal variances.

We can do this test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 2-Sample t --> Each sample is in its own column --> Sample1 : Operator1 --> Sample2 : Operator 2 --> Options --> COnfidence level : 95.0 --> Hypothesized difference : 0.0 --> Alternative hypothesis: not equal --> Assume equal variances --> Ok --> OK

Two-Sample T-Test and CI: Operator1, Operator2

Two-sample T for Operator1 vs Operator2

N Mean StDev SE Mean
Operator1 8 1.172 0.151 0.053
Operator2 8 1.174 0.149 0.053

Difference = μ (Operator1) - μ (Operator2)
Estimate for difference: -0.0018
95% CI for difference: (-0.1629, 0.1594)
T-Test of difference = 0 (vs ≠): T-Value = -0.02 P-Value = 0.982 DF = 14
Both use Pooled StDev = 0.1503

t =-0.02

P-value = 0.982

DF = 14

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that there is no difference between operator1 and operator2.

(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.)

Now we have to find confidence interval for mu1-mu2.

We can find confidence interval in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> One or more samples each in a column --> --> Variables : select difference variable --> Perform hypothesis test --> Hypothesized mean : 0 --> Options --> COnfidence level : 95.0 --> Alternative hypothesis : not equal --> ok --> ok

One-Sample T: operator1-operator2

Test of μ = 0 vs ≠ 0

Variable N Mean StDev SE Mean 95% CI T P
operator1-operator2 8 -0.00175 0.01021 0.00361 (-0.01028, 0.00678) -0.48 0.643

95% confidence interval for population mean difference is (-0.01028, 0.00678).