In: Statistics and Probability
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 | 1.336 | 1.074 | 1.229 | 0.938 | 1.005 | 1.180 | 1.286 |
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 = _______ (x bar equals)
s = _______
Describe the distribution of these differences using words.
The sample is too small to make judgments about skewness or symmetry.
The distribution is Normal.
The distribution is right skewed.
The distribution is left 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.)
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 cannot reject H0 based on this sample.
We can 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 machine. 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 may be representative of future subjects, but the test results and confidence interval are suspect because this is not a random sample.
The subjects from this sample, test results, and confidence interval are representative of future subjects.
Operator 1 | Operator 2 | Difference |
1.328 | 1.323 | 0.005 |
1.336 | 1.322 | 0.014 |
1.074 | 1.073 | 0.001 |
1.229 | 1.233 | -0.004 |
0.938 | 0.934 | 0.004 |
1.005 | 1.019 | -0.014 |
1.180 | 1.184 | -0.004 |
1.286 | 1.304 | -0.018 |
a) Sample mean of the difference using excel function AVERAGE(), x̅d = -0.0020
Sample standard deviation of the difference using excel function STDEV.S() sd = 0.0104
Sample size, n = 8
The distribution is Normal.
b) Null and Alternative hypothesis:
Ho : µd = 0
H1 : µd ≠ 0
Test statistic:
t = (x̅d)/(sd/√n) = (-0.002)/(0.0104/√8) = -0.544
df = n-1 = 7
p-value :
Two tailed p-value = T.DIST.2T(ABS(-0.5436), 7) = 0.6036
Decision:
p-value > α, Do not reject the null hypothesis
Conclusion:
We cannot reject H0 based on this sample.
c) 95% Confidence interval :
At α = 0.05 and df = n-1 = 7, two tailed critical value, t-crit = T.INV.2T(0.05, 7) = 2.365
Lower Bound = x̅d - t-crit*sd/√n = -0.002 - 2.365 * 0.0104/√8 = -0.0107
Upper Bound = x̅d + t-crit*sd/√n = -0.002 + 2.365 * 0.0104/√8 = 0.0067
-0.0107 < µd < 0.0067
d) 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.