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Chapter 9: Q39 E (page 389)

Scientists and engineers frequently wish to compare two different techniques for measuring or determining the value of a variable. In such situations, interest centers on testing whether the mean difference in measurements is zero. The article "Evaluation of the Deuterium Dilution Technique Against the Test Weighing Procedure for the Determination of Breast Milk Intake" (Amer: J. of Clinical Nutr., \(1983: 996 - 1003\)) reports the accompanying data on amount of milk ingested by each of\(14\)randomly selected infants.

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;\;\;Infant\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;\;\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;5}\\{D method\;\;\;\;\;\;\;\;\;\;\;1509\;\;\;\;\;1418\;\;\;\;\;1561\;\;\;\;\;1556\;\;\;\;\;2169}\\{W method\;\;\;\;\;\;\;\;\;\;1498\;\;\;\;\;1254\;\;\;\;\;1336\;\;\;\;\;1565\;\;\;\;\;2000}\\{jifference\;\;\;\;\;\;\;\;\;\;\;\;11\;\;\;\;\;\;\;\;\;\;164\;\;\;\;\;\;\;225\;\;\;\;\;\;\; - 9\;\;\;\;\;\;\;\;\;\;169}\end{array}\)

\(\begin{array}{*{20}{c}}{}&{Infant}&{}&{}&{}&{}\\{}&6&7&8&9&{10}\\{DD method}&{1760}&{1098}&{1198}&{1479}&{1281}\\{TW method}&{1318}&{1410}&{1129}&{1342}&{1124}\\{Difference}&{442}&{ - 312}&{69}&{137}&{157}\end{array}\)

\(\begin{array}{*{20}{c}}{}&{Infant}&{}&{}&{}\\{}&{11}&{12}&{13}&{14}\\{DD method}&{1414}&{1954}&{2174}&{2058}\\{TW method}&{1468}&{1604}&{1722}&{1518}\\{Difference}&{ - 54}&{350}&{452}&{540}\end{array}\)

a. Is it plausible that the population distribution of differences is normal?

b. Does it appear that the true average difference between intake values measured by the two methods is something other than zero? Determine the\(P\)-value of the test, and use it to reach a conclusion at significance level . \(05\).

Short Answer

Expert verified

Reject null hypothesis.

Step by step solution

01

Step 1:

The Infant data is given in the following table:

\(\begin{array}{*{20}{c}}i&{{\rm{ DD method, }}{x_i}}&{{\rm{ TW method, }}{y_i}}&{{\rm{ Difference, }}{d_i} = {x_i} - {y_i}}\\1&{1509}&{1498}&{11}\\2&{1418}&{1254}&{164}\\3&{1561}&{1336}&{225}\\4&{1556}&{1565}&{ - 9}\\5&{2169}&{2000}&{169}\\6&{1760}&{1318}&{442}\\7&{1098}&{1410}&{ - 312}\\8&{1198}&{1129}&{69}\\9&{1479}&{1342}&{137}\\{10}&{1281}&{1124}&{157}\\{11}&{1414}&{1468}&{ - 54}\\{12}&{1954}&{1604}&{350}\\{13}&{2174}&{1722}&{452}\\{14}&{2058}&{1518}&{540}\\{}&{}&{}&{}\end{array}\)

02

 Step 1: Mapping the graph

The normal probability plot indicated that the differences are normally distributed. It is plausible that the population distribution of differences is normal.

03

b)Step 2: Find the test statistic value

Because the experiment performed on the same\(14\)randomly selected infants, the pair\(t\)test should be used.

The Paired\(t\)Test:

When\(D = X - Y\)(difference between observations within a pair),\({\mu _D} = {\mu _1} - {\mu _2}\)

and null hypothesis

\({H_0}:{\mu _D} = {\Delta _0},\)

the test statistic value for testing the hypotheses is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }}\)

where \(\bar d\) and\({s_D}\) are the sample mean and the sample standard deviation of differences\({d_i}\), respectively. In order to use this test, assume that the differences \({D_i}\) are from a normal population. Depending on alternative hypothesis, the \(P\) value can be determined as the corresponding area under the \({t_{n - 1}}\)curve.

04

Find the mean and standard deviation

The hypotheses of interest are\({H_o}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} \ne 0\). The value of test statistic can be obtained when the sample mean and the sample standard deviation of the differences is found.

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean\(\bar d\)for the differences is

\(\bar d = \frac{1}{{14}} \cdot (11 + 164 + \ldots + 540) = 167.2\)

The delta value is

\({\Delta _0} = 0\)

The only missing value to compute the test statistic value is the sample standard deviation of differences\({d_i}\).

The Sample Variance\({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation\(s\)is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{14 - 1}} \cdot \left( {{{(11 - 167.2)}^2} + {{(164 - 167.2)}^2} + \ldots + {{(540 - 167.2)}^2}} \right)\\ = 52,080.18\end{array}\)

and the sample standard deviation\({s_D} = \sqrt {52,080.18} = 228\)

05

Find the test statistic value

The test statistic value is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{167.2 - 0}}{{228/\sqrt {14} }} = 2.74\)

The test is two-sided where the alternative hypothesis is\({H_a}:{\mu _D} \ne 0\), therefore the\(P\)value is two times the area under the the\({t_{n - 1}} = {t_{14 - 1}} = \$ \$ {t_1}3\)curve to the right of\(|t|\)value

\(P = 2 \cdot P(T > 2.74) = 2 \cdot 0.009 = 0.018\)

where\(T\)has student distribution with\(13\)degrees of freedom, and the probability as computed using software (you can use table in the appendix of the book). At significance level\(0.05\), because

\(P = 0.018 < 0.05\)

reject null hypothesis

at given significance level. There is enough evidence to conclude that the true average difference between the values of the two methods is different than \(0.\)

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Most popular questions from this chapter

Persons having Reynaud’s syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output \((cal/c{m^2}/min)\) was measured. For \(m = 10\) subjects with the syndrome, the average heat output was \(\bar x = .64\), and for \(n = 10\) non-sufferers, the average output was \(2.05\). Let \({\mu _1}\) and \({\mu _2}\) denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with \({\sigma _1} = .2\) and \({\sigma _2} = .4\).

a. Consider testing \({H_0}:{\mu _1} - {\mu _2} = - 1.0\) versus \({H_2}:{\mu _1} - {\mu _2} < - 1.0\)at level . \(01\). Describe in words what \({H_a}\) says, and then carry out the test.

b. What is the probability of a type II error when the actual difference between \({\mu _1}\) and \({\mu _2}\) is \({\mu _1} - {\mu _2} = - 1.2?\)

c. Assuming that \(m = n\), what sample sizes are required to ensure that \(\beta = .1\) when \({\mu _1} - {\mu _2} = - 1.2?\)

Teen Court is a juvenile diversion program designed to circumvent the formal processing of first-time juvenile offenders within the juvenile justice system. The article "An Experimental Evaluation of Teen Courts" (J. of Experimental Criminology, 2008: 137-163) reported on a study in which offenders were randomly assigned either to Teen Court or to the traditional Department of Juvenile Services method of processing. Of the \(56TC\) individuals, 18 subsequently recidivated (look it up!) during the 18 -month follow-up period, whereas 12 of the 51 DJS individuals did so. Does the data suggest that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so? State and test the relevant hypotheses using a significance level of 0.10.

The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of \(43\) chronic schizophrenics, resulting in \(\bar x = 2.69\) and \({s_1} = 2.30,\), as well as for \(45\) normal subjects, resulting in \(\bar y = 6.35\) and \({s_2} = 4.03.\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha = .01\)

. (Hint: \({H_0}\) and \({H_a}\) here have a different form from the three standard cases. Let \({\mu _1}\) and \({\mu _2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta = 2{\mu _1} - {\mu _2}\). Write \({H_0}\) and \({H_a}\) in terms of \(\theta \), estimate \(\theta \), and derive \({\hat \sigma _{\tilde \theta }}\) (“Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients,” Nature, July 28, 1972: 225–226).) \(\alpha = .01\)

The article "Evaluating Variability in Filling Operations" (Food Tech., 1984: 51-55) describes two different filling operations used in a ground-beef packing plant. Both filling operations were set to fill packages with \(1400\;g\)of ground beef. In a random sample of size 30 taken from each filling operation, the resulting means and standard deviations were \(1402.24\;g\) and \(10.97\;g\) for operation 1 and \(1419.63\;g\) and \(9.96\;g\) for operation 2.

a. Using a .05 significance level, is there sufficient evidence to indicate that the true mean weight of the packages differs for the two operations?

b. Does the data from operation 1 suggest that the true mean weight of packages produced by operation 1 is higher than\(1400\;g\)? Use a \(.05\)significance level.

McNemar's test, developed in Exercise 56, can also be used when individuals are paired (matched) to yield n pairs and then one member of each pair is given treatment 1 and the other is given treatment 2 . Then \({X_1}\)is the number of pairs in which both treatments were successful, and similarly for\({H_0}\)\({X_2},{X_3}\), and\({X_4}\). The test statistic for testing equal efficacy of the two treatments is given by\(\left( {{X_2} - {X_3}} \right)/\sqrt {\left( {{X_2} + {X_3}} \right)} \), which has approximately a standard normal distribution when \({H_0}\)is true. Use this to test whether the drug ergotamine is effective in the treatment of migraine headaches.

The data is fictitious, but the conclusion agrees with that in the article "Controlled Clinical Trial of Ergotamine Tartrate" (British Med, J., 1970: 325-327).
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