6.5 Test of Hypothesis about a Population Mean: Student’s t-Statistic

Most water-treatment facilities monitor the quality of their drinking water on an hourly basis. One variable monitored is pH, which measures the degree of alkalinity or acidity in the water. A pH below 7.0 is acidic, one above 7.0 is alkaline, and a pH of 7.0 is neutral. One water-treatment plant has a target pH of 8.5. (Most try to maintain a slightly alkaline level.) The mean and standard deviation of 1 hour’s test results, based on 17 water samples at this plant, are

x=8.42s=.16

Does this sample provide sufficient evidence that the mean pH level in the water differs from 8.5?

This inference can be placed in a test-of-hypothesis framework. We establish the target pH as the null hypothesized value and then utilize a two-tailed alternative that the true mean pH differs from the target:

H0:μ=8.5(Mean pH level is 8.5.)Ha:μ8.5(Mean pH level differs from 8.5.)

Recall from Section 7.3 that when When we are faced with making inferences about a population mean from the information in a small sample, two problems emerge:

  1. The normality of the sampling distribution for x does not follow from the Central Limit Theorem when the sample size is small. We must assume that the distribution of measurements from which the sample was selected is approximately normally distributed in order to ensure the approximate normality of the sampling distribution of x.

  2. If the population standard deviation σ is unknown, as is usually the case, then we cannot assume that s will provide a good approximation for σ when the sample size is small. Instead, we must use the t-distribution rather than the standard normal z-distribution to make inferences about the population mean μ.

Therefore, as the test statistic of a small-sample test of a population mean, we use the t-statistic:

Test statistic:t=x¯μ0s/n=x¯8.5s/n

In this equation, μ0 is the null-hypothesized value of the population mean μ. In the example here, μ0=8.5.

To find the rejection region, we must specify the value of α, the probability that the test will lead to rejection of the null hypothesis when it is true, and then consult the t table (Table III of Appendix B). With α=.05, the two-tailed rejection region is

Rejection region:tα/2=t.025=2.120withn1=16degreesoffreedomRejectH0ift<2.120ort>2.120

The rejection region is shown in Figure 6.15.

Figure 6.15

Two-tailed rejection region for small-sample t-test

We are now prepared to calculate the test statistic and reach a conclusion:

t=x¯μ0s/n=8.428.50.16/17=.08.039=2.05

Since the calculated value of t does not fall into the rejection region (Figure 6.15), we cannot reject H0 at the α=.05 level of significance. Thus, the water-treatment plant should not conclude that the mean pH differs from the 8.5 target on the basis of the sample evidence.

It is interesting to note that the calculated t value, 2.05, is less than the .05-level z-value, 1.96. The implication is that if we had incorrectly used a z statistic for this test, we would have rejected the null hypothesis at the .05 level and concluded that the mean pH level differs from 8.5. The important point is that the statistical procedure to be used must always be closely scrutinized and all the assumptions understood. Many statistical lies are the result of misapplications of otherwise valid procedures.

The technique for conducting a small-sample test of hypothesis about a population mean is summarized in the following boxes:

Small-Sample Test of Hypothesis about μ Based on a Student’s t-Statistic

Test statistic: tc=(x¯μ0)(sn)
One-Tailed Tests Two-Tailed Test
H0:μ=μ0 H0:μ=μ0
Ha:μ>μ0 Ha:μμ0
Rejection region: tc>tα |tc|>tα/2
p-value: P(t>tc) 2P(t>tc) if tc is positive
2P(t<tc) if tc is negative
Decision: Reject H0 if α>p-value or if test statistic (tc) falls in rejection region where P(t>tα)=α,P(t>tα/2)=α/2,α=P(TypeIerror)=P(RejectH0|H0true), and t is based on (n1) degrees of freedom.

[Note: The symbol for the numerical value assigned to μ under the null hypothesis is μ0.]

Conditions Required for a Valid Small-Sample Hypothesis Test for μ

  1. A random sample is selected from the target population.

  2. The population from which the sample is selected has a distribution that is approximately normal.

ENGINE Example 6.8 A Small-Sample Test for μ—Does a New Engine Meet Air-Pollution Standards?

Problem

  1. A major car manufacturer wants to test a new engine to determine whether it meets new air-pollution standards. The mean emission μ of all engines of this type must be less than 20 parts per million of carbon. Ten engines are manufactured for testing purposes, and the emission level of each is determined. The data (in parts per million) are listed in Table 6.5. Do the data supply sufficient evidence to allow the manufacturer to conclude that this type of engine meets the pollution standard? Assume that the manufacturer is willing to risk a Type I error with probability α=.01.

    Table 6.5 Emission Levels for Ten Engines

    Alternate View
    15.6 16.2 22.5 20.5 16.4 19.4 19.6 17.9 12.7 14.9

    Data Set: ENGINE

Solution

  1. The manufacturer wants to support the research hypothesis that the mean emission level μ for all engines of this type is less than 20 parts per million. The elements of this small-sample one-tailed test are as follows:

    H0:μ=20(Mean emission level is 20 ppm.)Ha:μ<20(Mean emission level is less than 20 ppm.)Test statistic:t=x¯20s/n

    Rejection region: For α=.01 and df=n1=9, the one-tailed rejection region (see Figure 6.16) is t<t.01=2.821.

    Figure 6.16

    A t-distribution with 9 df and the rejection region for Example 6.8

    Assumption: The relative frequency distribution of the population of emission levels for all engines of this type is approximately normal. Based on the shape of the MINITAB stem-and-leaf display of the data shown in Figure 6.17, this assumption appears to be reasonably satisfied.

    To calculate the test statistic, we analyzed the ENGINE data with MINITAB. The MINITAB printout is shown at the bottom of Figure 6.17. From the printout, we obtain x=17.57 and s=2.95. Substituting these values into the test statistic formula and rounding, we get

    t=x¯20s/n=17.57202.95/10=2.60

    Since the calculated t falls outside the rejection region (see Figure 6.16), the manufacturer cannot reject H0. There is insufficient evidence to conclude that μ<20 parts per million and that the new type of engine meets the pollution standard.

Figure 6.17

MINITAB analysis of 10 emission levels, Example 6.8

Look Back

Are you satisfied with the reliability associated with this inference? The probability is only α=.01 that the test would support the research hypothesis if in fact it were false.

Now Work Exercise 6.57ab

Example 6.9 The p-Value for a Small-Sample Test of μ

Problem

  1. Find the observed significance level for the test described in Example 6.8. Interpret the result.

Solution

  1. The test of Example 6.8 was a lower-tailed test: H0:μ=20 versus Ha:μ<20. Since the value of t computed from the sample data was t=2.60, the observed significance level (or p-value) for the test is equal to the probability that t would assume a value less than or equal to 2.60 if in fact H0 were true. This is equal to the area in the lower tail of the t-distribution (highlighted in Figure 6.18).

    Figure 6.18

    The observed significance level for the test of Example 6.8

    One way to find this area (i.e., the p-value for the test) is to consult the t-table (Table III in Appendix B). Unlike the table of areas under the normal curve, Table III gives only the t values corresponding to the areas .100, .050, .025, .010, .005, .001, and .0005. Therefore, we can only approximate the p-value for the test. Since the observed t value was based on nine degrees of freedom, we use the df=9 row in Table III and move across the row until we reach the t values that are closest to the observed t=2.60. [Note: We ignore the minus sign.] The t values corresponding to p-values of .010 and .025 are 2.821 and 2.262, respectively. Since the observed t value falls between t.010 and t.025, the p-value for the test lies between .010 and .025. In other words, .010<p-value<.025. Thus, we would reject the null hypothesis H0:μ=20 parts per million for any value of α larger than .025 (the upper bound of the p-value).

    A second, more accurate, way to obtain the p-value is to use a statistical software package to conduct the test of hypothesis. Both the test statistic (2.60) and the p-value (.014) are highlighted on the MINITAB printout of Figure 6.17.

    You can see that the actual p-value of the test falls within the bounds obtained from Table III. Thus, the two methods agree and we will reject H0:μ=20 in favor of Ha:μ<20 for any α level larger than .025.

Now Work Exercise 6.57c

Small-sample inferences typically require more assumptions and provide less information about the population parameter than do large-sample inferences. Nevertheless, the t-test is a method of testing a hypothesis about a population mean of a normal distribution when only a small number of observations is available.

What Can Be Done If the Population Relative Frequency Distribution Departs Greatly from Normal?

Answer: Use the nonparametric statistical method described in Optional Section 6.8.

Exercises 6.51–6.72

Understanding the Principles

  1. 6.51 In what ways are the distributions of the z-statistic and t-statistic alike? How do they differ?

  2. 6.52 Under what circumstances should you use the t-distribution in testing a hypothesis about a population mean?

Learning the Mechanics

  1. 6.53 For each of the following rejection regions, sketch the sampling distribution of t and indicate the location of the rejection region on your sketch:

    1. t>1.440,wheredf=6

    2. t<1.782,wheredf=12

    3. t<2.060ort>2.060,wheredf=25

  2. 6.54 For each of the rejection regions defined in Exercise 6.53 , what is the probability that a Type I error will be made?

  3. 6.55 A random sample of n observations is selected from a normal population to test the null hypothesis that μ=10. Specify the rejection region for each of the following combinations of Ha,α, and n:

    1. Ha:μ10;α=.05;n=14

    2. Ha:μ>10;α=.01;n=24

    3. Ha:μ>10;α=.10;n=9

    4. Ha:μ<10;α=.01;n=12

    5. Ha:μ10;α=.10;n=20

    6. Ha:μ<10;α=.05;n=4

  4. 6.56 The following sample of six measurements was randomly selected from a normally distributed population: 1,3,1,5,1,2.

    1. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis, μ<3. Use α=.05.

    2. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis, μ3. Use α=.05.

    3. Find the observed significance level for each test.

  5. 6.57 A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: x=4.8,s=1.3.

    1. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, μ<6. Use α=.05.

    2. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, μ6. Use α=.05.

    3. Find the observed significance level for each test.

  6. 6.58 Suppose you conduct a t-test for the null hypothesis H0:μ=1,000 versus the alternative hypothesis Ha:μ>1,000, based on a sample of 17 observations. The test results are t=1.89,p-value=.038.

    1. What assumptions are necessary for the validity of this procedure?

    2. Interpret the results of the test.

    3. Suppose the alternative hypothesis had been the two-tailed Ha:μ1,000. If the t-statistic were unchanged, what would the p-value be for this test? Interpret the p-value for the two-tailed test.

Applying the Concepts—Basic

  1. TRAPS 6.59 Lobster trap placement. Refer to Consider the Bulletin of Marine Science (Apr. 2010) observational study of lobster trap placement by teams fishing for the red spiny lobster in Baja California Sur, Mexico, Exercise 5.41 (p. 273). Trap spacing measurements (in meters) for a sample of seven teams of red spiny lobster fishermen are reproduced in the accompanying table. Let μ represent the average of the trap spacing measurements for the population of red spiny lobster fishermen fishing in Baja California Sur, Mexico.The In Exercise 5.41 you computed the mean and standard deviation of the sample measurements to be x=89.9 meters and s=11.6 meters, respectively. Suppose you want to determine if the true value of μ differs from 95 meters.

    Alternate View
    93 99 105 94 82 70 86

    Based on Shester, G. G. “Explaining catch variation among Baja California lobster fishers through spatial analysis of trap-placement decisions.” Bulletin of Marine Science, Vol. 86, No. 2, April 2010 (Table 1), pp. 479–498.

    1. Specify the null and alternative hypotheses for this test.

    2. Since x=89.9 is less than 95, a fisherman wants to reject the null hypothesis. What are the problems with using such a decision rule?

    3. Compute the value of the test statistic.

    4. Find the approximate p-value of the test.

    5. Select a value of α, the probability of a Type I error. Interpret this value in the words of the problem.

    6. Give the appropriate conclusion, based on the results of parts d and e.

    7. What conditions must be satisfied for the test results to be valid?

    8. In Exercise 5.41Suppose you found a 95% confidence interval for μ. Does the interval support your conclusion in part f?

  2. PAI 6.60 Music performance anxiety. Refer to Consider the British Journal of Music Education (Mar. 2014) study of performance anxiety by music students, Exercise 5.37 (p. 272). Recall that theThe Performance Anxiety Inventory (PAI) was used to measure music performance anxiety on a scale from 20 to 80 points. The table below gives PAI values for participants in eight different studies. A MINITAB printout of the data analysis is shown.

    Alternate View
    54 42 51 39 41 43 55 40

    Source: Patston, T. “Teaching stage fright?—Implications for music educators.” British Journal of Music Education, Vol. 31, No. 1, Mar. 2014 (adapted from Figure 1).

    1. Set up the null and alternative hypotheses for determining whether the mean PAI value, μ, for all similar studies of music performance anxiety exceeds 40.

    2. Find the rejection region for the test, part a, using α=.05.

    3. Compute the test statistic.

    4. State the appropriate conclusion for the test.

    5. What conditions are required for the test results to be valid?

    6. Locate the p-value for the test on the MINITAB printout and use it to make a conclusion. (Your conclusion should agree with your answer in part d.)

    7. How would your conclusion change if you used α=.01?

  3. 6.61 Dental anxiety study. Refer to Consider the BMC Oral Health (Vol. 9, 2009) study of adults who completed the Dental Anxiety Scale, presented in Exercise 4.85 (p. 207). Recall that scores range from 0 (no anxiety) to 25 (extreme anxiety). Summary statistics for the scores of 15 adults who completed the questionnaire are x=10.7ands=3.6. Conduct a test of hypothesis to determine whether the mean Dental Anxiety Scale score for the population of college students differs from μ=11. Use α=.05.

  4. SPIDER 6.62 Crab spiders hiding on flowers. Refer toConsider the Behavioral Ecology (Jan. 2005) experiment on crab spiders’ use of camouflage to hide from predators (e.g., birds) on flowers, presented in Exercise 2.42 (p. 51). Researchers at the French Museum of Natural History collected a sample of 10 adult female crab spiders, each sitting on the yellow central part of a daisy, and measured the chromatic contrast between each spider and the flower. The data (for which higher values indicate a greater contrast, and, presumably, an easier detection by predators) are shown in the accompanying table. The researchers discovered that a contrast of 70 or greater allows birds to see the spider. Of interest is whether the true mean chromatic contrast of crab spiders on daisies is less than 70.

    Alternate View
    57 75 116 37 96 61 56 2 43 32

    Based on Thery, M., et al. “Specific color sensitivities of prey and predator explain camouflage in different visual systems.” Behavioral Ecology, Vol. 16, No. 1, Jan. 2005 (Table 1).

    1. Define the parameter of interest, μ.

    2. Set up the null and alternative hypotheses of interest.

    3. Find x and s for the sample data, and then use these values to compute the test statistic.

    4. Give the rejection region for α=.10.

    5. State the appropriate conclusion in the words of the problem.

  5. MOLARS 6.63 Cheek teeth of extinct primates. Consider Refer to the American Journal of Physical Anthropology (Vol. 142, 2010) study of the characteristics of cheek teeth (e.g., molars) in an extinct primate species, Exercise 2.38 (p. 50). Recall that theThe researchers recorded the dentary depth of molars (in millimeters) for a sample of 18 cheek teeth extracted from skulls. These depth measurements are listed in the accompanying table. Anthropologists know that the mean dentary depth of molars in an extinct primate species—called Species A—is 15 millimeters. Is there evidence to indicate that the sample of 18 cheek teeth come from some other extinct primate species (i.e., some species other than Species A)? Use the SPSS printout at the bottom of the page to answer the question.

    18.12 16.55
    19.48 15.70
    19.36 17.83
    15.94 13.25
    15.83 16.12
    19.70 18.13
    15.76 14.02
    17.00 14.04
    13.96 16.20

    Based on Boyer, D. M., Evans, A. R., and Jernvall, J. “Evidence of dietary differentiation among Late Paleocene–Early Eocene Plesiadapids (Mammalia, Primates).” American Journal of Physical Anthropology, Vol. 142, © 2010.

    SPSS Output for Exercise 6.63

  6. TOMBS 6.64 Radon exposure in Egyptian tombs. Refer toConsider the Radiation Protection Dosimetry (Dec. 2010) study of radon exposure in Egyptian tombs, Exercise 5.39 (p. 272). The radon levels—measured in becquerels per cubic meter (Bq/m3)—in the inner chambers of a sample of 12 tombs are listed in the table. For the safety of the guards and visitors, the Egypt Tourism Authority (ETA) will temporarily close the tombs if the true mean level of radon exposure in the tombs rises to 6,000 Bq/m3. Consequently, the ETA wants to conduct a test to determine if the true mean level of radon exposure in the tombs is less than 6,000 Bq/m3, using a Type I error probability of .10. A SAS analysis of the data is shown on p. 399. Specify all the elements of the test: H0, Ha, test statistic, p-value, α, and your conclusion.

    Alternate View
    50 910 180 580 7800 4000
    390 12100 3400 1300 11900 1100

    SAS Output for Exercise 6.64

Applying the Concepts—Intermediate

  1. 6.65 Yield strength of steel connecting bars. To protect against earthquake damage, steel beams are typically fitted and connected with plastic hinges. However, these plastic hinges are prone to deformations and are difficult to inspect and repair. An alternative method of connecting steel beams—one that uses high-strength steel bars with clamps—was investigated in Engineering Structures (July 2013). Mathematical models for predicting the performance of these steel connecting bars assume the bars have a mean yield strength of 300 megapascals (MPa). To verify this assumption, the researchers conducted material property tests on the steel connecting bars. In a sample of three tests, the yield strengths were 354, 370, and 359 MPa. Do the data indicate that the true mean yield strength of the steel bars exceeds 300 MPa? Test using α=.01.

  2. 6.66 Pitch memory of amusiacs. Amusia is a congenital disorder that adversely impacts one’s perception of music. Refer toConsider the Advances in Cognitive Psychology (Vol. 6, 2010) study of the pitch memory of individuals diagnosed with amusia, Exercise 5.45 (p. 273). Recall that eachEach in a sample of 17 amusiacs listened to a series of tone pairs and was asked to determine if the tones were the same or different. In the first trial, the tones were separated by 1 second; in a second trial, the tones were separated by 5 seconds. The difference in accuracy scores for the two trials was determined for each amusiac (where the difference is the score on the first trial minus the score on the second trial). The mean score difference was .11 with a standard deviation of .19.

    1. In theory, the longer the delay between tones, the less likely one is to detect a difference between the tones. Consequently, the true mean score difference should exceed 0. Set up the null and alternative hypotheses for testing the theory.

    2. Carry out the test, part a, using α=.05. Is there evidence to support the theory?

  3. RECALL 6.67 Free recall memory strategy. Psychologists who study memory often use a measure of “free recall” (e.g., the number of correctly recalled items in a list of to-be-­remembered items). The strategy used to memorize the list—for example, category clustering—is often just as important. Researchers at Central Michigan University developed an algorithm for computing measures of category clustering in Advances in Cognitive Psychology (Oct. 2012). One measure, called ratio of repetition, was recorded for a sample of 8 participants in a memory study. These ratios are listed in the table. Test the theory that the average ratio of repetition for all participants in a similar memory study differs from .5. Select an appropriate Type I error rate for your test.

    Alternate View
    .25 .43 .57 .38 .38 .60 .47 .30

    Source: Senkova, O., & Otani, H. “Category clustering calculator for free recall.” Advances in Cognitive Psychology, Vol. 8, No. 4, Oct. 2012 (Table 3).

  4. 6.68 Increasing hardness of polyester composites. Polyester resins reinforced with fiberglass are used to fabricate wall panels of restaurants. It is theorized that adding cement kiln dust (CKD) to the polyester composite will increase wall panel hardness. In a study published in Advances in Applied Physics (Vol. 2, 2014), hardness (joules per squared centimeters) was determined for three polyester composite mixtures that used a 40% CKD weight ratio. The hardness values were reported as 83, 84, and 79 j/cm2. Research has shown that the mean hardness value of polyester composite mixtures that use a 20% CKD weight ratio is μ=76j/cm2. In your opinion, does using a 40% CKD weight ratio increase the mean hardness value of polyester composite mixtures? Support your answer statistically.

  5. SKID 6.69 Minimizing tractor skidding distance. Refer toConsider the Journal of Forest Engineering (July 1999) study of minimizing tractor skidding distances along a new road in a European forest, presented in Exercise 5.48 (p. 274). The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are repeated in the accompanying table. Recall that aA logger working on the road claims that the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use α=.10.

    Alternate View
    488 350 457 199 285 409 435 574 439 546
    385 295 184 261 273 400 311 312 141 425

    Based on Tujek, J., and Pacola, E. “Algorithms for skidding distance modeling on a raster Digital Terrain Model.” Journal of Forest Engineering, Vol. 10, No. 1, July 1999 (Table 1).

  6. LAKES 6.70 Dissolved organic compound in lakes. The level of dissolved oxygen in the surface water of a lake is vital to maintaining the lake’s ecosystem. Environmentalists from the University of Wisconsin monitored the dissolved oxygen levels over time for a sample of 25 lakes in the state (Aquatic Biology, May 2010). To ensure a representative sample, the environmentalists focused on several lake characteristics, including dissolved organic compound (DOC). The DOC data (measured in grams per cubic meters) for the 25 lakes are listed in the table on p. 400. The population of Wisconsin lakes has a mean DOC value of 15grams/m3. Use a hypothesis test (at α=.10) to make an inference about whether the sample is representative of all Wisconsin lakes for the characteristic dissolved organic compound.

    Data for Exercise 6.70

    Lake Doc
    Allequash 9.6
    Big Muskellunge 4.5
    Brown 13.2
    Crampton 4.1
    Cranberry Bog 22.6
    Crystal 2.7
    EastLong 14.7
    Helmet 3.5
    Hiawatha 13.6
    Hummingbird 19.8
    Kickapoo 14.3
    Little Arbor Vitae 56.9
    Mary 25.1
    Muskellunge 18.4
    Northgate Bog 2.7
    Paul 4.2
    Peter 30.2
    Plum 10.3
    Reddington Bog 17.6
    Sparkling 2.4
    Tenderfoot 17.3
    Trout Bog 38.8
    Trout Lake 3.0
    Ward 5.8
    West Long 7.6

    Based on Langman, O. C., et al. “Control of dissolved oxygen in northern temperate lakes over scales ranging from minutes to days.” Aquatic Biology, Vol. 9, May 2010 (Table 1).

  7. CIRCLES 6.71 Walking straight into circles. When people get lost in unfamiliar terrain, do they really walk in circles, as is commonly believed? To answer this question, researchers conducted a field experiment and reported the results in Current Biology (Sept. 29, 2009). Fifteen volunteers were blindfolded and asked to walk as straight as possible in a certain direction in a large field. Walking trajectories were monitored every second for 50 minutes using GPS and the average directional bias (degrees per second) recorded for each walker. The data are shown in the table below. A strong tendency to veer consistently in the same direction will cause walking in circles. A mean directional bias of 0 indicates that walking trajectories were random. Consequently, the researchers tested whether the true mean bias differed significantly from 0. A MINITAB printout of the analysis is shown below.

    1. Interpret the results of the hypothesis test for the researchers. Use α=.10.

    2. Although most volunteers showed little overall bias, the researchers produced maps of the walking paths showing that each occasionally made several small circles during the walk. Ultimately, the researchers supported the “walking in circles” theory. Explain why the data in the table are insufficient for testing whether an individual walks in circles.

    Alternate View
    4.50 1.00 0.50 0.15 0.00 0.01 0.02 0.05 0.15
    0.20 0.50 0.50 1.00 2.00 3.00

    Based on Souman, J. L., Frissen, I., Sreenivasa, M. N., and Ernst, M. O. “Walking straight into circles.” Current Biology, Vol. 19, No. 18, Sept. 29, 2009 (Figure 2).

    MINITAB Output for Exercise 6.71

Applying the Concepts—Advanced

  1. SHARK 6.72 Lengths of great white sharks. One of the most feared predators in the ocean is the great white shark. It is known that the white shark grows to a mean length of 21 feet; however, one marine biologist believes that great white sharks off the Bermuda coast grow much longer owing to unusual feeding habits. To test this claim, some full-grown great white sharks were captured off the Bermuda coast, measured, and then set free. However, because the capture of sharks is difficult, costly, and very dangerous, only three specimens were sampled. Their lengths were 24, 20, and 22 feet. Do these data support the marine biologist’s claim at α=.10?

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