The article “Nutrient Deprivation Improves Field Performance of Woody Seedlings in a Degraded Semi-arid Shrub-land” (R. Trubata, J. Cortina, and A. Vilagrosaa, Ecological Engineering. 2011:1164–1173) presents a study that investigated the effect of varying the; type of fertilizer on the height of certain Mediterranean woody tree species. In one experiment, three samples, each consisting of ten trees, were grown with three different fertilizers. One, the control group, was grown with a standard fertilizer. Another was grown with a fertilizer containing only half the nutrients of the standard fertilizer. The third was grown with the standard fertilizer to which a commercial slow-release fertilizer had been added. Following are the heights of the trees after one year. These data are consistent with the means and standard deviations reported in the article.
- a. Construct an ANOVA table. You may give a
range for the P-value. - b. Can you conclude that the heights differ among the types of fertilizer?
Want to see the full answer?
Check out a sample textbook solutionChapter 9 Solutions
STATISTICS FOR ENGINEERS+SCI.-ACCESS
Additional Math Textbook Solutions
Introductory Statistics (2nd Edition)
Business Analytics
Statistical Techniques in Business and Economics
Essentials of Statistics (6th Edition)
Introduction to Statistical Quality Control
- An article in the Journal of Quality Technology (Vol. 13, No. 2, 1981, pp. 111–114) describes an experimentthat investigates the effects of four bleaching chemicals on pulp brightness. These four chemicals wereselected at random from a large population of potential bleaching agents. The data are as follows:a. Test the significance of these chemical types with α=0.05.b. If proven significant, perform a multiple comparison method using Fisher’s LSDarrow_forwardA random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and on 18 of these helmets some damage was observed.arrow_forwardFollowing are the protein contents measured in two types of species:Species 1: 0.72 1.12 0.81 0.89 0.72 0.81 1.01 0.75 0.83Species 2: 1.21 0.93 0.80 1.12 1.22 0.94 0.87 i) Assuming normality, test the hypothesis that the two species have the sameaverage protein contents by using 5-step hypothesis testing procedure at 5 %level of significance, and using the critical values approach.ii) Calculate the p-value of this test and make decision.iii) Write down the standard error of this test and calculate its numerical value ?arrow_forward
- Suppose a researcher is interested inthe effectiveness in a new childhood exercise program implemented in a SRS of schools across a particular county. In order to test the hypothesis that the new program decreases BMI (Kg/m2), the researcher takes a SRS of children from schools where the program is employed and a SRS from schools that do not employ the program and compares the results. Assume the following table represents the SRSs of students and their BMIs. Student intervention group BMI (kg/m2) Student control group BMI (kg/m2) A 18.6 A 21.6 B 18.2 B 18.9 C 19.5 C 19.4 D 18.9 D 22.6 E 24.1 F 23.6 A) Assuming that all the necessary conditions are met (normality, independence, etc.) carry out the appropriate statistical test to determine if the new exercise program is effective. Use an alpha level of 0.05. Do not assume equal variances.B) Construct a 95% confidence interval about your estimate for the average difference in BMI between the groups.arrow_forward6.4.10. In a study of factors thought to be responsible for the adverse effects of smoking on human reproduction, cadmium level determinations (nanograms per gram) were made on placenta tissue of a sample of 14 mothers who were smokers and an independent random sample of 18 nonsmoking mothers. The results were as follows: Nonsmokers: 10.0.8.4. 12.8, 25.0, 11.8.9.8. 12.5, 15.4.23.5, 9.4. 25.1, 19.5. 25.5.9.8. 7.5, 11.8, 12.2, 15.0 Smokers: 30.0, 30.1, 15.0, 24.1, 30.5, 17.8, 16.8, 14.8, 13.4, 28.5, 17.5, 14.4, 12.5, 20.4 Does it appear likely that the mean cadmium level is higher among smokers than nonsmokers? Why do you reach this conclusion?arrow_forwardThe article “The Effects of a Low-Fat, Plant-Based DietaryIntervention on Body Weight, Metabolism, and InsulinSensitivity in Postmenopausal Women” (Amer. J. of Med.,2005: 991–997) reported on the results of an experiment inwhich half of the individuals in a group of 64 postmenopausaloverweight women were randomly assigned to a particularvegan diet, and the other half received a diet based on NationalCholesterol Education Program guidelines. The sample meandecrease in body weight for those on the vegan diet was 5.8kg, and the sample SD was 3.2, whereas for those on the control diet, the sample mean weight loss and standard deviationwere 3.8 and 2.8, respectively. Does it appear the true averageweight loss for the vegan diet exceeds that for the control dietby more than 1 kg? Carry out an appropriate test of hypotheses at significance level .05 based on calculating a P-valuearrow_forward
- A sample of men and women who had passed their driver's test either the first time or the second time were surveyed, with the following results: Results of the driving testGender First time Second timeMen 126 211Women 135 178a) Do these data suggest that there is a relationship between gender and the passing of their driver’s test from which the present sample was drawn? Let alpha=.05arrow_forwardThe dry shear strength of birch plywood bonded with different resin glues was studied with a completely randomized designed experiment. Here are the data: Glue A (102; 58; 45; 79; 68; 63; 117) Glue C (100; 102; 80; 119) Glue F (220; 243; 189; 176; 176). SSE = 8167.55; MSE = 628.27; and F stat = 37.99. What is the critical value at the 5% significance level (rounded off 2 decimals)?arrow_forwardThe article “Treadmill Exercise and Resistance Training in Patients With Peripheral Arterial Disease With and Without Intermittent Claudication. A Randomized Controlled Trial” (M. McDermott, P. Ades, et al., Journal of the American Medical Association, 2009:165–174) reported the results of a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A sample of 48 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 348 m, with a standard deviation of 80 m. For a control group of 46 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Find a 95% confidence interval for the difference in mean distance walked between the two groups of patients.arrow_forward
- The article “Treadmill Exercise and Resistance Training in Patients With Peripheral Arterial Disease With and Without Intermittent Claudication. A Randomized Controlled Trial” (M. McDermott, P. Ades, et al., Journal of the American Medical Association, 2009:165–174) reported the results of a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A sample of 48 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 348 meters, with a standard deviation of 80 m. For a control group of 46 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Can you conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls? Use the α = 0.05 level of significance.arrow_forwardSuppose that, as part of a research methods class, Bailey was asked to write a summary of a research paper on the topic of the effects of oil contamination in soil on seed germination rates. Identify the explanatory and the response variables.arrow_forwardOn snow-covered roads, winter tires enable a car to stop in a shorter distance than if summer tires were installed. In terms of the additive model for one-way ANOVA, and for an experiment in which the mean stopping distances on a snow-covered road are measured for each of four brands of winter tires. If the data are as shown in Sheet 48, what conclusion would be reached at the 0.01 level of significance? Shett 48 Supplier A 517 484 463 452 502 447 481 500 485 566 Supplier B 479 499 488 430 482 457 424 488 526 455 Supplier C 435 443 480 465 435 430 465 514 463 510 Supplier D 526 537 443 505 468 533 481 477 490 470 Select one: a) p-value = 0.28 greater than 0.05, the average distance is different for at list two tires b) F stat = 1.86, F crit = 4.38, not enough evidence to claim that the average distance is different for at list two tires c) F ratio = 4.38, not enough evidence to claim that the average distance is different for at list two tires d) F stat = 0.68, F…arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman