Essentials of Statistics (6th Edition)
6th Edition
ISBN: 9780134685779
Author: Mario F. Triola
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 11.2, Problem 15BSC
In Exercises 5–18, test the given claim.
15. Clinical Trial of Echinacea In a clinical trial of the effectiveness of echinacea for preventing colds, the results in the table below were obtained (based on data from “An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections,” by Turner et al., New England Journal of Medicine, Vol. 353, No. 4). Use a 0.05 significance level to test the claim that getting a cold is independent of the treatment group. What do the results suggest about the effectiveness of echinacea as a prevention against colds?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Not sure how to do this
A random sample of 15 families representing three different Scandinavian countries has been observed for the number of weekend trips they take per year. Are the differences significant (use p=.05, F(critical) = 3.88)? Use the five step model as a guide and write a sentence or two of interpretation for your results.
Sweden
Norway
Denmark
10
11
7
9
10
5
4
5
2
2
2
0
I need help explaining the answer using the table.
Chapter 11 Solutions
Essentials of Statistics (6th Edition)
Ch. 11.1 - Cybersecurity The table below lists leading digits...Ch. 11.1 - 2. Cybersecurity When using the data from Exercise...Ch. 11.1 - Cybersecurity The accompanying Statdisk results...Ch. 11.1 - Cybersecurity What do the results from the...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...
Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - In Exercises 520, conduct the hypothesis test and...Ch. 11.1 - Ben fords Law. According to Benfords law, a...Ch. 11.1 - Ben fords Law. According to Benfords law, a...Ch. 11.1 - Ben fords Law. According to Benfords law, a...Ch. 11.1 - Ben fords Law. According to Benfords law, a...Ch. 11.1 - Assumed mid-point x=fxn=39825180=221.25...Ch. 11.2 - Handedness and Cell Phone Use The accompanying...Ch. 11.2 - Ear Preference for Cell Phone Use 2. Hypotheses...Ch. 11.2 - Hypothesis Test The accompanying TI-83/84 Plus...Ch. 11.2 - Right-Tailed, Left-Tailed, Two-Tailed Is the...Ch. 11.2 - Prob. 5BSCCh. 11.2 - In Exercises 5-18, test the given claim. 6. Splint...Ch. 11.2 - In Exercises 5-18, test the given claim. 7....Ch. 11.2 - Prob. 8BSCCh. 11.2 - In Exercises 5-18, test the given claim. 9. Four...Ch. 11.2 - In Exercises 5-18, test the given claim. 10....Ch. 11.2 - In Exercises 5-18, test the given claim. 11....Ch. 11.2 - In Exercises 5-18, test the given claim. 12. Nurse...Ch. 11.2 - Soccer Strategy In soccer, serious fouls in the...Ch. 11.2 - In Exercises 5-18, lest the given claim. 14. Is...Ch. 11.2 - In Exercises 518, test the given claim. 15....Ch. 11.2 - In Exercises 5-18, test the given claim. 16....Ch. 11.2 - Prob. 17BSCCh. 11.2 - In Exercises 5-18, test the given claim. 18....Ch. 11.2 - In Exercises 5-18, lest the given claim. 19. Car...Ch. 11.2 - Is the Home Field Advantage Independent of the...Ch. 11.2 - Equivalent Tests A X2 test involving a 2 2 table...Ch. 11.2 - Using Yatess Correction for Continuity The...Ch. 11.3 - In Exercises 1-4, use the following listed arrival...Ch. 11.3 - In Exercises 1-4, use the following listed arrival...Ch. 11.3 - In Exercises 1-4, use the following listed arrival...Ch. 11.3 - In Exercises 1-4, use the following listed arrival...Ch. 11.3 - In Exercises 5-16, use analysis of variance for...Ch. 11.3 - In Exercises 5-16, use analysis of variance for...Ch. 11.3 - In Exercises 5-16, use analysis of variance for...Ch. 11.3 - In Exercises 5-16, use analysis of variance for...Ch. 11.3 - In Exercises 5-16, use analysis of variance for...Ch. 11.3 - Prob. 10BSCCh. 11.3 - Triathlon Times Jeff Parent is a statistics...Ch. 11.3 - Arsenic in Rice Listed below are amounts of...Ch. 11.3 - Prob. 13BSCCh. 11.3 - Speed Dating Listed below are attribute ratings of...Ch. 11.3 - Tukey Test A display of the Bonferroni test...Ch. 11.3 - Two-Way ANOVA The pulse rates in Table 12-3 from...Ch. 11 - Exercises 1-5 refer to the sample data in the...Ch. 11 - Exercises 15 refer to the sample data in the...Ch. 11 - Exercises 15 refer to the sample data in the...Ch. 11 - Prob. 4CQQCh. 11 - Exercises 15 refer to the sample data in the...Ch. 11 - Questions 610 refer to the sample data in the...Ch. 11 - Questions 610 refer to the sample data in the...Ch. 11 - Questions 6-10 refer to the sample data in the...Ch. 11 - Questions 6-10 refer to the sample data in the...Ch. 11 - Motor Vehicle Fatalities The table below lists...Ch. 11 - Tooth Fillings The table below shows results from...Ch. 11 - American Idol Contestants on the TV show American...Ch. 11 - Clinical Trial of Lipitor Lipitor is the trade...Ch. 11 - Weather-Related Deaths For a recent year, the...Ch. 11 - Weather-Related Deaths Review Exercise 5 involved...Ch. 11 - Chocolate and Happiness In a survey sponsored by...Ch. 11 - Chocolate and Happiness Use the results from part...Ch. 11 - Chocolate and Happiness Use the results from part...Ch. 11 - One Big Bill or Many Smaller Bills In a study of...Ch. 11 - 6. Probability Refer to the results from the 150...Ch. 11 - Car Repair Costs Listed below are repair costs (in...Ch. 11 - Forward Grip Reach and Ergonomics When designing...Ch. 11 - Use Statdisk, Minitab, Excel, StatCrunch, a...Ch. 11 - FROM DATA TO DECISION Critical Thinking: Was...Ch. 11 - Cola Weights Data Set 26 Cola Weights and Volumes...Ch. 11 - Speed Dating Data Set 18 Speed Dating in Appendix...Ch. 11 - Author Readability Pages were randomly selected by...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- The report "Comparative Study of Two Computer Mouse Designs"t included the following description of the subjects used in an experiment. Twenty-four Cornell University students and staff (12 males and 12 females) volunteered to participate in the study. Three groups of 4 men and 4 women were selected by their stature to represent the 5th percentile (female 152.1 ± 0.3 cm, male 164.1 ± 0.4 cm), 50th percentile (female 162.4 ± 0.1 cm, male 174.1 ± 0.7 cm), and 95th percentile (female 171.9 ± 0.2 cm, male 185.7 ± 0.6 cm) ranges.... All subjects reported using their right hand to operate a computer mouse. This experimental design incorporated direct control and blocking. (a) Is the potential effect of the extraneous variable stature (height) addressed by blocking or direct control? O blocking direct control (b) Whether the right or left hand is used to operate the mouse was considered to be an extraneous variable. Is the potential effect of this variable addressed by blocking or direct…arrow_forwardWe wish to test of hypothesisH0 : μ1 = μ2 = μ3vsH1 : Not all group means are the same.Use the significance level α = 0.01.a. Find the value of the test statistic and its distribution.b. Find the rejection region.c. Find the p-value of the test.d. Make a conclusion all subparts please! thank youarrow_forwardThe authors of the paper "Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement"† compared two different instruments for measuring a person's ability to breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people participated in the study, and for each person air flow was measured once using the Wright meter and once using the mini-Wright meter. Subject Mini-WrightMeter WrightMeter Subject Mini-WrightMeter WrightMeter 1 512 494 10 445 433 2 430 395 11 432 417 3 520 516 12 626 656 4 428 434 13 260 267 5 500 476 14 477 478 6 600 557 15 259 178 7 364 413 16 350 423 8 380 442 17 451 427 9 658 650 (a) Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce different…arrow_forward
- The authors of the paper "Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement"† compared two different instruments for measuring a person's ability to breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people participated in the study, and for each person air flow was measured once using the Wright meter and once using the mini-Wright meter. Subject Mini-WrightMeter WrightMeter Subject Mini-WrightMeter WrightMeter 1 512 494 10 445 433 2 430 395 11 432 417 3 520 516 12 626 656 4 428 434 13 260 267 5 500 476 14 477 478 6 600 557 15 259 178 7 364 413 16 350 423 8 380 442 17 451 427 9 658 650 (a) Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce…arrow_forwardYou wish to test the following claim (Ha) at a significance level of a = Ho: P₁ = P2 Ha: P₁ P2 You obtain a sample from the first population with 120 successes and 403 failures. You obtain a sample from the second population with 39 successes and 278 failures. The test statistic is... in the critical region O not in the critical region critical value = ± test statistic = This test statistic leads to a decision to... reject the null hypothesis fail to reject the null hypothesis = 0.005. O There is not sufficient evidence to support second population proportion. [three decimal accuracy] [three decimal accuracy] As such, the final conclusion is that... There is sufficient evidence to support that the first population proportion is not equal to the second population proportion. first population proportion not equal to thearrow_forwardDani is writing this at the top so she can find this question if someone kindly answers it! ;) You wish to test the following claim (HaHa) at a significance level of α=0.001α=0.001. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1<μ2Ha:μ1<μ2You obtain the following two samples of data. Sample #1 Sample #2 87 77.9 68.6 81.6 75.4 72.3 67.5 71.3 67.8 72.3 74.3 76.8 72.6 77.4 75.2 80 65.5 63.3 77.1 73.9 84.6 71.6 77.2 61.8 74.5 73 72.6 68.3 71.7 67.2 74.3 63.3 74 77.4 87 71.7 70.8 74.7 73 80.2 78 70.9 75.5 68.8 83.9 78.9 72.5 68.3 74.4 74.5 73.9 67.5 74.3 83.9 80.7 75.5 79 89 76.4 76 71.4 82.2 76 69.2 73 77.2 70.7 85.8 65.6 79.8 79.4 72.2 64.3 62.2 76.4 86.8 67.9 69.4 82.2 79.4 73.2 81.7 75 78.6 69.7 65 76.7 67.1 77.4 84.9 68.2 79 84.5 87.4 67.1 71.4 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees…arrow_forward
- The yield of alfalfa from a random sample of six test plots is 1.4, 1.6, 0.9, 1.9, 2.2,and 1.2 tons per acre. Assume the data can be looked upon as a sample from anormal population. Test at the 0.05 level of significance whether this supports thecontention that the average yield for this kind of alfalfa is 1.5 tones per acre.arrow_forwardIi part b?) independent-measures study comparing three t WO ment conditions. 24. The me M 2 M = 3 a. M = 4 =10 n = 10 n = 10 T = 20 T = 30 T = 40 s2 = 2.67 s2 = 2.00 s2 = 1.33 a. Use an ANOVA with oa .05 to determine || whether there are any significant differences among the three treatment means. Note: Because the samples are all the same size, MS average of the three sample variances. is the within b. Calculate n² to measure the effect size for this study. the data we with the same То problemarrow_forwardYou wish to test the following daim (Ha) at a significance level of a = 0.01. H.:P1 = P2 Ha:P1 + P2 You obtain a sample from the first population with 258 successes and 28 failures. You obtain a sample from the second population with 362 successes and 25 failures. critical value = [three decimal accuracy] test statistic = [three decimal accuracy] The test statistic is... O in the critical region O not in the critical region This test statistic leads to a decision to... O reject the null hypothesis O fail to reject the null hypothesis As such, the final condusion is that... O There is sufficient evidence to support that the first population proportion is not equal to the second population proportion. O There is not sufficient evidence to support that the first population proportion is not equal to the second population proportion.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillCollege Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
Introduction to experimental design and analysis of variance (ANOVA); Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=vSFo1MwLoxU;License: Standard YouTube License, CC-BY