Statistics for Business & Economics, Revised (MindTap Course List)
12th Edition
ISBN: 9781285846323
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
Publisher: South-Western College Pub
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Chapter 13.2, Problem 4E
In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST = 400. SSTR = 150. nT = 19. Set up the ANOVA table and test for any significant difference between the
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Chapter 13 Solutions
Statistics for Business & Economics, Revised (MindTap Course List)
Ch. 13.2 - The following data are from a completely...Ch. 13.2 - In a completely randomized design, seven...Ch. 13.2 - Refer to exercise 2. a. what hypotheses are...Ch. 13.2 - In an experiment designed to test the output...Ch. 13.2 - In a completely randomized design, 12 experimental...Ch. 13.2 - Develop the analysis of variance computations for...Ch. 13.2 - Three different methods for assembling a product...Ch. 13.2 - Refer to the NCP data in Table 13.4. Set up the...Ch. 13.2 - To study the effect of temperature on yield in a...Ch. 13.2 - Auditors must make judgments about various aspects...
Ch. 13.2 - Four different paints are advertised as having the...Ch. 13.2 - The Consumer Reports Restaurant Customer...Ch. 13.3 - The following data arc from a completely...Ch. 13.3 - The following data are from a completely...Ch. 13.3 - To test whether the mean time needed to mix a...Ch. 13.3 - Refer to exercise 15. Use Fishers LSD procedure to...Ch. 13.3 - The following data are from an experiment designed...Ch. 13.3 - To lest for any significant difference in the...Ch. 13.3 - Refer to exercise 18. Use the Bonferroni...Ch. 13.3 - The International League of Triple-A minor league...Ch. 13.4 - Consider the experimental results for the...Ch. 13.4 - The following data were obtained for a randomized...Ch. 13.4 - An experiment has been conducted for four...Ch. 13.4 - An automobile dealer conducted a test to determine...Ch. 13.4 - The price drivers pay for gasoline often varies a...Ch. 13.4 - The Scholastic Aptitude Test (SAT) contains three...Ch. 13.4 - A study reported in the Journal of the American...Ch. 13.5 - A factorial experiment involving two levels of...Ch. 13.5 - The calculations for a factorial experiment...Ch. 13.5 - A mail-order catalog firm designed a factorial...Ch. 13.5 - An amusement park studied methods for decreasing...Ch. 13.5 - As part of a study designed to compare hybrid and...Ch. 13.5 - A study reported in The Accounting Review examined...Ch. 13 - In a completely randomized experimental design,...Ch. 13 - A study reported in the Journal of Small Business...Ch. 13 - The U.S. Environmental Protection Agency (EPA)...Ch. 13 - Prob. 37SECh. 13 - Prob. 38SECh. 13 - In a study conducted to investigate browsing...Ch. 13 - A research firm tests the miles-per-gallon...Ch. 13 - The compact car market in the United States is...Ch. 13 - Major League Baseball franchises rely on...Ch. 13 - A factorial experiment was designed to test for...Ch. 13 - A manufacturing company designed a factorial...Ch. 13 - Wentworth Medical Center As part of a long-term...Ch. 13 - Compensation for Sales Professionals Suppose that...
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- Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). Using the hypothetical data provided below, test whether the mean pressure applied to the driver’s head during a crash test is equal for each types of car. Use α = 5%. Compact cars Midsize cars Full-size cars 643 469 484 655 427 456 702 525 402 Mean 666.67 473.67 447.33 Standard deviation 31.18 49.17 41.68arrow_forwardOne group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x¯1x¯1 = 9.3 x¯2x¯2 = 10.4 Sample std. dev. s1 = 2.4 s2 = 2.3 Number of students n1 = 15 n2 = 15 Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Find the test statistic tcalc. Use Excel to find the p-value.arrow_forwardFrom the ANOVA table, we see that the p-value is less than 0.0001. We can interpret this to mean: Question 11 options: The mean gene expression in at least one group is significantly different from the mean gene expression in at least one other group. The control group has a mean gene expression that is significantly different from at least 1 other group mean. There is not a significant difference in mean gene expression between the groups. Treatment 2 has a significantly higher mean expression level than at least one of the other 2 groups.arrow_forward
- One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x¯1x¯1 = 9.7 x¯2x¯2 = 10.0 Sample std. dev. s1 = 2.7 s2 = 2.9 Number of students n1 = 23 n2 = 23 Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.) Use Excel to find the p-value. (Round your answer to 3 decimal places.)arrow_forwardSuppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). Using the hypothetical data provided below, test whether the mean pressure applied to the driver's head during a crash test is equal for each types of car. Use a = 0.05, ssw=6300 and SST=6350. 1) The null and the alternative hypotheses are : a) Ho: H1 = H2 = Hg. H: At most one of uj:j = 1,2,3 is different. H : At least one of u;:j = 1,2,3 is different. H: At least one of H;:j = 1,2,3 is different. H: 41 = 42 = Hz. vs b) Ho: 41 = H2 = Hg. Vs %3D Vs d) Ho: At least one of u;:j = 1,2,3 is different. %3D VS 2) The F test statistics= a) 0.0228 b) 0.0218 c) 0.0208 d) 0.0238 3) The critical value = a) 4.3433 b) 5.5433 c) 5.1433 d) 5.3433 4) The decision of the test is: a) Do not reject Ho. There is insufficient evidence that at least one of the population means is different. b) Reject Hg.…arrow_forwardA researcher wanted to examine whether type of dog food (freeze dried meat vs generic brand kibble) would affect dogs’ attention span. The researcher hypothesized that dogs who ate freeze dried meat would have longer attention spans than those who ate generic brand kibble. The researcher used a single sample of 13 subjects, that took part in both treatment conditions. The researcher set the alpha level at .01. The mean of the difference scores (MD)was 7.0 and the SSDfor the difference in attention spans was 144. What is the calculated test statistic to the nearest thousandth? Be sure to put the positive or negative sign in front of it. Note: SHOW ALL WORK!!!arrow_forward
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- The data below are from a study comparing three different treatment conditions: Treatment 1 Treatment 2 Treatment 3 0 1 6 1 4 5 0 1 6 3 2 3 T = 4 T = 8 T = 20 SS = 6 SS = 6 SS = 6 N = 12 G = 32 ƩX2= 138 Conduct a single-factor independent-measures ANOVA to test the hypothesis that there are significant differences in the mean scores among the three treatment conditions. Use α = .01. Your decision is Fail to reject the null hypothesis and conclude that there there is at least one significant mean difference Fail to reject the null hypothesis and conclude that there are no significant differences among the three treatment conditions Reject the null hypothesis and conclude that there is at least one significant mean difference Reject the null hypothesis and conclude that there are no significant differences in mean scores among the treatment conditions b. Calculate η2 (eta squared) as a measure of the…arrow_forwardDr. Palmer found a small treatment effect. What can she do in improve her changes of finding a statistically significant result?arrow_forwardA study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment The previous treatment regime had produced an average increase of μ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of μ in comparison to the previous treatment. The sample data yielded ?̅ = 2.17 and ? = 1.05: Using α = 05, what conclusions can you draw about the research hypothesis?arrow_forward
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