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12. Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 23 people who drank ethanol and another group of 23 people given a placebo. The errors for the treatment group have a standard deviation of 2.40, and the errors for the placebo group have a standard deviation of 0.86. Assume that the two populations are normally distributed. Use a 0.05 significance level to test the claim that both groups have the same amount of variation among the errors. Let sample 1 be the sample with the larger sample variance, and let sample 2 be the sample with the smaller sample variance. What are the null and alternative hypotheses? O A. Ho: 07 62 = 02 OB. Ho: 07 = 02 H1 : 09 2 07 H1:092 02 O C. Ho: 02 03 OD. HO: 07 = 05 H1: 09 = NN Hy: 02 Identify the test statistic. (Round to two decimal places as needed.) Use technology to identify the P-value. (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that both groups have the same amount of variation among the errors. O B. Fail to reject Ho. There is insufficient evidence to warrant rejection of the claim that both groups have the same amount of variation among the errors. O C. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that both groups have the same amount of variation among the errors. O D. Reject Ho. There is insufficient evidence to warrant rejection of the claim that both groups have the same amount of variation among the errors.The accompanying data table contains chest deceleration measurements (in g, where g is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. 1 Click the icon to view the table of chest deceleration measurements Click the icon to view the table of analysis of variance results. a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. The population means are approximately normal. O B. There are three samples of measurements. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test the claim that the three size categories have the same mean chest deceleration, why is the method referred to as analysis of variance? O A. The method is based on showing that the population variances are similar. O B. The method is based on showing that the population variances are different. O C. The method is based on estimates of multiple varying population variances. O D. The method is based on estimates of a common population variance 1: More Info Chest Deceleration Measurements (g) from a Standard Crash Test Small 44 39 37 54 39 44 42 Midsize 36 53 43 42 52 49 41 Large 32 45 41 38 37 38 33 2: More Info SPSS Results Sum of Squares df Mean Square F Sig. Between Groups 200.857 2 00.429 3.28 0.061 Within Groups 549.714 18 30.540 Total 750.571 2014. The accompanying data table lists measured amounts of chest compression (mm) from car crash tests. Also shown are the SPSS results from analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different car sizes have the same mean amount of chest compression. Click here to view the table of chest deceleration measurements." Click here to view the table of analysis of variance results. What is the value of the test statistic? The test statistic is (Round to two decimal places as needed.) What distribution is used with the test statistic? The (1) -distribution is used. 1: Chest Deceleration Measurements Small 29 31 35 33 26 32 21 26 25 34 26 34 Midsize 32 28 26 23 25 26 19 29 26 20 22 22 Large 27 32 39 27 31 26 34 30 34 26 24 31 SUV 24 31 31 25 30 39 22 33 34 35 29 26 2: ANOVA results ANOVA CHEST Sum of Squares df Mean Square F Sig. Between Groups 223.750 3 74.583 3.815 016 Within Groups 860.167 44 19.549 Total 1083.917 47 (1) Ox O O tThe accompanying table shows the wait times (minutes) from six rides located In three different Walt Disney World theme parks. Wait times were reported by Disney and wait times at 10 AM and 5 PM were recorded on 50 different days (ve rows are shown in the accompanying table). Use the 10 AM wait times for Space Mou ntain, Rock 'n' Roller Coaster, Tower of Terror, and Flight of Passage. Use a O 05 Signicance level to test the claim that the four rides have the same mean wait time at 10 AM. 1 Click the icon to View the data table of wait times. Determine the null hypothesis H0: (1) Determine the alternative hypothesis H1: (2) Determine the test statistic. The test statistic is (Round to two decimal places as needed.) Determine the P-value The P-value is (Round to three decimal places as needed ) What can you conclude? There (3) sufficient eVidence at a 0.05 signicance level to warrant rejection of the claim that the four rides have (4) mean wait time(s) 1: Disney World Wait'mes SMALL SMALL ROCKN ROCKN TOWER TOWER FLIGHT FLIGHT NAVI NAVI U; SPACEMTSPACE MT WORLD WORLD ROLLER ROLLER TERROR TERROR PASSAGE PASSAGE RIVER RIVER 10AM 5PM 10AM 5PM 10AM 5PM 10AM 5PM 10AM 5PM 1 0AM 5PM 50 75 15 20 65 60 50 35 180 95 75 80 30 70 10 25 65 60 45 25 150 125 50 30 105 110 20 75 140 105 75 90 180 175 130 120 20 40 5 10 20 35 15 10 105 55 50 40 30 35 15 1O 5O 35 60 65 135 45 135 105 (1) 0 p1 = p2 = p3 = M 0 All of the means are different (2) 0 At least one of the means is different from the others O All of the means are different (3) O is (4) O the same 0 At least one of the means is different from the others 0 P1 > I12 > P3 > I14 0 15 "0t 0 different 0 P1