What are the hypotheses? O HO: At least one Mu is different; H1: Mu1 = Mu2 = Mu3 %3D %3D O HO: Mu1 = Mu2 = Mu3; H1: At least one Mu is different %3D
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- The life expectancy of a male during the course of the past 100 years is approximately 27,739 days. Use the table to the right to conduct a test using a = 0.10 to determine whether the evidence suggests that chief justices live longer than the general population of males. Suggest a reason why the conclusion drawn may be flawed. Life Span (Days) 30,716 Chief Life Span O (Days) 30,815 32,785 Chief Justice Justice A 30,519 30,144 J K. 26,044 30,824 D 28,311 L 32,360 31,674 25,002 32,643 M 32,377 30,763 32,319 31,238 G State the appropriate null and alternative hypotheses. O A. Ho: H= 27,739 versus H,: u>27,739 O B. Ho: H=30,533 versus H,: u# 30,533 O C. Ho: = 30,533 versus H,: p> 30,533 OD. Ho: H= 27,739 versus H,: u# 27,739 O E. Ho H=27,739 versus H,: H M C V Rata Set 12 “Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. 1. Calculate the t test stastic using the following formula. 2. Find the P value using Excel (i.e., =T.DIST.RT(t,df). What is it? 3. Compare the P-value to the signifcance level. How do they compare?ata Set 12 “Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. 1. Find the degrees of freedom using the conversative estimate formula. What is it? 2. Find the critical value. What is it? 3. Sketch a normal distribution curve and highlight the region of significance.
- Please don’t rejectllustration 16.22 (a). There are six seats available in a compart- ment. In how many ways can six persons be seated? CS Scanned with CamScannerPlacebo Shot Injected overall totul Deformed 11 185 196 Not Oe formed 14539 Is,685 30224 Overall total 19,550 (,870 30,420 ) shot efficacy Deformed Risk of Placebo - Deformed Pisk Of Shot injected Deformed Risk of Placebo NO = 185 Snot =.0116572149 I5870 -O 116572149 - .0007560137457 VE = O110572149 Shots 14550 S60137457 =0007500137457 V2 =.93514628OS
- Moviegoers The average “moviegoer” sees 8.5 moviesa year. A moviegoer is defined as a person who sees atleast one movie in a theater in a 12-month period.A random sample of 40 moviegoers from a largeuniversity revealed that the average number of moviesseen per person was 9.6. The population standarddeviation is 3.2 movies. At the 0.05 level ofsignificance, can it be concluded that this represents adifference from the national average?A Oklahoma History 20-21-Rowan X + Vactivity/2001003/assessment D. A Oklahoma History 2. + Matti Thomason 9t. M Inbox - mt229440@. Reading Plus® Insi. 1 2.1.3 Qulz: What is a Variable? Question 3 of 8 Which of the following statements contain a variable? Check all that apply. M A. There are 12 inches in a foot. O B. She arrived at 2 o'clock. C. Half the height of the building. V D. The length of the rope. SUBMIT E PREVIOUScation.wiley.com/was/ui/v2/assessment-player/index.html?launchld%3Deef1e3c4-0de6-4778-8346-dec63c264926#/question/10 rk updated Question 11 of 16 - / 1 E View Policies Current Attempt in Progress Transgender Adults A study of transgender adults examines the age at which they began to transition and the age of their earliest memories of gender dysphoria. (In this exercise, we examine the age beginning transition and in Exercise 6.107 we examine the age of their earliest memories.) In the study of 210 transgender adults, the mean age at which they began transitioning was 32.6 with a standard deviation of 18.2. Find a 95% confidence interval for the mean age at which transgender adults begin transitioning. Zaliznyak M, Bresee C, and Garcia MM, "Age at First Experienceof Gender Dysphoria Among Transgender Adults Seeking Gender- Affirming Surgery," JAMA Network Open, March 16,2020. Some of the data has been approximated from informationin the paper. to Confidence interval (round to three…
- Exercise 13-60 (Algo) (LO13-5) Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30. claims for the last year, the director of the actuarial department determined the distance from the fire station (x) and the amount of fire damage, in thousands of dollars (). The MegaStat output is reported here: ANOVA table Source Regression Residual • Total df SS 1,835.5782 1 1,270.4934 28 3,106.0716 29 Regression output Variables: Intercept Distance-X Coefficients 13.6984 3.2931 F MS 1,835.5782 40.4537 45.3748 Std. Error 3.0467 0.5178 t(df=28) 2.427 6.36 Click here for the Excel Data File Required: a-1. Determine the regression equation. (Round your answers to 3 decimal places.)ID Gender Major Employ Age MBA_GPA BS GPA Hrs_Studying Works FT 1 1 No Major Unemployed 39 2.82 3.05 3 0 2 1 No Major Full Time 55 3.49 3.45 7 1 3 1 No Major Part Time 43 3.28 3.5 7 0 4 1 No Major Full Time 56 3.25 3.55 7 1 5 1 No Major Full Time 38 3.26 3.3 6 1 6 1 No Major Unemployed 54 2.87 3.05 4 0 7 1 No Major Full Time 30 3.16 3.35 6 1 8 1 No Major Full Time 37 3.4 3.35 6 1 9 1 No Major Part Time 38 2.84 3.05 3 0 10 1 No Major Full Time 42 3.72 3.7 7 1 11 1 No Major Part Time 52 3.22 3.5 7 0 12 1 No Major Full Time 35 3.44 3.55 7 1 13 1 No Major Full Time 37 3.65 3.9 8 1 14 1 No Major Full Time 53 3.02 3.3 6 1 15 1 No Major Part Time 51 3.03 3.25 6 0 16 1 No Major Full Time 40 3.8 3.8 8 1 17 1 No Major Full Time 33 3.23 3.5 7 1 18 1 No Major Part Time 53 3.26 3.5 7 0 19 1 No Major Full Time 43 3.53 3.75 8 1 20 1 No Major Unemployed 35 3.75 3.9 8 0 21 1 No Major Full Time 57 3.15 3.2 6 1 22 1 No Major Part Time 32 3.66 3.75 8 0 23 1 No…Q.6
