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Statistics Assignment (Example)

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Student Name: Professor: Course: Date: 13.6. a) Ho:μ=32Ha:μ≠32t=X-μhypSX=35.89-322.887=3.57. The P-Value of t=3.57 for a 2-tailed t-test = 0.011784 from statistical software. Since P-Value<0.05 we reject the null hypothesis. b) CI=X±Z×Sn=35.89±1.96×2.887=35.89±2.13 =38.02 33.76. c) Since 32 is outside our range we reject the null hypothesis. 13.8 a) Ho: μ=90Ha: μ=90t stat=X-μhypSX=88-90928=-21.70084 =-1.1759. T critical (0.05 27) =2.052. Since -1.1759>-2.052 we fail to reject the null hypothesis and conclude that drinking coffee just before going to sleep does not affect the amount of dream time at 5% significance level. b) It is not appropriate to construct the 95% confidence interval since the test is not statistically significant. 13.9 a) As a car manufacturer I would prefer a d) CI=51.33±1.799×63.3351.33±113.93-62.6.26 165e) Cohen's d=51.3366.33=0.7739Blood doping in athletes reduced their running times by approximately one standard deviation( this indicated a large effect) f) The test showed that blood doping by athletes reduced their running timest=1.799 d.f=11 cohen's d=0.7739 and P=0.0107. 15.14 a) Ho: d=0Ha:( d≠0)d=∑dn=42810=42.8. SD=∑d-dn-1=64079=26.6812SE(d)=26.681210=8.4374t stat=dSEd=42.88.4374d= 5.0727 T critical (0.05 9) =2.262Since 5.0727>2.262 we reject the null hypothesis. b) P=42.8±2.262×8.4374=42.8±19.09=23.71 61.17This confirms that we can be 95% sure that the true mean is increase is somewhere between 24 and 61 points. c) Cohen's d=dSD=42.826.6812=1.604This shows that the difference between the two means is larger than one standard deviation. d) The test showed that the population mean is not equal to zero t=5.0727 d.f=9 cohen's d=1.604 [...]

Order Description:

Show all relevant work; use the equation editor in Microsoft Word when necessary. Chapter 13, numbers 13.6, 13.8, 13.9, and 13.10 Chapter 14, numbers 14.11, 14.12, and 14.14 Chapter 15, numbers 15.7, 15.8, 15.10 and 15.14

Subject Area: Mathematics

Document Type: Dissertation Proposal

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02.01.2018

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