Can someone please check my work? I am unsure if i have done this correctly.

MTH 245 - CH 9 Homework

For each of these problems follow the 6-step Hypothesis Testing Guideline in the CH 9 Power Point. You can use the p-value or the critical value method. You can calculate the test statistic and p-value (Step 4) either by hand using the formulas and tables, with a graphing calculator, or by using StatCrunch.Include the details of your calculations, not just the values. If you use the critical value method, state the critical value instead of the p-value.

6-Step Hypothesis Testing Procedure:

1. Determine which type of hypothesis test to use

2. Check Requirements

3. State H0 & H1 and identify which is the claim

4. Calculate test statistic and p-value

5. Reject or fail to reject, state reason

6. Write conclusion in context of original claim

1. Listed below are the costs (in dollars) of repairing the front ends and the rear ends of 9 different cars when they were damaged in controlled low-speed crash tests (based on data from the Insurance Institute for Highway Safety). Test the claim that the mean of the differences is greater than zero so that the front repair costs are greater than the corresponding rear repair costs.Use a 0.10 significance level.

Front Repair Cost

936

978

2252

1032

3911

4312

3469

2598

4535

Rear Repair Cost

1480

1202

802

3191

1122

739

2769

3375

1787

1.Matched Pair t-test

2.SRS - Not given, N is not greater than 30.

3.H0:md=0H1:md>0 (Claim)

4.L1 = Front repair cost, L2= Rear repair cost, L3= l2-L1

T=-1.0p=.83a=0.10

5.Fail to reject H0, p>a, 0.83>0.10

6.There is not enough evidence at the 0.10 significance level to reject H0 and support the claim that the mean of the differences is greater than zero so that the front repair costs are greater than the corresponding rear repair costs, his conclusion may not be valid for all cars.

2.Does stress affect the recall ability of police eyewitnesses? This issue was studied in an experiment that tested eyewitness memory a week after a nonstressful interrogation of a cooperative suspect and a stressful interrogation of an uncooperative and belligerent suspect.The numbers of details recalled a week after the incident was recorded, and the summary statistics are given below (based on data from "Eyewitness Memory of Police Trainees for Realistic Role Plays," by Yuille, et al.,Journal of Applied Psychology, Vol. 79, No. 6). Use a 0.01 significance level to test the claim in the article that "stress decreases the amount recalled." USE THE CONFIDENCE INTERVAL METHOD FOR THIS PROBLEM!!

Nonstress:n = 40,= 53.3, s = 11.6

Stress:n = 40,= 45.3, s = 13.2

1.2 Sample t-test

2.SRS- assumed, Independent samples OK, n1>30 or normal (n=40), n2>30 or normal (n=40) OK.

3.H0:m1=m2H1:m1>m2 (Claim)

4.95% CI (2.648, 13.532)

5.Reject H0, 0 is not in the interval (as told in the DB of CH9 hints)

6.There is enough evidence at the 0.01 significance level to reject the null hypothesis and support the claim that the article that "stress decreases the amount recalled."

3.Among 13,200 submitted abstracts that were blindly evaluated (with authors and institutions not identified), 26.7% were accepted for publication.Among 13,433 abstracts that were not blindly evaluated, 29.0% were accepted (based on data from "Effect of Blinded Peer Review on Abstract Acceptance," by Ross, et al.,Journal of the American Medical Association,Vol. 295, No. 14).Use a 0.05 significance level to test the claim that the acceptance rate is the same with or without blinding.

1. 2-Prop Z-test

2. Both SRS - NO, both independent - OK, blindly evaluated: np=13,200>5,np=3,524>5, Not Blindly evaluated: np=13,433>5, np=3,896>5 - OK

3. H0: p1 = p2 (claim)H1: p1p2

4. z= -4.197p=2.70a=0.;05

5. P>a, 2.70>0.05Fail to reject H0

6. There is not enough evidence at the 0.05 significance level to support the claim thatthe acceptance rate is the same with or without blinding.