For each of these problems follow the 6-step Hypothesis Testing Guideline on Sli
ID: 3218809 • Letter: F
Question
For each of these problems follow the 6-step Hypothesis Testing Guideline on Slide 11 of 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.
1. Listed below are the costs (in dollars) of repairing the front ends and the rear ends of 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
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
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.
Front Repair Cost
936
978
2252
1032
3911
4312
3469
2598
4535
Rear Repair Cost
1480
1202
802
3191
1122
739
2769
3375
1787
Explanation / Answer
Q3.
Given that,
sample one, n1 =13200, p1= x1/n1=0.267
sample two, n2 =13433, p2= x2/n2=0.29
null, the acceptance rate is the same with or without blinding Ho: p1 = p2
alternate, the acceptance rate is diffrent same with or without blinding H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.267-0.29)/sqrt((0.279*0.721(1/13200+1/13433))
zo =-4.186
| zo | =4.186
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =4.186 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -4.1861 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -4.186
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have evidence that the acceptance rate is diffrent for both the given data
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