SOLUTION: In a study of 817 randomly selected medical malpractice lawsuits, it was found that 489 of them were dropped or dismissed.
Use a 0.01 significance level to test the claim that mos
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Question 1193685: In a study of 817 randomly selected medical malpractice lawsuits, it was found that 489 of them were dropped or dismissed.
Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed.
What is the hypothesis test to be conducted?
What is the Test statistic?
What is the P Value?
What is the conclusion about the null hypothesis?
A:Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a
B:Fail to reject the null hypothesis because the P-value is greater than the significance level, a .
C:Reject the null hypothesis because P-value is less than or equal to the significance level, a .
D: Reject the null hypothesis because the P -value is greater than the significance level, a
What is the final conclusion?
A:There is not sufficient evidence to warrant rejection of the claim that most medical malpractice lawsuits are dropped or dismissed
B:There is not sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed .
C: There is sufficient evidence to warrant rejection of the claim that most medical malpractice are dropped or dismissed .
D:There is sufficient evidence to support the claim that most medical malpractice lawsults are dropped or dismissed
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Ho: p<=0.50
Ha:p>0.50
alpha=0.01 p{reject Ho|Ho true}
test Is a 1-sample proportion
p hat is 489/817=0.5985
calculation is z=(0.5985-0.50)/sqrt(0.5*0.5/817)
=0.0985/0.0175
=5.63
reject Ho, there is sufficient evidence to support the claim D.
p-value is 0 (to 7 decimal places)
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