Medical researchers have developed a new artificial heart constructed primarily
ID: 3224758 • Letter: M
Question
Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient’s body, but the battery pack needs to be recharged about every four hours. A random sample of 100 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation = 0.2 hour.
a. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use = 0.01. Perform the fixed level test (7 steps procedure)
b. What is the P-value for the test in part 5-1 above?
c. Explain how the question in part (a) above could be answered by constructing a one-sided confidence bound on the mean life.
d. What sample size would be required to detect a true mean battery life of 4.5 hours if we wanted the power of the test to be at least 0.95?
Explanation / Answer
Part-a
Null hypothesis H0: µ<=4
Alternative hypothesis Ha:µ>4
Level of significance =0.01
Critical value z=2.33
Test statistic Z=(xbar-4)/(/sqrt(n))
=(4.05-4)/(0.2/sqrt(100))
=2.5
p-value=P(Z>2.5)=0.0062 using excel function =1-NORMSDIST(2.5)
As calculated z>2.33 and also p-value<0.01, we reject the null hypothesis
We conclude that mean battery life exceeds 4 hours
Part-b
p-value=P(Z>2.5)=0.0062 using excel function =1-NORMSDIST(2.5)
Part-c
Right sided 99% confidence interval= =(xbar+2.33*(/sqrt(n), +infinity)
=4.05+2.33*0.2/sqrt(100)
=4.0966 , +infnity
As the right sided confidence interval does not contains 4, s we reject the null hypothesi and conclude that mean life is grater than 4.
Part-d
Sample size n=3
Power and Sample Size
1-Sample Z Test
Testing mean = null (versus > null)
Calculating power for mean = null + difference
Alpha = 0.01 Assumed standard deviation = 0.2
Sample Target
Difference Size Power Actual Power
0.5 3 0.95 0.977453