Medical researchers have developed a new artificial heart constructed primarily
ID: 3232673 • 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 49 battery packs is selected and subjected to a life test. The average life of these batteries is 4.1 hours. Assume that battery life is normally distributed with standard deviation sigma = 0.3 hour. Someone is interested in test the following hypothesis on the mean battery life: H_0: mu = 4 hours versus H_1: mu notequalto 4 hours (a) Is there evidence to reject the null hypothesis? Use fixed significance level (alpha = 0.05) approach to conduct the hypothesis test. (b) What is the P-value for the test? Based on the P-value, do you draw the same conclusions as the one in part (a)? Why or why not? (c) Compute the type II error probability if the true mean battery life is 4.15 hours. (d) Answer the question in part (a) by constructing an appropriate confidence interval on the mean battery life mu.Explanation / Answer
Part-a
Test statistic Z=(xbar-4)/(/sqrt(n))=(4.1-4)/(0.3/sqrt(49))=2.33
Crititcal Z=1.96 at 0.05 level.
AS calculated Z>1.96 we reject the null hypothesis
Part-b
P-value=2*P(Z<-2.33)=0.0198 using excel =2*NORMSDIST(-2.33)
As p-value,0.05, we reject the null hypothesis and same conclusion as part-a
Part-c
P(TypeII)=P(do not reject H0/µ1=4.15)
=P(|xbar|<4/µ1=4.15)
=P(|(xbar-4.15)/(/sqrt(n))|<(4-4.15)/(0.3/sqrt(49)))
=P(|z|<-3.5)
=2*P(Z<-3.5)
=0.00047 using excel function =2*NORMSDIST(-3.5)
Part-d
95% confidence interval= xbar±1.96*(/sqrt(n))
=4.1±1.96*(0.3/sqrt(49))
=(4.016 4.184)
As confidence interval does not includes 4,we reject the null hypothesis.