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Medical researchers have developed a new artificial heart constructed primarily

ID: 3133318 • Letter: M

Question

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart 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 20 battery packs gave the average life of these batteries to be 4.05 hours with standard deviation of 0.2 hour. Assume the battery life is normally distributed.

a) Test your hypothesis using the P-value method.

b) Construct a 95% confidence interval (CI) for this case.

c) Is your answer in (a) validated by CI method?

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   4  
Ha:    u   =/   4  
              
As we can see, this is a    two   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    4.05          
uo = hypothesized mean =    4          
n = sample size =    20          
s = standard deviation =    0.2          
              
Thus, t = (X - uo) * sqrt(n) / s =    1.118033989          

df = n - 1 =    19          

              
Hence, the p value is, using technology,              
              
p =    0.2774901

As P > 0.05, we fail to reject Ho.

There is no significant evidence that the average life of the batteries is not 4.00 hours. [CONCLUSION]

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b)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    4.05          
t(alpha/2) = critical t for the confidence interval =    2.093024054          
s = sample standard deviation =    0.2          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    0.093602881          
Lower bound =    3.956397119          
Upper bound =    4.143602881          
              
Thus, the confidence interval is              
              
(   3.956397119   ,   4.143602881   ) [ANSWER]

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c)

Yes, as 4.00 is inside the interval.