Question

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once implanted in the patient’s body, but the battery pack needs to be recharged every 4 hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.10 hours. Assume that battery life is normally distributed with a standard deviation of 0.2 hours. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use a significance level α = 0.01 what is the parameter of interest. Are the assumptions satisfied? Explain why. State hypotheses Calculate the test statistic. What is/are the critical value(s)? What is your decision? What is the conclusion/interpretation of this decision? Calculate the p-value for this test. What is your decision based on the p-value? Construct a 99% CI for true mean battery life, and interpret your CI. Construct a 95% CI for true mean battery life. Now, which of the CI is narrower and why? Please show all steps on calculating P-value and CI values!!

Answer 1

(a) Parameter of interest: mean battery life

(b) The assumptions are satisfied.

Explanation: Since Population SD = = 0.2 is given and Sample Size = n = 50 > 30 Large Sample and so, Central Limit Theorem is applicable.

(c) H0: Null Hypothesis: = 4

HA: Alternative Hypothesis: > 4

(d)

SE = /

= 0.2/

= 0.0283

Test Statistic is given by:
Z = (4.10 - 4)/0.0283

= 3.5355

So,

Test Statistic is: 3.5355

(e)

= 0.01

From Table, critical value of Z = 2.33

(f)

Decision:

Since calculated value of Z = 3.5355 is greater than critical value of Z = 2.33, the difference is significant. Reject null hypothesis.

(g)

Conclusion:

The data support the claim that mean battery life exceeds 4 hours.

(h)

Z score = 3.5355

By technology, area under standard normal curve = 0.4998

So,

P - Value = 0.5 - 0.4998 = 0.0002

(i)

Since P - value is less than , the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that mean battery life exceeds 4 hours.

(j)

= 0.01

Table gives critical values of Z = 2.576

Confidence Interval:

4.10 (2.576 X 0.0283)

= 4.10 0.0729

= ( 4.0271 ,4.1729)

(k)

= 0.05

Table gives critical values of Z = 1.96

Confidence Interval:

4.10 (1.96 X 0.0283)

= 4.10 0.0555

= ( 4.0445 ,4.1555)

(l) The CI is narrower for 95% because a narrow confidence interval enables more precise population estimate.