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SMAM 314 Exam 2 Name_____________ 1. The flow of current in milliamperes in wire strips of a certain type can be modeled as a normal distribution with mean 20 and standard deviation 1. A. What proportion of strips have a current flow between 18.5 and 22 milliamps?(10 points) " 18.5 ! 20 22 ! 20 % P(18.5 < X < 22) = P $ <Z< 1 1 '& # = P(!1.5 < Z < 2) = .9772 ! .0668 = .9104 normalcdf(18.5,22,20,1) = .9104 B. How large must a current flow be to be among the largest 5% of all flows?(10 points) P(X < c) = .95 " c ! 20 % P$ Z < = .95 1 '& # c ! 20 = 1.645 1 c = 21.645 invNorm(,95,20,1) = 21.645 2. Suppose that only 40% of all drivers in a certain state wear a seatbelt. A random sample of 500 drivers is selected. Use the normal approximation to the binomial distribution with the continuity correction to determine the probability that the number of drivers in the sample that wear a seat belt is at most 190. (15 points) µ = 500(.4) = 200 ! = 500(.4)(.6) = 10.95 $ 190.5 # 200 ' P(X " 190) = P & Z " = P(Z < #.87) = .1922 10.95 )( % normalcdf(-10^99,190.5,200,10.95)=.1928 3. The Rockwell hardness of certain metal pins is known to have a mean of 50 and a standard deviation of 1.5. What is the approximate probability that the average hardness of a random sample of thirty-six pins is at least 50.4?Use the Central Limit Theorem. (15 points) µ x = 50,! x = 1.5 / 36 = .25 $ 50.4 # 50 ' P(x " 50.4) = P & Z > + P(Z > 1.6) .25 )( % = 1# .9452 = .0548 normalcdf(50.4,10^99,50, .25)=.0548 4. The modulus of rupture (MOR) for a particular grade of pencil lead is known to have a standard deviation of 250 psi. A random sample from a normal population of 16 pencil leads yielded a sample mean of 6490. A. Construct a 95% confidence interval for the true mean MOR. (10 points) x± z! n 6490 ± 1.96(250) 16 6490 ± 122.5 (6367.5,6612.5) B. Interpret the confidence interval obtained in part A. (4 points) The interval in A has probability .95 of containing the true MOR. C. Find the sample size required to estimate the true MOR to within ±100 using a 99% confidence interval. (8 points) 2.576(250) n = 100 100 n = 2.576(250) n = 6.44 n = 41.47 n = 42 D. Suppose a test of hypothesis is performed on the population mean MOR where H 0 µ=6300 H1 µ ! 6300 Based only on the confidence interval in part A tell whether you would reject or not reject H0 at α =.05. Explain your answer.(5 points) You would reject the null hypothesis at α =.05 because 6300 is not in the confidence interval. 5. 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 every about four hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume the battery life is normally distributed with standard deviation σ = 0.2 hour. A. Perform a test of hypothesis at α =.05 to determine whether the true mean battery life is significantly greater than 4 hours. Write the complete report. (15 points) H0 µ =4 H1 µ >4 Assumptions Normal population with known standard deviation Region of rejection Z >1.645 Z= Computation Z= x!µ "/ n 4.05 ! 4 .2 / 50 = 1.7677 Reject or do not reject H0 reject H0 at α =.05 Conclusion and statement There is reason to believe that the life of these batteries is significantly greater than 4 hours. B. What is the p value of the test? (3 points) Pvalue = P(Z>1.77)=.0384 C. Would you reject or not reject H0 at α =.01?. Explain your answer. (4 points) You would not reject the null hypothesis at α =.01 for two reasons. First the p value is .0384 >.01. Second for a test at α =.01 the critical value is 2.33<1.77.